Question

Consider the following hypothesis test: H0: u>=71 Ha:u<71 A sample of 110 is used and the population standard deviation is 11. Compute the -value and state your conclusion for each of the following sample results. Use a=0.02. Round value to two decimal places and p-value to four decimal places. If your answer is zero, enter "0". Enter the negative value as a negative number. x^- =68.5 z-values =? p-value=? x^- =67.5 z-values=? p-value=? c. x^- =66 z-values =? p-value=? d. x^- =73 z-values =? p-value=?

CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed news, short features, and advertisements. The length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8.5 minutes. A sample of actual waiting times will be used to test this assumption and determine whether actual mean waiting time differs from this standard.

1. Formulate the hypotheses for this application.

Ho:u=8.5

Ho:=/8.5

3. A sample of 100 shoppers showed a sample mean waiting time of 8.6 minutes. Assume a population standard deviation of  o=3.8 minutes. What is the p-value?

z values =? To 2 decimals

p-value=? To 4 decimals

d. Compute a 98% confidence interval for the population mean. Does it support your conclusion?

  ,   (to 2 decimals)

Answer 1

Solution:-

1)

a) x = 68.5

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 71
Alternative hypothesis: u < 71

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.02. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 1.0488

z = (x - u) / SE

z = - 2.38

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of - 2.38

Thus the P-value in this analysis is 0.0087

Interpret results. Since the P-value (0.0087) is less than the significance level (0.02), we have to reject the null hypothesis.

b) x = 67.5

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 71
Alternative hypothesis: u < 71

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.02. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 1.0488

z = (x - u) / SE

z = - 3.34

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of - 3.34

Thus the P-value in this analysis is 0.0004

Interpret results. Since the P-value (0.0004) is less than the significance level (0.02), we have to reject the null hypothesis.

c) x = 66

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 71
Alternative hypothesis: u < 71

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.02. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 1.0488

z = (x - u) / SE

z = - 4.77

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of - 4.77

Thus the P-value in this analysis is 0.0000

Interpret results. Since the P-value (0.0000) is less than the significance level (0.02), we have to reject the null hypothesis.

d) x = 73

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 71
Alternative hypothesis: u < 71

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.02. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 1.0488

z = (x - u) / SE

z = 1.91

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of 1.91

Thus the P-value in this analysis is 0.97

Interpret results. Since the P-value (0.97) is greater than the significance level (0.02), we failed to reject the null hypothesis.