{"id":13465,"date":"2019-12-20T17:01:53","date_gmt":"2019-12-20T11:31:53","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=13465"},"modified":"2021-07-02T15:35:44","modified_gmt":"2021-07-02T10:05:44","slug":"2nd-puc-statistics-question-bank-chapter-6","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-statistics-question-bank-chapter-6\/","title":{"rendered":"2nd PUC Statistics Question Bank Chapter 6 Statistical Inference"},"content":{"rendered":"

You can Download Chapter 6 Statistical Inference Questions and Answers, Notes, 2nd PUC Statistics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Statistics Question Bank Chapter 6 Statistical Inference<\/h2>\n

2nd PUC Statistical Inference One and Two Marks Questions With Answers<\/h3>\n

Question 1.
\nWhat are parameter and statistic?\/Define parameter and statistic.
\nAnswer:
\nStatistical constants of the population such as Mean (\u00b5), S.D (\u03c3) are called parameter.
\nEx:- The population has mean = \u00b5, S.D = \u03c3, population proportion P0
\n<\/sub>Statistical measures computed from the samples such as Mean (x\u0304), S.D(\u03c3) are called statistic.
\nEx:- A random sample of size ‘n’ has mean=x\u0304 and S.D= s, sample proportion p<\/p>\n

Question 2.
\nDefine Statistical Hypothesis
\nAnswer:
\nIt is a statement or an assertion made regarding the parameters<\/p>\n

Question 3.
\nWhat are null and alternative hypothesis
\nAnswer:
\nNull Hypothesis is a hypothesis, which is being tested for possible rejection under the assumption that it is true. Denoted by H0<\/sub>
\nThe hypothesis, which is accepted when the null hypothesis is rejected, is called alternative hypothesis. Denoted by H1<\/sub>.<\/p>\n

Question 4.
\nDefine Simple and Composite Hypothesis.
\nAnswer:
\nA hypothesis, which completely specifies the parameter of the distribution, is a simple hypothesis.
\nEx:- H: \u00b5 = \u00b50<\/sub> (25) is a is a simple hypothesis,
\nH: The population is normally distributed with mean \u00b5 = 25 and. \u03c3 = 2
\nA hypothesis, which does not completely specify the parameter of the distribution, is a composite hypothesis.
\nEx:- H: The population is normally distributed with mean \u00b5 = 25<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nDefine Standard Error
\nAnswer:
\nThe standard deviation of the sampling distribution of sample statistic is called S.E of statistic<\/p>\n

Question 6.
\nGive any two uses of Standard Error.
\nAnswer:<\/p>\n

    \n
  1. It is used in interval estimation, to write down the confidence intervals.<\/li>\n
  2. It is used in testing of hypothesis, to test whether the difference between the sample statistic and the population parameter is significant or not.<\/li>\n<\/ol>\n

    Question 7.
    \nDefine Type I\/Type II Error
    \nAnswer:
    \nType I Error is taking a wrong decision to reject the null hypothesis when it is actually true.
    \nType II Error is taking a wrong decision to accept the null hypothesis when it is actually not true.3<\/p>\n

    Question 8.
    \nWhat is the Power of a test.
    \nAnswer:
    \nIt is the probability of rejecting the null hypothesis when it is not true. (1 – \u03b2 ),
    \nHere \u03b2 = P (Type II Error) which is referred as ‘consumers risk<\/p>\n

    Question 9.
    \nDefine size of a test.
    \nAnswer:
    \nProbability of occurrence of type I Error is called level of significance, denoted by ‘\u03b1’. It is also called as Size of the test.<\/p>\n

    Question 10.
    \nDefine critical Region and Critical value.
    \nAnswer:
    \nThe set of all those values of the test statistic, which lead to the rejection of the null hypothesis, is called critical region (\u03c9).
    \nThe Critical Region is also called as Rejection Region.
    \nThe value of test statistic, which separates the critical region (ie. rejection region) and . acceptance regions is called the Critical value, denoted by + k or -k or – k1<\/sub> \/ + k2<\/sub><\/p>\n

    Question 11.
    \nDefine level of significance, confidence level and confidence co-efficient.
    \nAnswer:
    \nProbability of occurrence of type I Error is called level of significance, denoted by ‘\u03b1’
    \nThe confidence interval is the interval within which the unknown population parameter is expected to lie.
    \nThe prob. that the confidence interval contains the parameter is called confidence coefficient.<\/p>\n

    Question 12.
    \nDefine Point estimation and Interval estimation.
    \nAnswer:
    \n‘While estimating the unknown parameter, if a specific’value is proposed as an estimate, which is called Point estimation’.<\/p>\n

    ‘While estimating the unknown parameter instead of a specific value, an interval is proposed, which is likely to contain the parameter is called Interval estimation’<\/p>\n

    Question 13.
    \nWhat is the best estimator of the population mean?
    \nAnswer:
    \nx\u0304<\/p>\n

    Question 14.
    \nDefine sample space and parameter space.
    \nAnswer:
    \nThe set of all samples of size ‘n’ that can be drawn from population is called sample space(S)
    \nThe set of all the admissible values of the population parameters is called parameter space.<\/p>\n

    Question 15.
    \nFor a S.N.V. test statistic Z, in left tail test the calculated value is (- 2.63) and the table value is (-1.90).
    \nIndicate suitable inference to be drawn based on this data.
    \nAnswer:
    \nHere Zcal<\/sub> < Ztab<\/sub> H0<\/sub> is rejected and H1<\/sub> is accepted.<\/p>\n

    Question 16.
    \nMention any two conditions while applying %2 – test in Testing the Independent of attributes
    \nAnswer:
    \n(a) Observations should be independent,
    \n(b) Total frequency N should be large,
    \n(c) Each cell frequencies ie ,a, b, c, d are all \u2265 5<\/p>\n

    Question 17.
    \nMention any two conditions of test for goodness of fit
    \nAnswer:
    \n(a) Observations should be independent,0.6
    \n(b) Total frequency N should be large,
    \n(e) Each Ei<\/sub>‘s \u2265 5.<\/p>\n

    Question 18.
    \nFor the application of %2 – distribution, what is the condition for the expected cell frequency?when pooling is made done in testing of goodness of fit?
    \nAnswer:
    \nEach cell frequency ie. Ei<\/sub>‘s \u2265 5<\/p>\n

    Question 19.
    \nIn a chi-square test for goodness of fit if there are 8 classes and if two parameters are’ estimated, that is the degrees of freedom of the test statistic.
    \nAnswer:
    \n(n – 1 – 2) = (n – 3)d.f.<\/p>\n

    Question 20.
    \nA random sample of size 25 is drawn from a population whose standard deviation is 3. Find the standard error of the sample mean.
    \nAnswer:
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 21.
    \nGiven H, : < p2, Write H0<\/sub>.
    \nAnswer:
    \n\u00b51<\/sub> = \u00b52 .<\/sub><\/p>\n

    Question 22.
    \nThe proportion of vegetarians in a city is 0.48 Find the standard Error of the proportion of vegetarians in a random sample of size 20.
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 23.
    \nGiven P = 2 \/ 3, n = 40, find SE (P).
    \nAnswer:
    \nSE(p) = \\(\\sqrt{\\frac{\\mathrm{PQ}}{\\mathrm{n}}}=\\sqrt{\\frac{2 \/ 3 \\times 1 \/ 3}{20}}\\) = 0.07<\/p>\n

    Question 24.
    \nThe proportion of vegetarians in village A is 0.42.The proportion of vegetarians in village B is 0.37. Among 70 randomly selected people from village A, if PI is the proportion of vegetarians and among 60 randomly selected people from village B, if P2<\/sub> is the proportion of vegetarians, find Standard Error of (P1<\/sub> = P2<\/sub>)
    \nAnswer:
    \n\"2nd<\/p>\n

    25.
    \nIn paired t-test if n = 10, d = 4.3 and Sd<\/sub> = 1.3 What would be tobs<\/sub>?
    \nAnswer:
    \n\"2nd<\/p>\n

    Standard Errors:<\/p>\n

    Question 26.
    \nWrite down the standard of Proportion.
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 27.
    \nWrite down the standard of difference of sample means of two samples
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 28.
    \nWrite down the standard of difference of proportion of two samples.
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 29.
    \nWrite down the test statistics for test for mean for large sample:
    \nAnswer:
    \nLarge sample tests: ( n \/ n1<\/sub> \/ n2<\/sub> >30):
    \n\"2nd<\/p>\n

    Question 30.
    \nWrite down the test statistics for test for equality of two means for large samples:
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 31.
    \nWrite down the test statistics for test for proportion:
    \nAnswer:
    \nTest for proportion
    \n\"2nd<\/p>\n

    Question 32.
    \nWrite down the test statistics for test for equality of two proportions.
    \nAnswer:
    \nTest for equality of proportion:
    \n\"2nd<\/p>\n

    Question 33.
    \nWrite down the chi-square test statistics for testing the variance with Hypothesis.
    \nAnswer:
    \nChi-square tests:
    \nTest for population variance:
    \nH0<\/sub>: \u03c32<\/sup> = \u03c32<\/sup>0<\/sub>; H1<\/sub> : \u03c32<\/sup> \u2260 \u03c32<\/sup>0<\/sub>; OR H,: \u03c32<\/sup> > \u03c32<\/sup>0<\/sub>; OR H1<\/sub>: \u03c32<\/sup> < \u03c32<\/sup>0<\/sub>;
    \n\"2nd<\/p>\n

    Question 34.
    \nWrite down the chi-square test statistics for testing the Goodness of fit with Hypothesis.
    \nAnswer:
    \nTest for good ness of fit: H0<\/sub> : Oi<\/sub> and Ei<\/sub> are good fit; H1<\/sub>: Oi<\/sub> and Ei <\/sub>are not good fit
    \n\"2nd
    \nwhen any of the parameter (p \/ \u03bb) is estimate d.f = (n – 1 – 1) = (n – 2).<\/p>\n

