me ID 4. Sample of 17 items provides a sample standard deviation of 9.0 (sample
ID: 3329877 • Letter: M
Question
me ID 4. Sample of 17 items provides a sample standard deviation of 9.0 (sample variance is 81) a. Compute 90% confidence interval estimate of the population variance. b. Test the following hypotheses using a-0.05. What is your conclusion? (Using the critical value Hy:2>50 is the population variance. Chi Square Distribution table Upper Tail Area 0.1 005 a 0025 0.9 79 21.064 23.685 26.119 2934 0.975 0.95 0.99 df 14 15 16 17 18 4.66 5.629 6571 5229 6262 7261 8547 22307 24996 274883057 5.812 6408 75648.672 10.085 24769 27587 30191 33.409 7.015 8231 9.39 10.865 25.989 28.869 31 52634.80 6.908 7962 9.312 23.542 26296 28.845 32Explanation / Answer
Q.4 (a) Sample standard deviation s= 9.0
Sample variance s2 = 81.0
sample size = 17 so dF = 16
90% lower confidence ineterval = (n-1)s2 / x2/2 = (17 -1) * 81/ x20.05 = 16 * 81 / 28.845 = 44.93
90% Upper confidence ineterval =(n-1)s2 / x21-/2 = (17 -1) * 81/ x20.95 = 16 * 81 / 7.962 = 162.77
so 90% CI = (44.93, 162.77)
b. We have to test the hypothesis at alpha = 0.05 level.
Hypothesis are
2 < 50
2 > 50
Test statistic
X2 = (n-1)s2 /2 = (17-1) * 81/ 50 = 25.92
Critical value for dF = 16
X2cr = 26.296 for alpha = 0.05 and df = 16
so we cannot reject the null hypothesis.
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