Question

Problems 6-10: In 2015 about 60% of college students used Twitter. In a recent study of 100 students, 80 now use Twitter. Is this evidence to think the percent has changed? We wish to conduct a hypothesis test with 5% significance.

6) State the null and alternative hypothesis.

7) Find the test statistic and P-value and report those values here:

8) What is the decision and conclusion?

9) Now, construct a 95% confidence interval using the sample of 100 students and 80 saying they use twitter.

10) Does the Confidence Interval support your decision for #8? Why or why not?

11 (Bonus): What type of error could we have made in this anaylsis? (Type 1 or Type 2)

Answer 1

using excel>addin>phstat>one sampel test

we have

Z Test of Hypothesis for the Proportion
Data
Null Hypothesis            p =	0.6
Level of Significance	0.05
Number of Items of Interest	80
Sample Size	100
Intermediate Calculations
Sample Proportion	0.8
Standard Error	0.0490
Z Test Statistic	4.0825
Two-Tail Test
Lower Critical Value	-1.9600
Upper Critical Value	1.9600
p-Value	0.0000
Reject the null hypothesis

6) the null and alternative hypothesis.

Ho:p=0.60

Ha:p 0.60

7) the test statistic is 4.0825 and P-value 0.0000

8) since p value is less than 0.05 so reject Ho  and conclude the percent has changed

using excel

we have

Confidence Interval Estimate for the Proportion
Data
Sample Size	100
Number of Successes	80
Confidence Level	95%
Intermediate Calculations
Sample Proportion	0.8
Z Value	-1.9600
Standard Error of the Proportion	0.0400
Interval Half Width	0.0784
Confidence Interval
Interval Lower Limit	0.7216
Interval Upper Limit	0.8784

9) a 95% confidence interval is 0.7216,0.8784

10)yes the Confidence Interval support the decision

11 Type 2.