    Question 35.
    \nWrite down the chi-square test statistics for testing the Independence of attributes.
    \nAnswer:
    \nTo test for Independence of Attributes:
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 36.
    \nWrite down the t- test statistics for testing the mean.
    \nAnswer:
    \nSmall sample tests
    \n\"2nd<\/p>\n

    Question 37.
    \nWrite down the t- test statistics for testing the equality of meAnswer:
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 38.
    \nWrite down the t- test statistics for testing the equality of means when observation are paired.
    \nAnswer:
    \nPaired t-test for equality of means:
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 39.
    \nWhat do you mean by an Estimator?
    \nAnswer:
    \nThe Statistic used for the purpose of estimation of unknown population parameter is called Estimator.<\/p>\n

    Question 40.
    \nWhat is Estimate?
    \nAnswer:
    \nEstimate is a numerical value of the estimator computed from a given set of sample values.<\/p>\n

    Question Question 41.
    \nWhat are Confidence interval?
    \nAnswer:
    \nA Confidence interval is a interval within which the unknown population parameter is expected to lie.<\/p>\n

    Question 42.
    \nWhat is Test Statistic? ,
    \nAnswer:
    \nTest Statistic is the statistic based on whose distribution testing is conducted.
    \n\"2nd<\/p>\n

    Question 43.
    \nDefine sampling distribution.
    \nAnswer:
    \nThe distribution of values of the statistic for different samples of the same size is called sampling distribution of a statistic.<\/p>\n

    Question 44.
    \nGiven X\u0304 = 22 gm, \u03bc0<\/sub> = 20 gm, \u03c3=10 gm and n = 64 .Calculate test statistic z
    \nAnswer:
    \nHere statistic x\u0304 and parameter \u03bc0<\/sub> are given, the test statistic is
    \n\"2nd<\/p>\n

    Question 45.
    \nA hospital has 150 doctors out of whom 78 are ladies. When a sample of 35 doctors are taken, find mean and standard error of proportion of ladies.
    \nAnswer:
    \nGiven N=150, X = 78, n = 35
    \nPopulation proportion = P0<\/sub> = \\(\\frac { X }{ N }\\) = \\(\\frac { 78 }{ 150 }\\) = 0.52; Q0<\/sub> = 1 -P0<\/sub>= 0.48
    \nSample proportion = p = p0<\/sub> = 0.52
    \nStandard error of sample proportion is
    \nSE = \\(\\sqrt{n}=\\sqrt{\\frac{0.52 \\times 0.48}{35}}\\) = 0.084<\/p>\n

    Question 46.
    \nSize of two samples are 40 and SO. Population standard deviations are 10 and 20. Compute S.E. (x\u03041<\/sub> – x\u03042<\/sub>)
    \nAnswer:
    \nGiven : n1<\/sub> = 40, n2<\/sub> = 50, \u03c31<\/sub> = 10, \u03c32<\/sub>= 20
    \n\"2nd<\/p>\n

    Question 47.
    \nWhat are meant by one tailed test and two tailed test in testing of hypothesis?
    \nAnswer:<\/p>\n

      \n
    • if the critical region is considered at one tail of the null distribution of the test statistic, the test is one tailed.<\/li>\n
    • if the critical region is considered at both the tails of the null distribution of the test statistic, the test is two tailed.<\/li>\n<\/ul>\n

      Question 48.
      \nIf Zcal<\/sub> = 2.089 and Ztab<\/sub> = 2.58, then write down the decision on null hypothesis.
      \nAnswer:
      \nHere Zcal<\/sub> < Ztab<\/sub>. (H0<\/sub>) Null hypothesis is accepted.<\/p>\n

      Question 49.
      \nFor a S.N.V. test statistics Z, the calculated value Z is 2.63 and the table value of Z is 2.58. Indicate suitable inference to be drawn based on this data.
      \nAnswer:
      \nGiven Z = 2.63 and Ztab<\/sub> = 2.58.
      \nHere Zcal<\/sub> > Ztab<\/sub>. Null hypothesis is rejected.<\/p>\n

      Question 50.
      \nThe standard deviation of weight of children is 2 kg. What is the standard error of mean weight of 16 children?
      \nAnswer:
      \nGiven \u03c3 =2 kg, n= 16.
      \n\"2nd<\/p>\n

      Question 51.
      \nFrom the following data find the S.E. of difference between means of samples A&B.
      \n\"2nd
      \nAnswer:
      \nGiven n1<\/sub> = 1000 n2<\/sub> = 500
      \nx\u03041<\/sub> = 150 x\u03042<\/sub> = 146
      \nS1<\/sub> = 40 S2<\/sub> = 30
      \n\"2nd<\/p>\n

      Question 52.
      \nWrite down any two conditons for X test for goodness of fit.
      \nAnswer:
      \n(a) If any parameters are estimated from the data one degrees of freedom should be lessened.
      \n(b) All E; \u2265 5 if any E; < 5, it should be pooled with the adjacent frequencies.<\/p>\n

      Question 53.
      \nIf \"2nd\u00a0are cell frequencies, write the \u03c72<\/sup> test statistics with Yate’s correction.
      \nAnswer:
      \n\"2nd<\/p>\n

      Question 54.
      \nIt (X\u03041<\/sub>– X\u03042<\/sub>) = 2.7 and S.E (X\u03041<\/sub> – X\u03043<\/sub>) = 1.3, find Z.
      \nAnswer:
      \nIf (X\u03041<\/sub> -X\u03042<\/sub>) = 2.7 and S.E (X\u03041<\/sub>-X\u03042<\/sub>) = 1.3, find Z.
      \n\"2nd<\/p>\n

      Question 55.
      \nWrite down the mean and standard deviation of at variate with n degrees of freedom.
      \nAnswer:
      \nMean = 0, S.D. = \\(\\sqrt{\\text { variance }}=\\sqrt{\\frac{n}{n-2}}\\) ; n >2.<\/p>\n

      Question 56.
      \nIf p1<\/sub> = 0.3 and p2<\/sub> = 0.2, find mean of (p1<\/sub> – p2<\/sub>).
      \nAnswer:
      \nMean of (p1<\/sub> – p2<\/sub>) = E(p1<\/sub> – p2<\/sub>) = 0.3 – 0.2 = 0.1.<\/p>\n

      \"KSEEB<\/p>\n

      2nd PUC Statistical Inference Five Marks Question With Answers<\/h3>\n

      Large Sample Test<\/strong><\/p>\n

      Question 1.
      \nA certain brand soap is known to have weight with variance25. A random sample of 256 soaps had mean weight 122 gms. Can we conclude that the mean weight of the soaps manufactured by a firm is 125 gms. (Use 5% level of significance)
      \nAnswer:
      \nGiven; n-256, \u03c32<\/sub> = 25, \u2234 \u03c3 = 5. x\u0304= 122, \u00b5 = 125 ; \u03b1 = 5%.
      \nH0<\/sub>: Mean Weight of the soaps manufactured is 125 gms (i.e., H0<\/sub>: \u00b5 = 125)
      \nH1<\/sub>: Mean Weight of the soaps manufactured is differ from 125 gms.
      \n(i.e., H1<\/sub> : \u00b5 \u2260 125) {two tailed test}
      \n\"2nd
      \nThe test statistics is –
      \n\"2nd
      \nAt \u03b1 = 2.5% the two fail critical values are K1<\/sub> = -1.96 and K2<\/sub> = 1.96
      \nHere Zcal<\/sub>, = -9.6 lies outside -K1<\/sub> and K2<\/sub>
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion: Mean Weight of the soaps manufactured by a firm differs from 125 gms.<\/p>\n

      Question 2.
      \nThe standard deviation of marks scored by PUC students in an examination is 10.5. The mean marks scored by 64 students of a college is 38. Can we conclude at 1% level of significance that mean marks of PUC students is 40 ?
      \nAnswer:
      \nGiven: \u03c3=10.5, n = 64 x\u0304=38, p = 40, \u00b5=l%
      \nH0<\/sub> = Mean marks of PUC students is 40. ( i.e., H0<\/sub> = : \u00b5 = 40)
      \nH1<\/sub> = Mean marks of PUC students differs from 40.( i.e,. H1<\/sub> = : \u00b5 \u2260 40)
      \n\"2nd
      \nThe test statistics is :
      \n\"2nd
      \nAt \u03b1 = 1% the two tail critical values are -K1<\/sub> = 2.58 and + K2<\/sub> = + 2.58
      \nHere, Zeal = -1.52 is in between -K1<\/sub> and K2<\/sub>.
      \nH0<\/sub> is accepted.
      \nConclusion : Mean marks of PUC students is 40.<\/p>\n

      \"KSEEB<\/p>\n

      Question 3.
      \nA random sample of 200 cans containing baby food has mean height 999.5 gms and standard deviation 15 gms. Test whether the contents of the cans can be considered as 1 kg. (Test at 1% level of significance.)
      \nAnswer:
      \nGiven: n = 200, x\u0304 = 999.5 gms, s = 15 gms, \u00b5 = 1 kg i.e., \u00b5 = 1000 gms.
      \nH0<\/sub>: The contents of the can is 1 kg. (i.e., H0<\/sub>: \u00b5 = 1000 gms.)
      \nH1<\/sub>: The contents of the can is less than 1 kg (i.e., H1<\/sub> : \u00b5 < 1000 gms)
      \n\"2nd
      \nThe test statistic is:
      \n\"2nd
      \nAt \u03b1 = 1% the left\/lower tail critical value is k = -2.33
      \nHere, Zcal<\/sub> = -0.47 is in acceptance region.
      \n\u2234 H0<\/sub> is accepted;
      \nConclusion: The contents of the can be considered as 1 kg.<\/p>\n

