Question

A research center claims that less than 20% of Internet users in the United States have a wireless network in their home. In a random sample of 100 adults, 15% say “yes have a wireless network in their home”. At alpha=0.01, specifically follow and address the questions below to determine if there enough evidence to support the researcher’s claim.

1. Verify that np>=5 and nq>=5
2. Identify the claimed distribution and state Ho and Ha
3. Specify the level of significance, alpha
4. Find the critical values z, and identify the rejection region
5. Use the z-test to find the standardized test statistic z
6. Decide whether to reject the null hypothesis
7. Interpret the decision in the context of the original claim

Answer 1

a.

n.p = 100 * 0.20 = 20 > 5

n.q = 100 * 0.80 = 80 > 5

b.

Ho:p = 0.20.

The proportion of Internet users in the United States have a wireless network in their home is 0.20

Ha:p < 0.20.

The proportion of Internet users in the United States have a wireless network in their home is less than 0.20

c.

the level of significance is 0.01.

d.

The critical values z and the rejection region are zc​=−2.33 and R = { z : z < −2.33} respectively.

e.

z statistics is -1.25.

f.

Do not reject the null hypothesis.

g.

Therefore, there is not enough evidence to claim that the proportion of Internet users in the United States have a wireless network in their home is less than 0.20

......................................................................................................................................................................................

SOLUTION

The sample size is N = 100, the number of favorable cases is X = 15, and the sample proportion is

, and the significance level is α=0.01

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:p = 0.20.

The proportion of Internet users in the United States have a wireless network in their home is 0.20

Ha:p < 0.20.

The proportion of Internet users in the United States have a wireless network in their home is less than 0.20

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

(2) Rejection Region

The significance level is α = 0.01, and the critical value for a left-tailed test is zc​=−2.33.

The rejection region for this left-tailed test is R = { z : z < −2.33}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z = −1.25 ≥ zc ​=−2.33, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.1056, and since p = 0.1056 ≥ 0.01, it is 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 proportion p is less than p0​, at the α=0.01 significance level.

Therefore, there is not enough evidence to claim that the proportion of Internet users in the United States have a wireless network in their home is less than 0.20

Confidence Interval

The 99% confidence interval for p is: 0.058<p<0.242.