Question

1.Test the hypothesis that the proportion of students who have the Wuhan flu is .3. Use a .10 significance level, a two-tail test and the following data: A sample of 100 students has 40 with the virus.

2 Test the hypothesis that the mean number of hours the Wuhan flu can live on a cell phone is more than 20. Use a .01 significance level and a one tail test and the following data: a sample of 50 phones has a sample mean =21.5 and a variance=9.

3 Test the null hypothesis that the proportion of students who believe TRUMP policy is appropriate dealing with the TRUMP virus is .6. The alternative hypothesis is the proportion >.6. Use a .05 significance level. A sample of 25 students has 19 think TRUMP policy is effective.

4) Are students avoiding events that average more than 20 people? A sample of 25 event has an average attendance of 23.3 people and a variance of 4 people. Use a significance level of 5% and a one-tail test.

5. Is the vaccine effective for the Wuhan flu? A sample of 30 students were given the vaccine and 12 got the virus. Without the vaccine, the rate of infection is 50%. Use a 1 tail test and a 5% significance level. What is the p value. [Khan academy has a good discussion of the P value]

6. Do students spend more than 6 ours a day on the phones? A sample of 150 students has an average usage of 8 hours and a variance of 6 hours. Use a 10% significance level.

Answer 1

SOLUTION 1: p= 0.3 level of significance=0.1 Two tailed test. X=40 and N=100

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested:

Ho:p=0.3

Ha:p≠​0.3

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 isα=0.1, and the critical value for a two-tailed test is zc​=1.64.

The rejection region for this two-tailed test is R={z:∣z∣>1.64}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis: Since it is observed that |z| = 2.182 > z_c = 1.64∣z∣=2.182>zc​=1.64, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0291, and since p=0.0291<0.1, 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 pp is different than 0.3​, at the α=0.1 significance level.

NOTE: AS PER THE GUIDELINES I HAVE DONE THE FIRST PLEASE RE POST THE REST. THANK YOU.