      Question 4.
      \nA sample of 100 students is taken from a large population. The mean height of these students is 64″ and standard deviation 4″. Can it be reasonably regarded that, the population mean height is more than 66″. (Take a = 5%)
      \nAnswer:
      \nGiven: n = 100, x\u0304 = 64″, S = 4″, \u00b5 = 66″, \u03b1 = 5%
      \nH0<\/sub>: The mean height of the population is 66″ (i.e., H0<\/sub>: \u00b5 = 66″)
      \nH1<\/sub> : The mean height of the population is more than (i.e., H1<\/sub> : \u00b5 > 66″)
      \n\"2nd
      \nThe test statistic is –
      \n\"2nd
      \nAt \u03b1 = 5% the right\/upper tail critical value K = 1.645.
      \nHere Zeal = -5 is in acceptance region
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : The mean height of the populations is 66″.<\/p>\n

      Question 5.
      \nA random sample of 1000 apples has mean weight of 187 gms and S.D. 8 gms. Another random sample of 800 apples has mean weight of 188.4 gms and S.D. of 10 gms. Test the hypothesis that the mean weights differ significantly at 1% level of significance.
      \nAnswer:
      \nGiven:
      \nn1<\/sub> = 1000; x\u03041<\/sub> =187gms, S1<\/sub> = 8 gms
      \nn2<\/sub> = 800; x\u03042<\/sub> = 188gms, S2<\/sub> = 10 gms
      \nH0<\/sub> : means are same (i.e., H0<\/sub> : \u00b5<sub.1 = \u00b52<\/sub>)
      \nH, : means are not same.( i.e., H, : \u00b51<\/sub> \u2260 \u00b52<\/sub>)
      \n\"2nd
      \nThe test statistics is –
      \n\"2nd
      \nAt \u03b1 = 5% the critical values are – K1<\/sub> = – 1.96, K2<\/sub> = 1.96.
      \nHere Zeal = – 3.22 is out side of -K1<\/sub> and K2<\/sub>
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion: Mean weight of apples are same.<\/p>\n

      Question 6.
      \nFrom the following data, test at 1% level of significance whether the sample means differ significantly.
      \n\"2nd
      \nAnswer:
      \nGiven:
      \nn1<\/sub>=200; x\u03041<\/sub> =219.3, S1<\/sub>2<\/sup> = 3,
      \nn2<\/sub> = 300; x\u03042<\/sub> =220.1, S2<\/sub>2<\/sup> = 2, a=l%
      \nH0<\/sub>: Means are equal\/ means does not differ significantly (i.e., H1<\/sub> : \u00b51<\/sub> = \u00b52<\/sub>)
      \nH1<\/sub>: Means are not equal\/means differ significantly (i.e. H1<\/sub> > : \u00b5<sub>1<\/SUB> \u2260 \u00b5<sub>2<\/sub>)
      \n\"2nd
      \nThe test statistic is –
      \n\"2nd
      \nZcal<\/sub> = – 5.43
      \nAt \u03b1 = 1 %, the two Tailed critical values are -K1<\/sub> = -2.58 and K2<\/sub> = 2.58
      \nHere Zeal = -5.43 is in rejection region\/ Zcal<\/sub> is outside the interval -k1<\/sub> and k2<\/sub>
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion : means are not equal.<\/p>\n

      Question 7.
      \nSamples of electric lamps manufactured by two firms gave the following results:
      \n\"2nd
      \nTest whether the average life of bulbs manufactured by the firm A is more than firm B. Use 5% level of significance
      \nAnswer:
      \nGiven:
      \nn1<\/sub> = 60, x\u03041<\/sub> =1400, S1<\/sub>= 106
      \nn2<\/sub> = 70, x\u03042<\/sub> = 1300, S2<\/sub> =100, \u03b1 = 5%.
      \nH0<\/sub> : Average life of bulbs of firm A and B are same.( i.e., H0<\/sub>: \u00b51<\/sub>,, = \u00b5,sub>2)
      \nH1<\/sub> : Average life of bulbs of firm A is more than firm B. (i.e., H1<\/sub> : \u00b51<\/sub> > \u00b52<\/sub>)
      \n\"2nd
      \nThe test statistics is –
      \n\"2nd
      \nZcal = 5.50
      \nAt \u03b1 = 5% the upper tail critical value K = 1.645
      \nHere Zcal<\/sub> = 5.50 is in Rejection region \/ zcal<\/sub> > K
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion : The mean life of bulbs of firm A is more than firm B.<\/p>\n

      \"KSEEB<\/p>\n

      Question 8.
      \nThe following data gives pass percentage in two colleges :
      \n\"2nd
      \nTest at 5% level of significance. Whether the pass percentage of college A is more than college B.
      \nAnswer:
      \nGiven:
      \nx\u03041<\/sub> =80, S1<\/sub> = 5, n1<\/sub>=100
      \nx\u03042 = 70, S2<\/sub> = 4, n2<\/sub> = 90, \u03b1 = 5%.
      \nH0<\/sub> : The pass percentage of college A and B is same, i.e., H0<\/sub> : \u00b51<\/sub> = \u00b52<\/sub>
      \nH1<\/sub> : The pass percentage of college A is more than college B. (i.e., H1<\/sub> : \u00b51<\/sub> > \u00b52<\/sub>)
      \n\"2nd
      \nThe test statistic is –
      \n\"2nd
      \nAt \u03b1 = 5%, the upper\/right Tail critical value K = 1.645.
      \nHere Zeal = 15.29 is in rejection region\/Zcal<\/sub> > K.
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion: The pass percentage of college A is more than college B.<\/p>\n

      Question 9.
      \nIn an election to the parliament the leaders of a specific party contend that they would secure 40% of the votes. A pre-poll survey of 250 voters revealed that the percentages is 32. Does the survey support the leaders claim? (Test at 5% level of significance).
      \nAnswer:
      \nGiven: n = 250, P0<\/sub> = 0.40 P = 0.32. \u2234 Q0<\/sub> = 1 – P0<\/sub> = 0.60
      \nH0<\/sub>: The leader would secure 40% votes, (i.e, H0<\/sub> : P1<\/sub>= 0.40)
      \nH1<\/sub>: The leader would not secure 40% votes (i.e., H1<\/sub> : P \u2260 0.40)
      \n\"2nd
      \nThe test statistic is –
      \n\"2nd
      \nAt \u03b1 = 5% the two tail critical values are -K1<\/sub> = 1.96 and K2<\/sub> = 1.96
      \nHere Zcal = 2.58 lies in rejection region\/ Zcal<\/sub> > Kr
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion: The Leader of a specific party does not secure 40% votes or The survey does not support the claim.<\/p>\n

      Question 10.
      \nA manufacturer claims that less than 2% of his products are defective. A retailer buys a batch of 250 articles from the manufacturer and find that 13 are defectives. Test at 1% level of significance that whether the manufacturer\u2019s claim is justifiable.
      \nAnswer:
      \nGiven P0<\/sub> = 0.02, n = 250, x=13, \u03b1 = l%
      \n\u2234 P = \\(\\frac{x}{n}=\\frac{13}{250}\\) = 0.052; Qo<\/sub> = 1 – p0<\/sub> =1 – 0.02 = 05.98
      \nH0<\/sub> : proportion of defective products are 2%
      \n(i.e. H0<\/sub>: P0<\/sub> = 2% = 0.02)
      \nH1<\/sub> : proportion of defective products are less than 2%……
      \n(i.e H<sub.1 : p0<\/sub> < 2% )
      \n\"2nd
      \nThe test Statistic is
      \n\"2nd
      \nZcal<\/sub> =3.636
      \nAt \u03b1 = 5% the lower tail critical value – k = – 2.33
      \nHere Zcal<\/sub> > – k \/ Zcal<\/sub> lies in (AR) acceptance region.
      \n\u2234 H0<\/sub> is accepted
      \nConclusion : proportion of defective products are 0.02 or 2%.<\/p>\n

      Question 11.
      \nFrom a large consignment of mangoes, a random sample of 50 mangoes were examined and among them four were found to be bad. Test whether it is reasonable to assume that more than 10% of mangoes were bad. [The critical value at 5% level of significance is 1.645].
      \nAnswer:
      \nGiven n = 4 x = 4 \u2234p = \\(\\frac{x}{n}=\\frac{4}{50}\\) = 0.08
      \n\u2234 Q0<\/sub> = 1 – p0<\/sub> = 0.9
      \nH0<\/sub>: The proportion of bad mangoes is 10% = 0.1 (i.e., H0<\/sub>: P = 0.1)
      \nH1<\/sub> : The proportion of bad mangoes is more than 10% = 0.1 (i.e., H1<\/sub> : P > 0.1)
      \n\"2nd
      \nThe test statistic is
      \n\"2nd
      \nZcal = -0.47
      \nAt 5% level of the upper tail critical value. K = 1.645.
      \nHere Zeal = -0.47 is acceptance region\/ Zcal<\/sub> < K.
      \n\u2234 H0<\/sub> is accepted
      \nConclusion : The proportion of bad mangoes is 10%.<\/p>\n

      \"KSEEB<\/p>\n

      Question 12.
      \nIn a village, 200 out of 300 men are smokers. Does this data support the argument that majority of men in the village are smokers ?
      \nAnswer:
      \nGiven : n = 300 ; x = 200
      \n\u2234 p = \\(\\frac{x}{n}=\\frac{200}{300}\\) = 0.67
      \nH0<\/sub>: Proportion of smokers is 0.5
      \n(i.e., H0<\/sub>: P0<\/sub> = 0.5) .
      \nH1<\/sub> : Proportion of smokers is more than 0.5
      \n(i.e., H1<\/sub> : P0<\/sub> > 0.5) {upper tail test}
      \nHere it is assumed that proportion of smokers and non-smokers are equal. Therefore P0<\/sub> = 0.5.
      \n\"2nd
      \nThe test statistic is :
      \n\"2nd
      \nAt \u03b1 = 5% the upper tail critical value is K = 1.645.
      \nHere Zcal<\/sub> > K.
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion : Majority of men in the village are smokers i.e., proportion of smokers is more than 0.5.<\/p>\n

