The following data are the daily number of minutes of smartphone use for a random sample of 8 students at your institution: 117, 156, 89, 72, 116, 125, 101, 100. (a) Test the null hypothesis according to which the true mean is 100 min against the alternative that it is different than 100. Use a significance level of 5% (state the null and the alternative hypothesis, the test statistic and the conclusion). (10 points)
(b) Compute a 95% confidence interval for the true mean number of minutes per day a student uses his or her smartphone. Is your finding consistent with your answer in (a) (Explain why or why not)?

Step-by-step explanation:

Number of students = 8

Mean of sample = 876/8

= 109.5

We solve for standard deviation using formula:

S = √(Xi -barX)/n-1

S = 25.3264

A.

H0: u = 100

H1: u not equal to 100

T test = barX - u/s/√n

t = 109.5-100/(25.3264/√8)

= 9.5/8.9542

= 1.0609

Df = 8-1 = 7

P value = 0.953141 at 5% significance level. P-value > 0.05 so we fail to reject H0

Conclusion: insufficient evidence that mean = 100

B.

95% = bar x +-t*s/√n

t = 2.365 at 95% and df = 7

109.5 +- 2.365(25.3264/√8)

= 109.5 +- 2.365(8.9542)

= 109.5-21.1768, 109.5+21.1768

= (88.3232, 130.6768)

In conclusion true mean is in the confidence interval, so it is consistent.

1.5%

step-by-step explanation:

a. -0.3 > -0.33