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.

Required:
a. Test the null hypothesis according to which the true mean is greater or equal to 100 min against the alternative that it is less than 100. Use a significance level of 5% (state the null and the alternative hypothesis, the test statistic and the conclusion).
b. Compute a 90% 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)

We fail to reject the Null ;

(88.323, 130.677)

Step-by-step explanation:

Given the data :

117, 156, 89, 72, 116, 125, 101, 100

H0 : μ ≥ 100

H1: μ < 100

From the data: using calculator :

Mean, x = 109.5

Standard deviation, s = 25.33

The test statistic:

(x - μ) ÷ s/Sqrt(n)

(109.5 - 100) ÷ 25.33/sqrt(8)

Test statistic = 1.06

Since, same size is small, we use t test ;

Using the Pvalue calculator :

T at df = 8 - 1 = 7, α = 0.05

Pvalue = 0.161

Pvalue > α

0.161 > 0.05 ; Hence, we fail to reject H0.

B.)

C. I = mean ± Tcritical *s/√n

Tcritical = 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)

(88.323, 130.677)

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

step-by-step explanation:

i think you should think about the question and not look online.

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50