      Question 13.
      \nIn a random sample of 200 plastic bags manufactured by machine A, 16 were found to be defective. In a random sample of 300 plastic bags produced by a machines B, 30 were found to be defective. Test at 5% level of significance the hypothesis, that the proportion of the defective plastic bags produced by the machines are same.
      \nAnswer:
      \nGiven:
      \nn1<\/sub> =200, x1<\/sub> = 16 \u2234p1<\/sub> = \\(\\frac{x_{1}}{n_{1}}=\\frac{16}{200}\\) = 0.08
      \nn2<\/sub> = 300, x2<\/sub> = 30 \u2234P2<\/sub>= \\(\\frac{x_{2}}{n_{2}}=\\frac{30}{300}\\) = 0.1
      \nH0<\/sub>: The Proportion of defective plastic bags produced by machines are same.
      \n(i.e\u201e H0<\/sub>: P1<\/sub> = P2<\/sub>) .
      \nH1<\/sub> : The proportion of defective plastic bags produced by machines are not same.
      \n(i.e., H1<\/sub> = P1<\/sub> \u2260 P2<\/sub>) {Two tailed test}
      \n\"2nd
      \nUnder H0<\/sub>, The test statistic is –
      \n\"2nd
      \n\u2234 Zcal = -0.758
      \nAt \u03b1 = 5% the two tail critical values are -K1<\/sub> = -1.96 and K2<\/sub> = 1.96.
      \nHere Zcal = -0.758 is in acceptance region\/ Zcal<\/sub> is in between -K1<\/sub> and K2<\/sub>
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : The proportion of defective plastic bags produced by machines are same.<\/p>\n

      Question 14.
      \nAmong 40 randomly selected people from urban area, 5 are interested in viewing football match. Among 50 randomly selected people from rural area, 4 are interested in viewing football match. Test st 5% level of significance, whether the proportion of viewers in urban and rural areas are the same.
      \nAnswer:
      \nGiven:
      \nn1<\/sub> = 40 ; x1<\/sub> = 5, \u2234p1<\/sub> = \\(\\frac{x_{1}}{n_{1}}=\\frac{5}{40}\\) =0.125
      \nn2<\/sub> = 50: x2<\/sub> = 4 \u2234P2<\/sub> = \\(\\frac{x_{2}}{n_{2}}=\\frac{4}{40}\\) = 0.08
      \nH0<\/sub>: The proportion of viewers in urban and rural areas are same.
      \n(i e., H0<\/sub>: P1<\/sub>= P2<\/sub>)
      \nH1<\/sub> : The proportion of viewers in urban and rural areas are not same,
      \n(i.e., H1<\/sub> : P1<\/sub> \u2260 P2<\/sub>) {two Tailed test}
      \n\"2nd
      \nThe test Statistic is
      \n\"2nd
      \nZcal = 0.7075
      \nAt \u03b1 = 5%, the two tail critical values are -K1<\/sub>= -1.96 and K2<\/sub> = 1.96.
      \nHere Zeal = 0.7075 is in acceptance region\/ Zcal<\/sub> is in between K1<\/sub> and K2<\/sub>
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : The proportion of viewers of football match in urban and rural areas are the same.<\/p>\n

      Question 15.
      \nA machine produced 25 defective articles in a batch of 400. After over hauling it produced 15 defectives in a batch of 200. Test at 1% level of significance whether there is a reduction of defective articles after overhauling.
      \nAnswer:
      \nGiven:
      \nn1<\/sub> = 400, x1<\/sub> = 25 \u2234 P1<\/sub> = \\(\\frac{x_{1}}{n_{1}}=\\frac{25}{400}\\) = 0.0625.
      \nn2<\/sub> = 200 x2<\/sub> = 15 \u2234P2<\/sub> = \\(\\frac{x_{2}}{n_{2}}=\\frac{15}{200}\\) = 0.075
      \nH0<\/sub> : The proportion of defective articles before and after overhauling is the same
      \n(i.e H0<\/sub> : p1<\/sub> = p2<\/sub>)
      \nH1<\/sub> : The proportion of defective articles are more before overhauling
      \n(i.e H1<\/sub> : p1<\/sub> > p2<\/sub>) { Right upprt Tailed test}
      \n\"2nd
      \nThe test statistic is
      \n\"2nd
      \n\u2234 Zeal = -0.576
      \nAt \u03b1 = 1 % the upper\/right tail critical values K = 2.33
      \nHere Zcal = -0.576 is in acceptance region\/ Zcal<\/sub>
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : The proportion of defective articles before and after .over hauling is same.<\/p>\n

      Question 16.
      \nFrom the following data, test whether the difference between the proportions in the two samples is significant.
      \n\"2nd
      \nAnswer:
      \nGiven:
      \nn1<\/sub> = 1000, P1<\/sub> = 0.02
      \nn2<\/sub> = 1200, P2<\/sub> = 0.01
      \n\"2nd
      \np0<\/sub> = 0.0145 \u2234Q0<\/sub> = 0.9855
      \n\"2nd
      \nH0<\/sub>: The proportions are equal.
      \n(i e., H0<\/sub> = P1<\/sub> = P2<\/sub>)
      \nH, : The proportions are not equal,
      \n(i.e., H1<\/sub> : P1<\/sub> \u2260 P2<\/sub>)
      \n\"2nd
      \nZcal = 1.953
      \nAt a= 5% the two Tail critical values are -K1<\/sub> = -1.96 and K2<\/sub> – 1.96
      \nHere Zcal = 1.953 is in acceptance region.
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion: The proportions are same.<\/p>\n

      \"KSEEB<\/p>\n

      Chi Square Tests<\/strong><\/p>\n

      Question 1.
      \nThe weights of 10 babies at the time of birth are as follows :
      \nWeight (Kgs) : 2.8, 3.4, 2.6, 3.0, 2.4, 2.7, 1.6, 3.1, 2.9, 2.4
      \nTest the hypothesis that the standard deviation differs significantly from lkg. (use \u03b1 = 1%)
      \nAnswer:
      \nGiven : n = 10, \u03c3 = population SD = lkg.
      \nH0<\/sub>: Sample SD does not differ significantly from population SD
      \ni.e., H0<\/sub>: \u03c3 = 1 kg
      \nH1<\/sub> : Sample SD differs significantly from population SD.
      \ni.e., H1<\/sub> : \u03c3 \u2260 1 kg. {Two Tailed .test}
      \nThe test statistics –
      \n\"2nd
      \nWhere s2<\/sup> = sample variance
      \n\"2nd
      \nLet \u2018x\u2019 be the weight
      \n\"2nd
      \ns2<\/sub> = \\(\\frac { 14.03 }{ 10 }\\) – \\(\\left(\\frac{10.9}{10}\\right)^{2}\\) = 0.2149
      \n\u2234 \u03c72<\/sup> = \\(\\frac{10 \\times 0.2149}{1}\\)\u00a0 \u00a0\u03c72<\/sup>cal<\/sub> = 2.149
      \nFor (n – 1) = (10 – 1) = 9 d.f at \u03b1 = 1% the two tail critical values (from the \u03c72<\/sup> – table) are + K1<\/sub>= 1.73 and K2<\/sub> = 23.59
      \n\"2nd
      \nHere \u03c72<\/sup>cal<\/sub> = 2.149 is in acceptance region.
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : The sample SD doesn\u2019t differs from population S.D. = 1 kg.<\/p>\n

      Question 2.
      \nWeight of 10 jack fruits are as follows :
      \nWeight (kgs) : 6, 5, 8, 7, 4, 5, 8, 6, 5, 6
      \nTest at 5% level of significance that SD differs significantly from 2 kgs.
      \nAnswer:
      \nGiven: n=10, \u03c3 = 2 kgs.
      \nH0<\/sub>: The sample SD doesnot differs significantly from population S.D. = 2 kgs.
      \ni.e., H0<\/sub>: \u03c3 = 2 kgs.
      \nH1<\/sub> : The sample SD differs significantly from population S.D. = 2 kgs.
      \ni.e., H1<\/sub> : \u03c3 \u2260 2 kgs
      \nThe test statistics is
      \n\"2nd
      \nLet x be the weight
      \n\"2nd
      \n\"2nd
      \n\u2234 \u03c72<\/sup> = \\(\\frac{10 \\times 1.6}{2^{2}}\\) = 4
      \n\u03c72<\/sup>cal<\/sub> = 4
      \nFor (n – 1) = (10 – 1) = 9 d.f at \u03b1 = 5 % the two Tail critical values (from the table)
      \nK1<\/sub> = 2.70 and K2<\/sub> = 19.02. .
      \nHere \u03c72<\/sup>cal<\/sub> = 4 is in acceptance region.
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : The sample SD doesn\u2019t differs significantly from population SD=2 kgs.
      \n\"2nd<\/p>\n

