A random sample of 100 observations from a population with s

A random sample of 100 observations from a population with standard deviation 64 yielded a sample mean of 112. Complete parts a through c. a.Test the null hypothesis that mu = 100 against the alternative hypothesis that mu> 100, using alpha = 0.05.lnterpret the results of the test. .H0 is not rejected. .H0 is rejected. Interpret the results of the test. Choose the correct interpretation below. . There is sufficient evidence to indicate the true population mean is not equal to 100 at alpha = 0.05. There is sufficient evidence to indicate the true population mean is greater than 100 at alpha = 0.05. There is sufficient evidence to indicate the true population mean is smaller than 100 at alpha = 0.05. b. Test the null hypothesis that mu= 100 against the alternative hypothesis that alpha = 100, using a 0.05. Interpret the results of the test. H0 is rejected. H0 is not rejected. Interpret the results of the test, Choose the correct interpretation below, . there is sufficient evidence to indicate mu is greater than 100 at alpha = 0.05, . There is sufficient evidence to indicate mu is not equal to 100 at alpha = 0.05. .There is insufficient evidence to indicate mu is smaller than 100 at alpha = 0.05. c. Compare the results of the two tests you conducted. Explain why the results differ. Choose the correct answer below.

Solution

(a) The test statistic is

Z=(xbar-mean)/(s/vn)

=(112-100)/(64/sqrt(100))

=1.88

It is a right-tailed test.

Given a=0.05, the critical value is Z(0.05) = 1.645 (from standard normal table)

The rejection region is if Z>1.645, we reject the null hypothesis.

Since Z=1.88 is larger than 1.645, we rejct Ho.

Answer: Ho is rejected.

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A

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(b)It is a two-tailed test.

Give a=0.05, the critical values are Z(0.025) = -1.96 or 1.96 (from standard normal table)

The rejection regions are if Z<-1.96 or Z>1.96, we reject Ho.

Since Z=1.88 is between -1.96 and 1.96, we do not reject Ho.

Answer: Ho is not rejected

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B