Question

Question 1 2.5 pts Use the following to answer all of the questions. According to the Pew Internet and American Life Project 75% of American adults use the Internet. A survey of 600 adults aged 50-64 found 420 used the Internet (70%). We want to determine whether the proportion of Internet users age 50-64 differs from the overall proportion of 0.75. What is the alternative hypothesis? OUz O.75 p 0.75 Op 0.70 Op 0.75 2.5pts Question 2 Compute the test statistic. Give your answer to 2 decimal places.

Answer 1

The following information is provided: The sample size is N = 600, the number of favorable cases is X = 420, and the sample proportion is $\bar p = \frac{X}{N} = \frac{ 420}{ 600} = 0.7$ , 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 = 0.75

Ha: p $\neq$ 0.75

This corresponds to a two-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 two-tailed test is z_c = 1.96.

(3) Test Statistics

The z-statistic is computed as follows:

$z = \frac{\bar p - p_0}{\sqrt{p_0(1-p_0)/n}}$

$z = \frac{ 0.7 - 0.75 }{\sqrt{ 0.75(1- 0.75)/600}}$

z = -2.828

(4) Decision about the null hypothesis

Since it is observed that |z| = 2.828 > z_c = 1.96, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0047, and since p = 0.0047 < 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 different than p0​, at the α = 0.05 significance level.

Question1

B) p $\neq$ 0.75

Question 2

z = -2.828

Question 3

p = 0.0047

Question 4

Reject Ho