      Question 3.
      \n25 randomly selected resistors have S.D. of resistance of 0.016 ohms. Test the hypothesis that the resistors in the lot have SD of resistance less than 0.01 ohm.
      \nAnswer:
      \nGiven: n = 25, S = 0.016ohms, \u03c3 = 0.01ohms.
      \nH0<\/sub> : The SD of resistance of resistors is 0.01 ohms,
      \ni.e., H0<\/sub> : \u03c3 = 0.01 ohm.
      \nH1<\/sub> : The SD of resistance of resistors is less than 0.01 ohm.
      \ni.e., H1<\/sub> : \u03c3 < 0.01 ohm [left\/lower tail test]
      \n\"2nd
      \n\"2nd
      \n\u03c72<\/sup> = \\(\\frac{25(0.016)^{2}}{(0.01)^{2}}\\)
      \n\u03c72<\/sup>cal<\/sub> = 64
      \nFor (n – 1) = (25 – 1) = 24, d.f at a = 5% the left\/lower tail critical value is K = 13.85.
      \nHere \u03c72<\/sup>cal<\/sub> = 64 is in acceptance region \u03c72<\/sup>cal<\/sub> > K2<\/sub>
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : The SD of resistance of resistors is 0.01 ohm.<\/p>\n

      \"KSEEB<\/p>\n

      Question 4.
      \nA normal variate has standard deviation 2.6. Twenty sample observations of the variate have standard deviation 2. Test at 1% level of significance that population standard deviation is less than 2.6.
      \nAnswer:
      \nGiven: \u03c30<\/sub> = 2.6, n = 20, S = 2, a=l%
      \nH0<\/sub>: \u03c30<\/sub> =2.6(Standard deviation is 2.6)
      \nH1<\/sub> : \u03c30<\/sub> < 2.6 (Standard deviation less than 2.6) {Lower tailed test}
      \n\"2nd
      \nUnder H0<\/sub>, the test statistic is :
      \n\"2nd
      \n\\(=\\frac{20 \\times 2^{2}}{(2.6)^{2}}\\)
      \n\u03c72<\/sup>cal<\/sub> = 11.8343
      \nFor (n- 1) = (20 – 1) = 19 d.f the lower tail critical value at 5% level of significance is
      \nK1<\/sub> – 10.12;
      \nHere \u03c72<\/sup>cal<\/sub> > K1<\/sub>
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion: Standard deviation is 2.6.<\/p>\n

      Question 5.
      \nFrom the following data, test whether there is any significant difference between observed and expected frequencies, using \u03c72<\/sup> test at 5% level of significance.
      \n\"2nd
      \nAnswer:
      \nH0<\/sub> : There is no significant difference between observed and expected frequencies.
      \nH1<\/sub> : There is a significant difference between observed and expected frequencies.
      \nThe test statistics is –
      \n\"2nd
      \n\u03c72<\/sup>cal<\/sub> = 4.69
      \n\"2nd
      \nFor (n – 1) = (6 – 1) = 5 d.f at \u03b1 = 5% the upper Tail critical value K2\/K = 11.07.
      \nHere \u03c72<\/sup>cal<\/sub> = 4.69 is in acceptance region\/ \u03c72<\/sup>cal<\/sub> < K2<\/sub>.
      \nH0<\/sub> is accepted.
      \nConclusion: There is no significant difference between observed and expected frequencies. [Note: In the above table the first one and last two expected frequencies are less than 5, they are pooled with the adjacent frequencies such that finally all the frequencies are 5 or more.]<\/p>\n

      Question 6.
      \nFour coins are tossed 80 times. The distribution of number of heads is given:
      \n\"2nd
      \nApply \u03c72<\/sup> – test and test at 1% level of significant if the coin is unbiased.
      \nAnswer:
      \nH0<\/sub>: The coin is unbiased.
      \n\u2234 p = \\(\\frac { 1 }{ 2 }\\)
      \nH1<\/sub> : The coin is biased.
      \ni.e., p \u2260 \\(\\frac { 1 }{ 2 }\\) [upper Tail test only]
      \n\"2nd
      \nThe test statistics is –
      \n\"2nd
      \nLet O and E be the observed and expected frequencies, and let x be the number of heads.
      \n\"2nd
      \n\u2234 \u03c72<\/sup>cal<\/sub>=0.73
      \nFor (n-1) = (5-1) = 4 d.f, at 1% level of significance the upper Tail critical value {From table} K\/K2<\/sub> = 13.28
      \nHere \u03c72<\/sup>cal<\/sub> = 0.73 is in acceptance region\/ \u03c72<\/sup>cal<\/sub> < K2<\/sub>
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : The coin unbiased.
      \n[Note : In the above table none of the expected frequencies are less than 5. So, pooling is not necessary.]<\/p>\n

      \"KSEEB<\/p>\n

      Question 7.
      \nThe following is the data regarding day-wise accidents occured in Bangalore during 2000:
      \n\"2nd
      \nTest whether there is day-wise equidistribution of accidents
      \nAnswer:
      \nH0<\/sub> : There is day-wise equidistribution of accidents.
      \nH1<\/sub> : There is no day-wise equidistribution of accidents. [Upper tail test only]
      \n\"2nd
      \nThe test statistics is –
      \n\"2nd
      \nSince there is a day-wise equidistribution of accidents.
      \nTherefore the expected frequencies, each is \\(\\frac { 105 }{ 7 }\\) =15.
      \nLet O and E be the observed and expected number of accidents.
      \n\"2nd
      \n\u2234 \u03c72<\/sup> = 6.42
      \nFor (n – 1) = 7 – 1 = 6 d.f at 5% level of significance the upper tail critical value K\/K2<\/sub>= 12.59
      \nHere \u03c72<\/sup>cal<\/sub> = 6.46 is in acceptance region\/ \u03c72<\/sup>cal<\/sub> < K2<\/sub>
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion: There is day-wise equidistribution of accidents.<\/p>\n

      Question 8.
      \n70 accidents that have occurred in a state in a week are tabulated as follows:
      \n\"2nd
      \nTest whether accidents occur uniformly throughout the week-
      \n[\u03c72<\/sup>tab<\/sub> is 16.81]
      \nAnswer:
      \nH0<\/sub> : Accidents occur uniformly throughout the week
      \nH1<\/sub>\u00a0: Accidents does not occur uniformly throughout the week…(upper tail test)
      \nThe test statistic is :
      \n\"2nd
      \nLet \u2018O\u2019 and \u2018E\u2019 be the observed and expected number of accidents.
      \nSince accidents occur uniformly throughout the week.
      \nExpected frequencies each is \\(\\frac { 70 }{ 7 }\\) = 10
      \n\"2nd
      \n\"2nd
      \n\u2234 \u03c72<\/sup>cal<\/sub>= 6.8
      \nFor (n – 1) = (7 – 1) = 6 degrees of freedom at 1% level of significance, the upper tail critical value given k2<\/sub>= 16.81 (\u03c72<\/sup>tab<\/sub>)
      \nHere \u03c72<\/sup>cal<\/sub> < k2<\/sub>\/\u03c72<\/sup>cal<\/sub> lies on (AR-) acceptance region.
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion: Accidents occur uniformly throughout the week.<\/p>\n

      Question 9.
      \nA die is rolled 90 times and the fol lowing distribution is obtained:
      \n\"2nd
      \nTest whether the die is unbiased use 5% level of significance.
      \nAnswer:
      \nH0<\/sub> : The die is unbiased
      \nH1<\/sub> : The die is biased.
      \nThe test statistic is –
      \n\"2nd
      \nSince, the die is unbiased, the numbers of throws should be equally distributed on all 6 faces.
      \n\u2234 The expected frequencies each is \\(\\frac { 90 }{ 6 }\\) = 15.
      \nLet \u2018O\u2019 and \u2018E\u2019 be the observed number of throws and expected frequencies and let x be the face value.
      \n\"2nd
      \n\"2nd
      \n\u2234 \u03c72<\/sup>cal<\/sub> = 5.868
      \nFor (n – 1) = (6 – 1) = 5 d.f. At 5% level of sigificance the upper Tail critical value K\/K2<\/sub> = 11.07
      \nHere = 5.868 is in acceptance region.\/ \u03c72<\/sup>cal<\/sub> < K
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : The die is unbiased.<\/p>\n

      Question 10.
      \nAccording to Mendal’s law, the first generation of cross breeding of red and white 4 o’clock (flowers) will have 1:2:1 red, pink and white 4’oclocks. In an experiment, the frequencies happen to be 13, 21 and 10 respectively. Docs the data support Mendal’s law?
      \nAnswer:
      \nH0<\/sub>: The data support Mendal’s law.
      \nH1<\/sub> : The data does not support Mendal’s law.
      \nThe test statistics is –
      \n\"2nd
      \nLet \u2018O\u2019 be the experiment\/observed frequencies (i.e., 13, 21 and 10)
      \nHere total frequencies = 44
      \n\u2234 The Expected frequencies \u2018E\u2019 should be in ratio of 1:2:1
      \ni.e., divide 44 in to 3 out of 1 + 2 + 1 = 4 parts as :
      \nE1<\/sub> = \\(\\frac { 44 }{ 4 }\\) \u00d7 1 = 11 E2<\/sub> = \\(\\frac { 44 }{ 4 }\\) \u00d7 2 = 22 E3<\/sub> = \\(\\frac { 44 }{ 4 }\\) \u00d7 3 = 11
      \n\"2nd
      \n\"2nd
      \n\u2234 \u03c72<\/sup>cal<\/sub> = 0.496
      \nFor (n – 1) = (3 – 1) = 2 d.f at 5% level of significance upper tail critical value K2<\/sub> = 5.99
      \nHere\u03c72<\/sup>cal<\/sub> = 0.496 is in Acceptance region.\/ \u03c72<\/sup>cal<\/sub> < K2<\/sub>
      \n\u2234 H0<\/sub> is accepted
      \nConclusion : The data support Mendal’s Law.<\/p>\n

