Question

assume that when adults with smart phones are randomly selected, 54% of them and use them in meetings. If 12 adult smartphone users are randomly selected, find the probability that fewer than 4 of them use their smart phones in meetings or classes.

Answer 1

The following information is provided: The sample size is N=12, the number of favorable cases is X=4, and the sample proportion is  p = 4/12 = 0.3333, and the significance level is α=0.05

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: p = 0.54

Ha:p<0.54

This corresponds to a left-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 left-tailed test is zc​=−1.64.

The rejection region for this left-tailed test isR={z:z<−1.64}

(3) Test Statistics

The z-statistic is computed as follows:

z=p0​(1−p0​) / n​pˉ​−p0​​ = 0.54(1−0.54)/12​0.3333−0.54​ =−1.436

(4) Decision about the null hypothesis

Since it is observed that z=−1.436 ≥ zc​=−1.64, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The at fewer than 4 of them use their smart phones in meetings or classes. p=0.0754, and since p=0.0754≥0.05, 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 pp is less than α=0.05 significance level.

Confidence Interval

The 95% confidence interval for  fewer than 4 of them use their smart phones in meetings or classes. pp is: 0.067<p<0.6.