Question

A new article reported that college students who have part-time jobs work an average of 15 hour per week. The staff of a college from for newspaper thought that the average might be different from 15 hours per week for their college. Data were collected on the number of hours worked per week for a random sample of students at the college who have part-time jobs. The data were used to test the hypotheses H_o: mu = 15 H_a: mu notequalto 15. where mu is the mean number of hours worked per week for all students at the college with part-time jobs. The results of the test are shown in the table below. Assuming all conditions for inference were met, which of the following represents a 95 percent confidence interval for mu?

(A) 13.755 plusminus 0.244
(B) 13.755 plusminus 0.286
(C) 13.755 plusminus 0.707
(D) 13.755 plusminus 1.245
E) 13, 755 plusminus 1.456

Answer 1

Answer:

Option D) $13.755 \pm 1.245$  

Step-by-step explanation:

The table is attached in the question.

We are given the following in the question:

Population mean, μ = 15 hours per week

Sample mean = 13.755 hours per week

Degree of freedom = 25

Standard error = 0.707

Alpha, α = 0.05  

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 15\text{ hours per week}\\H_A: \mu \neq 15\text{ hours per week}$

95% confidence interval:

$\bar{x} \pm t_{critical}\text{Standard error}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 25 and}~\alpha_{0.05} = \pm 1.761$  

$13.755 \pm 1.761(0.707) = 13.755 \pm 1.245 = (1.9052 ,4.3952)$  

Thus, the correct answer is

Option D) $13.755 \pm 1.245$

Answer 2

Answer:

E (13.755 ± 1.456)

Step-by-step explanation:

with 95% confidence interval and df=25

we can find t score on the t table : t*=2.06

(because the table has already provided us standard error , we DON'T have to calculate it by using  σsample=S/$\sqrt{n}$) (standard error = standard deviation)

so the interval should be  13.755± 2.06(0.707)= 13.755± 1.456