      \"KSEEB<\/p>\n

      Question 11.
      \n5 coins are tossed 128 times. the result are tabulated as below: (10 Marks)
      \n\"2nd
      \nAnswer:
      \nLet x be the number heads obtained is a Binomial variable with the parameters n = 5 and
      \np = probability of getting head = \\(\\frac { 1 }{ 2 }\\) \u2234 q = \\(\\frac { 1 }{ 2 }\\)
      \nThe p.m.f is –
      \np (x) = nCn<\/sub> px<\/sup>qn-x<\/sup>; x = 0, 1, 2 ………….. n.
      \n= 5Cx<\/sub> (0.5)x<\/sup> (0.5)5-x<\/sup>; x = 0, 1,2 ………….. 5
      \nTheoretical\/expected frequency Tx<\/sub> = p(x) \u00d7 N
      \n\u2234 T(x = 0) = p(x = 0) \u2234 128
      \n= 5C0<\/sub>(0.5)\u00b0(0.5)5-0<\/sup>\u00a0\u00d7 128
      \nT (0) = 4
      \nUsing recurrence relation for theoretical frequency Tx<\/sub> = \\(\\frac{n+1-x}{x} \\frac{p}{q} T_{x-1}\\)
      \nT(x – 1) = \\(\\frac{5+1-1}{1} \\times \\frac{0.5}{0.5} \\times \\mathrm{T}_{(1-1)}\\)
      \n= 5 \u00d7 1 \u00d7 T0<\/sub>
      \n= 5 \u00d7 4
      \nT(1) = 20
      \nsimilarly,
      \n\"2nd
      \n\u2234 The fitted observed theoretical frequency distribution is :
      \n\"2nd
      \nChi-square Test is :
      \nH0<\/sub>: Binomial distribution is good fit
      \nH1<\/sub> : Binomial distribution is not a good fit
      \nThe Test statistic is –
      \n\"2nd
      \nLet \u2018O\u2019 and \u2018E\u2019 be the observed and expected frequencies.
      \n\"2nd
      \n\"2nd
      \nHere first two and last two expected frequencies are less than 4 and are pooled with adjacent frequencies :
      \n\u2234 \u03c72<\/sup>cal<\/sub> = 100.932
      \nFor (n – 1) = (4 – 1) = 3 d.f at 5% level of significance the upper Tail critical value K\/K2<\/sub> = 7.81
      \nHere \u03c72<\/sup>cal<\/sub> = 100.932 is in rejection region\/ \u03c72<\/sup>cal<\/sub> > K2<\/sub>
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion : Binomial distribution is not a good fit.<\/p>\n

      Question 12.
      \nFor the following data fit a Poisson distribution and test whether it is a good fit (test at 5% level of significance) (10 marks)
      \n\"2nd
      \nAnswer:
      \nLet x be the no. of T. V. Sold is a Poisson variate. The parameter \u03bb can be obtained as below:
      \nLet f be the no. of days.
      \n\"2nd
      \ni.e., \u03bb
      \n\"2nd
      \n\u03bb = 1.9
      \n\"2nd
      \nTheoretical\/expected frequency Tx<\/sub> = P(x) \u00d7 N.
      \n\u2234 T(x = o) = p(x= o) \u00d7 150
      \n\\(=\\frac{\\mathrm{e}^{1.9}(1.9)^{0}}{0 !} \\times 150\\)
      \n= 0.1496 \u00d7 150
      \nT0<\/sub> = 22.44
      \nusing recurring relation for theoretical frequency
      \n\"2nd
      \n\u2234 T(x – 1) = \\(\\frac { 1.9 }{ 1 }\\) T1-1<\/sub> = \\(\\frac { 1.9 }{ 1 }\\) \u00d7 22.44
      \nsimilarly
      \nT1<\/sub> = 42.64
      \n\"2nd
      \n\"2nd
      \n\u2234 The fitted poisson distribution is :
      \n\"2nd
      \nChi- square test:
      \nH0<\/sub>: Poisson distribution is good fit
      \nH1<\/sub> : Poisson distribution is not a good fit.
      \nThe test statistic is –
      \n\"2nd
      \nSince the parameter X is estimated so, one more d.f is lessened.
      \nLet \u2018O\u2019 and \u2018E\u2019 be the observed and expected frequencies.
      \n\"2nd
      \n\"2nd
      \n\u2234 \u03c72<\/sup>cal<\/sub> = 1.938
      \nFor (n – 2) = (6 – 2) = 4 d.f at 5% level of significance the upper tail critical value K2<\/sub> = 9.49
      \nHere \u03c72<\/sup>cal<\/sub> = 1.938 is in acceptance region. (\u03c72<\/sup>cal<\/sub> < K)
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : Poisson distributions is good fit.<\/p>\n

      Question 13.
      \nFrom the following data regarding the intelligence of fathers and sons, test at 1% level of significance, whether father\u2019s intelligence and son\u2019s intelligence are independent.
      \n\"2nd
      \nAnswer:
      \nH0<\/sub> : Father\u2019s intelligence and sons intelligence are independent
      \nH1<\/sub> : Father\u2019s intelligence and sons intelligence are dependent.
      \nThe test statistic is – {Right\/upper tail test}
      \n\"2nd
      \nThe given data can be represented in 2 \u00d7 2 table as
      \n\"2nd
      \n\"2nd
      \n\"2nd
      \nFor 1 d.f. at 5% level of significance the appear
      \nTail critical value K = 3.84
      \nHere \u03c72<\/sup>cal<\/sub> is in rejection region \u03c72<\/sup>cal<\/sub> > \u03c72<\/sup>tab<\/sub>
      \nH0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion: Father\u2019s intelligence and sons intelligence are dependent.<\/p>\n

      \"KSEEB<\/p>\n

      Question 14.
      \nTo test the effectiveness of inoculation against Cholera the following table was obtained:
      \n\"2nd
      \nTest at 5% level of significance whether inoculation and attack of Cholera are independent.
      \nAnswer:
      \nH0<\/sub> : Inoculation and attack of cholera are independent
      \nH1<\/sub> : Inoculation and attack of cholera are dependent
      \nThe given data can be presented in the form of 2 \u00d7 2 contingency table as below :
      \n\"2nd
      \n\"2nd
      \nThe test statistic is
      \n\"2nd
      \n\u03c72<\/sup>cal<\/sub> = 0.6349
      \nFor 1 d.f at 5% level of significance the upper tail critical value k= 3.84
      \nHere \u03c72<\/sup>cal<\/sub> < k \/ \u03c72<\/sup>cal<\/sub> lies in (AR) acceptance region.
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion: Inoculation and attack of cholera are independent.<\/p>\n

      Question 15.
      \nThe members of a driving school examined the result of 200 candidates who where taking their driving test for the first time. They found that, out of 90 men, 62 passed and out of 110 women, 49 passed. Do these results indicate at 5% level of significance a relationship between the sex of a candidate and the ability to pass the driving test for the first time?(Use \u03c72<\/sup> test).
      \nAnswer:
      \nH0<\/sub>: Sex and ability to pass the driving test are independent.
      \nH1<\/sub> : Sex and ability to pass the driving test are dependent. {upper tail test}
      \nThe test statistic is –
      \n\"2nd
      \nThe given data can be presented in 2 \u00d7 2 contigency table as below:
      \n\"2nd
      \nFor ld.f. at 5% level of significance the upper tail critical value K = 3.84
      \nHere \u03c72<\/sup>cal<\/sub> = 11.88 is in rejection region\/ \u03c72<\/sup>cal<\/sub> > \u03c72<\/sup>tab<\/sub>
      \n\"2nd
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted
      \nConclusion : sex and ability to pass the driving test are dependent.<\/p>\n

      Question 16.
      \nOut of the 75 patients admitted to a hospital with Hepatities symtoms, 40 were administered a certain new drug, of these 34 recovered. Of the remaining who were give the traditional drug, 19 recovered. Test at 5% and 1% level of significance that the new drug is an improvement over the traditional drug.
      \nAnswer:
      \nH0<\/sub>: Type of drug and recovery from the disease are independent.
      \nH1<\/sub> : Type of drug and recovery from the disease are dependent.
      \nThe test statistic is –
      \n\"2nd
      \nThe given data can be presented in 2 \u00d7 2 contigency table as below:
      \n\"2nd
      \n\u03c72<\/sup> = \\(\\frac{75(34 \\times 16-6 \\times 19)^{2}}{40 \\times 35 \\times 53 \\times 22}\\)
      \n\u03c72<\/sup>cal<\/sub> = 8.495
      \nAt 5% level: for 1 d.f at 5% level of significance the upper tail critical value is K = 3.84.
      \nHere \u03c72<\/sup>cal<\/sub> = 8.495 is in rejection region\/\u03c72<\/sup>cal<\/sub> > \u03c72<\/sup>tab<\/sub>
      \n\u2234 H0<\/sub> rejected and H1<\/sub> is accepted.
      \n\"2nd
      \nAt 1% level : Here K = 6.63
      \nHere \u03c72<\/sup>cal<\/sub> = 8.495 is in rejection region\/ \u03c72<\/sup>cal<\/sub> > \u03c72<\/sup>tab<\/sub>
      \n\u2234 H0<\/sub> rejected and H1<\/sub> is accepted.
      \nConclusion: type of drug and recovery from the disease are dependent, t – Tests.<\/p>\n

