17 In 1997 a survey of 940 households showed that 144 of the

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In 1997, a survey of 940 households showed that 144 of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level. Use this information to answer the following questions.

The test statistic is z=______

(Round to two decimal places as needed.)

The P-value is_______

(Round to four decimal places as needed.)

Solution

Using a formula for a proportional one-sample z-test with your data included, we have:
z = .16 - .15 -->test value (150/940) is .16) minus population value (.15) divided by
?[(.15)(.85)/940] --> .85 represents 1-.15 and 940 is sample size.
Finish the calculation.
Use a z-table to find the p-value. The p-value is the actual level of the test statistic.

Check a z-table for the critical value at .05 level of significance for a one-tailed test. Compare the test statistic you calculated to the critical value from the table. If the test statistic exceeds the critical value, reject the null and conclude p>0.15 (there is sufficient evidence to support the claim); if the test statistic does not exceed the critical value from the table, do not reject the null (there is not sufficient evidence to support the claim).