Question

the unknown Popular random sample of size 17 is obtained from a normal (15) 4.Malding a sample standard deviation of S. Teet the nuit that the unknown population variance is greater than or 169, versus the alternative hypothesis that the unknown lance is less than 169 using a 1% level of significance hypothesis that Set up the null and alternative hypotheses, clearly defining any unknown parameters. Note the value is always in the mull hypothesis m atatest statistie: (1) sensitive to the unknown parameter with mown distribution under the value of the unknown parameter. First state and clearly label the distribution fact that you are using and han state and clearly label the test statistic and its distribution under Ho. Estimate of unknown parameter in : Distribution Fact: under Ho Test Statistic: ___obs - (1) Look at the alternative hypothesis and decide what values of the estimate of the parameter in (1) would lead you to Reject Ho (general direction) Based on the answer state what values of the test statistics would lead you to Reiet Hot direction). Then draw and label a number line reflecting this answer. Using these results along with the level of significance, find the cutoff point(s) for the acceptance/rejection regions ) Clearly state your decision rules (Are they com You decision rules (Are they consistent with the number tine in step 3? (vi) Calculate your test statistic, in test statistic, indicate whether you will accept Ho or reject Ho. Inter reject Ho. Interpret your results.

Answer 1

n = 17

sample variance =s^2 =8^2 = 64

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 0² > 169 Ha:02 < 169 This corresponds to a left-tailed test, for which a Chi-Square test for one population variance will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the the rejection region for this left-tailed test is R= {x? : x2 <5.812}. (3) Test Statistics The Chi-Squared statistic is computed as follows: (n-1)s x = - (17 – 1)-64 E = 6.059 169 2 (4) Decision about the null hypothesis Since it is observed that x? = 6.059 > x2 = 5.812, it is then concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population variance o2 is less than 169, at the 0.01 significance level.