      Question 17.
      \nA random sample of 8 envelopes are taken from a letter box of a post office and their weights in grams are found to be 12.1, 11.9, 12.4, 12.3, 11.9, 12.1, 12.4, 12.1. Does this sample indicate an average weight of the envelopes received at that post office is less than 12.35 gms at 1% level of significance.
      \nAnswer:
      \nGiven: n=8, \u00b5 = 12.35, a= 1%
      \nHere n is small use t-test (i.e., n < 30)
      \nH0<\/sub> : Average weight of the envelops is less than 12.35 gms.
      \ni.e., H0<\/sub> : \u00b5 = 12.35 gms.
      \nH1<\/sub> : Average weight of the envelope is less than 12.53 gms. i.e.,
      \nH1<\/sub> : (\u00b5< 12.35 gms {left Tailed test}
      \n\"2nd
      \nThe test statistic is –
      \n\"2nd
      \nHere, S = Sample S.D.
      \nX\u0304 = Sample mean.
      \nLet X be the weight in gms
      \n\"2nd
      \n\"2nd
      \n\\(=\\sqrt{\\frac{0.78}{8}-\\left(\\frac{2}{8}\\right)^{2}}\\)
      \n\u2234 s = 0.187
      \n\\(t=\\frac{12.15-12.35}{0.187 \/ \\sqrt{8-1}}\\)
      \n\u2234 tcal<\/sub> = -2.83
      \nFor (n – 1) = (8 – 1) = 7. d.f the lower left tail crtical value, (from t-table) at 5% level of significance is -k1<\/sub> =-1.90
      \nHere tcal<\/sub> = -2.83 is in rejection region (tcal<\/sub> < ttab<\/sub>)
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion : Average weight of envelops is less than 12.35 gms.<\/p>\n

      \"KSEEB<\/p>\n

      Question 18.
      \nThe following table gives the lengths of 12 samples of Egyptian cotton taken from a consginment. 48, 46, 49, 46, 52, 45, 43, 47, 47, 46, 45, 50. Test if the mean length of the consignment can be taken as more than 46.
      \nAnswer:
      \nGiven: n=12, \u00b5 = 46, Here n is small use t-test (i.e., n < 30)
      \nH0<\/sub>: Mean length of consignment of cotton is 46.
      \ni.e., H0<\/sub>: \u00b5 = 46.
      \nH1<\/sub> : Mean length of consignment of cotton is more than 46.
      \ni.e., H1<\/sub> : \u00b5 > 46. {Upper\/right tailed test} .
      \nThe test statistic is –
      \n\"2nd
      \nWhere x\u0304 = sample mean and
      \ns = sample standard deviation.
      \n\"2nd
      \n\"2nd‘
      \n\"2nd
      \n\\(=\\sqrt{\\frac{0.78}{8}-\\left(\\frac{2}{8}\\right)^{2}}\\)
      \n\u2234 S =0.187.
      \n\\(t=\\frac{12.15-12.35}{0.187 \/ \\sqrt{8-1}}\\)
      \n\u2234 tcal<\/sub> = -283.
      \nFor (n – 1) = (8 – 1) = 7. d.f the lower left tail crtical value, (from t-table) at 5% level of significance is -k1<\/sub> = -1.90
      \nHere tcal<\/sub> = -2.83 is in rejection region (tcal<\/sub> < ttab><\/sub>)
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion : Average weight of envelops is less than 12.35 gms.<\/p>\n

      Question 19.
      \nThe following table gives the lengths of 12 samples of Egyptian cotton taken from a consginment. 48, 46, 49, 46, 52, 45, 43, 47, 47, 46, 45, 50. Test if the mean length of the consignment can be taken as more than 46.
      \nAnswer:
      \nGiven: n = 12, \u00b5 = 46, Here n is small use t-test (i.e., n < 30)
      \nH0<\/sub>: Mean length of consignment of cotton is 46.
      \ni.e., H0<\/sub> : \u00b5 = 46.
      \nH1<\/sub> : Mean length of consignment of cotton is more than 46.
      \ni.e., H, : \u00b5 > 46. {Upper\/right tailed test} .
      \nThe test statistic is –
      \n\"2nd
      \nWhere x\u0304 = sample mean and
      \ns = sample standard deviation
      \n\"2nd
      \n\"2nd
      \n= 43 + \\(\\frac { 48 }{ 12 }\\) = 47
      \n\"2nd
      \n\u2234 s = 2.345
      \n\\(t=\\frac{47-46}{2.345 \/ \\sqrt{12-1}}\\)
      \n\u2234 tcal<\/sub> = 1.41
      \n\"2nd
      \nFor (n – 1) = (12 – 1) = 11 d.f at 5% level of significance the upper\/right tail critical value
      \nk2<\/sub>=1.80
      \nHere tcal<\/sub> = 1.41 is in acceptance region (i.e., tcal<\/sub> < ttab<\/sub>)
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : Mean length of the consignment of cotton is 46.<\/p>\n

      Question 20.
      \nIt is required to test whether those who practice yoga have average blood sugar less than 120. A sample of consisting of 18 persons who practice yoga is observed. If their mean blood sugar is 108 and S.D is 8, what would you conclude ?
      \nAnswer:
      \nGiven: n=18, x\u0304 =108, S = 8, \u00b5 =120.
      \nH0<\/sub> : The average blood sugar is 120.
      \ni.e., H0<\/sub>: \u00b5 = 120.
      \nH1<\/sub> : The average blood sugar is less than 120.
      \ni.e H1<\/sub>: \u00b5 < 120 {left tailed test}
      \n\"2nd
      \nThe test statistic is –
      \n\"2nd
      \n\\(t=\\frac{108-120}{s \/ \\sqrt{18-1}}\\)
      \nt cal<\/sub> = -6.18
      \nFor (n- 1) = (18 – 1)= 17 d.f at 5% level of significance the left\/upper lower Tail critical value -k = -1.74
      \nHere tcal<\/sub> = -6.18 is in rejection region. (tcal<\/sub> < ttab<\/sub>)
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion : The average blood sugar is less than 120.<\/p>\n

      Question 20.
      \nIt was found that a machine has produced pipes having a thickness 0.50 mm. To determine whether the machine is in proper working order a sample of 10 pipes is chosen for which the mean thickness is 0.53 mm and S.D. is 0.03 mm. Test the hypothesis that the machine is in proper working order using a level of significance of 0.01.
      \nAnswer:
      \nGiven: n= 10, x\u0304 =0.53, \u03bc = 0.50, S = 0.03, \u03b1 = 0.01 = l%
      \nH0<\/sub>: The machine is in proper working order,
      \ni.e., H0<\/sub>: \u00b5 = 0.50
      \nH1<\/sub> : The machine is not in proper working order.
      \ni.e., H1<\/sub> : \u00b5 \u2260 0.50 {two tailed test}
      \n\"2nd
      \nThe test statistic is –
      \n\"2nd
      \n\\(t=\\frac{0.53-0.50}{0.03 \/ \\sqrt{10-1}}\\)
      \nFor (n-1) = (10 -1) = 9 d.f at 1% level of significance the two tail critical values are
      \nk1<\/sub> =-3.25 and k2<\/sub> = 3.25
      \nHere tcal<\/sub> = 3 is in acceptance region.
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : The machine is in proper working order.<\/p>\n

      \"KSEEB<\/p>\n

      Question 21.
      \nTo compare the prices of a certain product in two cities ten shops were selected at
      \n\"2nd
      \nTest whether the average prices can be said to be same in the towns.
      \nAnswer:
      \nGiven: n1<\/sub> = 10 n2<\/sub> = 10.;
      \n{Here n1<\/sub> n2<\/sub> are less than 30 use small sample test for equality of meAnswer:}
      \nH0<\/sub> : The average prices in two cities are same,
      \ni.e., H0<\/sub> \u00b51<\/sub> – \u00b52<\/sub>.
      \nH1<\/sub> : The average prices in two cities are not same.
      \ni.e., H1<\/sub> : \u00b51<\/sub> \u2260 \u00b52<\/sub> {two tailed test}
      \n\"2nd
      \nThe test statistic is –
      \n\"2nd
      \nwhere
      \nx\u03041<\/sub> , x\u03042<\/sub> – sample means of city A & B
      \nS1<\/sub>2<\/sup>, S2<\/sub>2<\/sup> – sample variances of city A & B
      \nLet x and y be the prices of city A and B. For City A
      \n\"2nd
      \n\"2nd
      \n\\(=45+\\frac{124}{10} \\quad=45+\\frac{107}{10}\\)
      \nx\u03041<\/sub> = 57.4 x\u03042<\/sub> = 55.7
      \n\"2nd
      \n\\(=\\frac{1786}{10}-\\left(\\frac{124}{10}\\right)^{2}\\)
      \nS1<\/sub>2<\/sup> = 24.84
      \nS2<\/sub>2<\/sup> = 25.01
      \n\"2nd
      \ntcal<\/sub> = 0.723
      \nFor (n1<\/sub> + n2<\/sub> – 2) = (10 + 10 -2) = 18 d.f at 5% level of sigficance the two Tail critical values are -k1<\/sub> = -2.10 and k2<\/sub>\u00a0= 2.10.
      \nHere tcal<\/sub> = 0.723 is in acceptance region.
      \n\u2234 H0<\/sub> is accepted
      \nConclusion.: The average prices in two cities are same.
      \nH1<\/sub> : The increase in weight due to diet A is more than diet B.
      \ni.e., : H1<\/sub> : \u00b51<\/sub> > \u00b52<\/sub>r {Right tailed test}
      \nThe test statistic is –
      \n\"2nd
      \nWhere x\u03041<\/sub>, x\u03042<\/sub> – Sample means of diet A and diet B and S1<\/sub>2<\/sup>, S1<\/sub>2<\/sup> – Sample variances of Diet A and B.
      \nLet x and y be the weights of diet A and B.
      \n\"2nd
      \n\"2nd
      \n= 2 + \\(\\frac { 54 }{ 10 }\\) = 2 + \\(\\frac { 32 }{ 8 }\\)
      \nx\u03041<\/sub> = 7.4 x\u03042<\/sub> = 6
      \n\"2nd
      \n\"2nd
      \n\\(=\\frac{394}{10}-\\left(\\frac{54}{10}\\right)^{2} \\quad=\\frac{208}{8}-\\left(\\frac{32}{8}\\right)^{2}\\)
      \nS1<\/sub>2<\/sup> = 10.24 s2<\/sub>2<\/sup> = 10.
      \n\"2nd<\/p>\n

