Test scores in STAT 1342 course are assumed to be normally distributed with unkn

Question

Test scores in STAT 1342 course are assumed to be normally distributed with unknown population summaries. A sample of 16 students was selected for a study of variability in the test scores. the sample summaries obtained from the sample are shown below. [Sample Mean] = 73.7 and [Sample Standard Deviation] = 8.0. A. Estimate the population variance with the 99 % confidence. Round the confidence limits to the 2nd decimal place. B. At the 5% significance level, do we have sufficient evidence that the population variance exceeded 30? Show the following elements of your decision. Critical Value(s) = Test Statistic = Rejection Rule states: Decision: YES (Reject) or NO (Do Not Reject) C. At the 5% significance level, do we have sufficient evidence that the population variance was below 128? Show the following elements of your decision. Critical Value(s) = Test Statistic = Rejection Rule: Reject the null hypothesis if... Decision: YES (Reject) or NO (Do Not Reject)

Explanation / Answer

a)

As              
              
df = n - 1 =    15          
alpha = (1 - confidence level)/2 =    0.005          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    32.80132065          
chi^2(alpha/2) =    4.600915572          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    29.26711428          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    208.6541222          
              
Thus, the confidence interval for the variance is              
              
(   29.26711428   ,   208.6541222   ) [ANSWER]

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b)

Formulating the null and alternative hypotheses,              
              
Ho:   sigma^2   <=   30
Ha:    sigma^2   >   30
              
As we can see, this is a    right   tailed test.      
              
Thus, getting the critical chi^2, as alpha =    0.05   ,      
alpha =    0.05          
df = N - 1 =    15          
chi^2 (crit) =    24.99579014   [ANSWER, CRITICAL VALUE]     

*********
              
Getting the test statistic, as              
s = sample standard deviation =    8          
sigmao = hypothesized standard deviation =    5.477225575          
n = sample size =    16          
              
              
Thus, chi^2 = (N - 1)(s/sigmao)^2 =    32   [ANSWER, TEST STATISTIC]
      
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We reject Ho when chi^2 > 24.996 [ANSWER]

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As chi^2 < 24.996, then we FAIL TO REJECT THE NULL HYPOTHESIS.   [ANSWER: NO]          

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