Question

In a certain city, 15% of senior citizens have used Uber. To see if an advertising campaign was effective, a researcher samples 200 senior citizens and finds that 40 have used Uber. Testing at significance level α = .05, find:

H₀ and H_a:

Sample proportion =

Standard error (S.E.) assuming H₀ is true =

z-value =

p-value =

Conclusion:

Group of answer choices

do not reject H₀

50-50 chance advertising campaign was effective

.05 chance advertising campaign was effective

reject H₀

Answer 1

The following information is provided: The sample size is N = 200 , the number of favorable cases is X = 40 and the sample proportion is and the significance level is \alpha = 0.05α=0.05

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

advertising campaign was not effective

Ho: p = 0.15

advertising campaign was effective

Ha: p > 0.15

This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is z_c = 1.64

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z = 1.98 >zc​=1.64, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0238 , and since p = 0.0238 <0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is greater than p0​, at the α=0.05 significance level.

reject H₀

Conclusion:advertising campaign was effective