      Question 22.
      \nFrom the following data regarding weights of randomly selected boys and girls of P.U.C. classes, test whether P.U.C. boys weigh more than P.U.C. girls. [Test at 5% level of significance]
      \n\"2nd
      \nAnswer:
      \nGiven n1<\/sub> = 10, n2<\/sub> = 12 (less than 30, use t test)
      \nx\u03041<\/sub> = 71, x\u03042<\/sub> = 69
      \ns1<\/sub>2<\/sup> = 3 s2<\/sub>2<\/sup> = 2
      \nH0<\/sub> :mean weight of boys and girls are same,
      \n(i.e. H0<\/sub>: \u00b51<\/sub>= \u00b52<\/sub>)
      \nH1<\/sub> : mean weight of boys is more than girls
      \n(i.e. H1<\/sub>: \u00b51<\/sub> > \u00b52<\/sub>) …(upper tail test)
      \n\"2nd
      \nThe test statistic is
      \n\"2nd
      \nFor (n1<\/sub> + n2<\/sub> – 2) = (10 + 12 -2) = 20 degrees of freedom the upper tail critical value for 5% level of significance k2<\/sub> = 1.73
      \nHerz tcal<\/sub> > k2<\/sub>\/tcal<\/sub> lies in (RR) rejection region.
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion : Mean weight of boys is more than girls.<\/p>\n

      \"KSEEB<\/p>\n

      Question 23.
      \nA group of 10 rabbits fed on diet A and another group of 8 rats fed on a different diet B recorded the following increase in weig it in gms.
      \n\"2nd
      \nTest whether diet A is superior to diet B.
      \nAnswer:
      \nGiven : n1<\/sub> =10, n2<\/sub> = 8. {Here n1<\/sub>,n2<\/sub> < 30 use t-test.}
      \nH0<\/sub> : There is no significant difference between mean weight due to diet A and B.
      \ni.e., : H0<\/sub> : \u00b51<\/sub> = \u00b52<\/sub>.
      \nH1<\/sub> : The increase in weight due to diet A is more than diet B.
      \ni.e., : H1<\/sub>: \u00b51<\/sub> > \u00b52<\/sub> {Right tailed test}
      \nThe test statistic is –
      \n\"2nd
      \nWhere x\u03041<\/sub>, x\u03042<\/sub> – Sample means of diet A and diet B and S1<\/sub>2<\/sup>, S2<\/sub>2<\/sup> – Sample variances of Diet A and B.
      \nLet x and y be the weights of diet A and B.
      \n\"\"
      \n\"2nd
      \n= 2 + \\(\\frac { 54 }{ 10 }\\) = 2 \\(\\frac { 32 }{ 8 }\\)
      \nx\u0304\u03041<\/sub> = 7.4 x\u03042<\/sub> = 6
      \n\"2nd
      \n\"2nd
      \n\\(=\\frac{394}{10}-\\left(\\frac{54}{10}\\right)^{2} \\quad=\\frac{208}{8}-\\left(\\frac{32}{8}\\right)^{2}\\)
      \ns1<\/sub>2<\/sup> = 10.24 s2<\/sub>2<\/sup> = 10
      \n\"2nd
      \ntcal<\/sub> =0.874
      \nFor (n1<\/sub> + n2<\/sub> – 2) = (10 + 8 – 2) = 16 d.f at 5% level of significance the upper tail critical
      \nvalue k= 1.75
      \nHere tcal<\/sub> = 0.874 is in acceptance region (tcal<\/sub> < ttab<\/sub>)
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : There is no siginificant difference betwee mean weight due to diet A and B.<\/p>\n

      Question 24.
      \nTwo types of batteries are tested for their length of life and the following data is obtained. Is there a significant difference in the two means?
      \n\"2nd
      \nAnswer:
      \nGiven:
      \nn1<\/sub> = 9, x\u03041<\/sub> =500, S1<\/sub>2<\/sup> = 100
      \nn2<\/sub> = 8, x\u03042<\/sub> =540, S2<\/sub>2<\/sup> = 121 :
      \nH0<\/sub> : There is no significant difference in the two means,
      \ni.e., : H0<\/sub>: \u00b51<\/sub>, = \u00b52<\/sub>
      \nH1<\/sub> : There is a significant difference in the two meAnswer:
      \ni.e., : H1<\/sub> : \u00b51<\/sub> \u2260 \u00b52<\/sub>.
      \n\"2nd
      \nThe test statistic is –
      \n\"2nd
      \nFor (n1<\/sub> + n2<\/sub> – 2) = (9 + 8 – 2) = 15 d.f. at 5%
      \nlevel of signficance the two tail critical values are -k1<\/sub> = -2.13 and k2<\/sub> = 2.13.
      \nHere tcal<\/sub> = -7.38 is in rejection region.
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion: There is no significant difference in the two meAnswer:<\/p>\n

      Question 25.
      \nThe following are the B.P. (Blood Pressure) of 8 persons before and after performing ‘yoga’.
      \n\"2nd
      \nCan we conclude that yoga reduces BP?
      \nAnswer:
      \nHere B.P. before and after yoga of the same persons are given, so use paired t-test for equality of means :
      \nH0<\/sub> : B.P. before and after yoga is same.
      \ni.e., H0<\/sub>: \u00b51<\/sub> = \u00b52<\/sub>
      \nH1<\/sub> : BP reduced after yoga.
      \ni.e., H1<\/sub> : \u00b51<\/sub> > \u00b52<\/sub> {Right Tailed test}
      \n\"2nd
      \nLet x and y be the BP before and after yoga.
      \n\"2nd
      \n\u2234 d\u0305 = \\(\\frac { 38 }{ 8 }\\)
      \n\\(S=\\sqrt{\\frac{252}{8}-\\left(\\frac{38}{8}\\right)^{2}}=2.99\\)
      \n\\(t=\\frac{4.75}{2.99 \/ \\sqrt{8-1}}\\)
      \n\u2234 tcal<\/sub> = 4.203
      \n\"2nd
      \nFor (n – 1) = 8 – 1 = 7 d.f. at 5% level of significance the upper tail critical value k = 1.90.
      \nHere tcal<\/sub> = 4.203 is in rejection region (tcal<\/sub> > ttab<\/sub>)
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> is accepted.
      \nConclusion : B.P reduced after yoga.<\/p>\n

      \"KSEEB<\/p>\n

      Question 26.
      \nThe following are the marks obtained by 5 students before and after attending coaching classes.
      \n\"2nd
      \nTest whether marks of students has improved after attending coaching classes.
      \nAnswer:
      \nHere marks of same students before and after attending coaching class has given so use paired t-test for equality of meAnswer:
      \nH0<\/sub> : There is no change in marks of the students before and after attending coaching class,
      \ni.e., H0<\/sub>: \u00b51<\/sub> = \u00b52<\/sub>
      \nH1<\/sub> : The marks of the students have improved after attending coaching class i.e.,
      \nH1<\/sub> : \u00b51<\/sub>, < \u00b52<\/sub> {Left Tail test}
      \nThe test statistic is –
      \n\"2nd
      \nLet x and y be the marks of students before and after attending coaching classes
      \n\"2nd
      \n\u2234 d\u0305 = \\(\\frac { -10 }{ 5 }\\) = 2
      \ns = \\(\\sqrt{\\frac{140}{5}-\\left(\\frac{-10}{5}\\right)^{2}}\\)
      \ns = 4.89
      \n\"2nd
      \n\\(t=\\frac{-2}{4.89 \/ \\sqrt{5-1}}\\)
      \n\u2234 tcal<\/sub> = -0.818
      \nFor (n – 1) = (5 – 1) = 4 d.f the lower critical value at 5% level of significance
      \n-k = -2.13
      \nHere tcal<\/sub> = -0.818 is in acceptance region.
      \n\u2234 H0<\/sub> is accepted.
      \nConclusion : There is no change in the marks of the students before and after attending coaching classes.<\/p>\n

      Question 27.
      \nThe manufacturer of a cattle feed claims that, cows fed on his product yield more milk. To substaintiate his claim, 8 cows were fed on the manufacturer’s feed and their milk yield in litres. When they were under usual feed were given below:
      \nAnswer:
      \nHere milk yield of the same set of cows are given, use paired t-test for equality of meAnswer:
      \nH0<\/sub> : Milk yield of cows before and after manufacturers feed is same.
      \ni.e., H0<\/sub>: \u00b51<\/sub>\u00a0= \u00b52<\/sub>.
      \nH1<\/sub> : Milk yield of cows after manufacturers feed is more.
      \ni.e., H1<\/sub> : \u00b51<\/sub> < \u00b52<\/sub> {left tailed test}
      \nThe test statistic is –
      \n\"2nd
      \n\"2nd
      \n\u2234 d\u0305 = \\(\\frac { -10 }{ 5 }\\) = -1.25
      \ns = \\(\\sqrt{\\frac{31.7}{8}-\\left(\\frac{-10}{8}\\right)^{2}}\\)
      \ns = 1.55
      \n\\(t=\\frac{1.25}{1.55 \/ \\sqrt{8-1}}\\)
      \ntcal<\/sub> = 2.13
      \n\"2nd
      \nFor (n – 1) = (8 – 1) = 7 d.f. The upper tail critical value at 5% level of significance is k = 1.90
      \nHere tcal<\/sub> = -2.13 is in rejection region.\/ tcal<\/sub> < K
      \n\u2234 H0<\/sub> is rejected and H1<\/sub> accepted.
      \nConclusion : Milk yield of cows after manufacturers feed has increased.<\/p>\n

      \"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

      You can Download Chapter 6 Statistical Inference Questions and Answers, Notes, 2nd PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Statistics Question Bank Chapter 6 Statistical Inference 2nd PUC Statistical Inference One and Two Marks Questions With …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13465"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=13465"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13465\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=13465"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=13465"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=13465"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}