Question

8.52 The weights (in pounds) of 30 randomly selected boy students from a high school are given below. 181 168 150 175 174 154 156 187 161 162 173 163 164 168 147 170 170 163 173 153 174 175 150 175 176 177 178 143 186 157 a. Find a 95% confidence interval for o. b. Based on the confidence interval from part (a), what would be your decision for the test H:0° = 100 versus H: 0° +100 at the level of a = 0.05? c. Find the p-value of the test.

Answer 1

Calculate the standard deviation of the given data. Date: ----24/11/2019 Mean (7) = 2x 181+168 +...+186 +157 30 5003 30 = 166.7667 = 166.77 (Round to 2 decimal place) (x, - F) Standard deviation(s) = n-1 |(181–166.77) +...+(157–166.77)? 30-1 =11.569

Calculate the degrees of freedom. df = n-1 = 30-1 = 29 Calculate the chi-square critical values. Xin = 2.05, 9) =(=CHIINV(0.05,29)) = 42.557 (Use MS Excel) KŘight = X{X-0.05,29) = 420.95.29) = (=CHIINV(0.95,29)) = 17.708 (Use MS Excel)

a) isht Calculate the 95% confidence interval for o? 95%CI for (62) = (n-1)s? co?- (n-1) s? Xiet . XRight _ (30–1)(11.569) (301)(11.569)? 42.557 17.708 =91.20499727<b> <219.1896922 =91.205 <o? <219.190 Hence, the required confidence interval is 91.205 <<219.190). Null hypothesis, H,:= 100 Alternative hypothesis, H,:0° +100 Level of significance, a = 0.05 Conclusion: It is observed that the 95% confidence interval for variance does not contain null value (zero), So, it is failing to reject null hypothesis and conclude that the population variance is equal to 100.

c) Calculate the test statistic value. Zāralian = (n-1) s? _(30 – 1)(11.569) 100 = 38.81411 = 38.81 (Round to 2 decimal place) Calculate the P-value. P-value = (=CHIDIST(38.81,29)) (Use MS Excel function) = 0.105357 = 0.1054 (Round to 4 decimal place) Hence, the required P-value is 0.1054)|