Question

Time
1.2
2.8
1.5
19.3
2.4
0.7
2.2
0.7
18.8
6.1
6
1.7
29.1
2.6
0.2
10.2
5.1
0.9
8.2

Company A is trying to sell its website to Company B. As part of the sale, Company A claims that the average user of their site stays on the site for 10 minutes. Company B is concerned that the mean time is significantly less than 10 minutes. Company B collects the times (in minutes) below for a sample of 19 users. Assume normality.

Conduct the appropriate hypothesis test for Company B using a 0.08 level of significance.

a) What are the appropriate null and alternative hypotheses?

H₀: μ = 10 versus H_a: μ > 10

H₀: μ = 10 versus H_a: μ < 10    

H₀: x = 10 versus H_a: x ≠ 10

H₀: μ = 10 versus H_a: μ ≠ 10

b) What is the test statistic? Give your answer to four decimal places.  
c) What is the critical value for the test? Give your answer to four decimal places.  
d) What is the appropriate conclusion?

Reject the claim that the mean time is 10 minutes because the test statistic is larger than the critical point.

Fail to reject the claim that the mean time is 10 minutes because the test statistic is larger than the critical point.  

Reject the claim that the mean time is 10 minutes because the test statistic is smaller than the critical point.

Fail to reject the claim that the mean time is 10 minutes because the test statistic is smaller than the critical point.

Answer 1

Solution:

Here, we have to use one sample t test for the population mean.

a) What are the appropriate null and alternative hypotheses?

H₀: μ = 10 versus H_a: μ < 10    

b) What is the test statistic?

The formula for test statistic is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

Xbar = 6.3

S = 7.897749328

n = 19

t = (Xbar - µ)/[S/sqrt(n)]

t = (6.3 - 10)/[ 7.897749328/sqrt(19)]

t = -3.7/1.8119

t = -2.0421

Test statistic = t = -2.0421

c) What is the critical value for the test?

We have

df = n – 1 = 19 – 1 = 18

α = 0.08

So, critical value by using t-table is given as below:

Critical value = -1.4656

d) What is the appropriate conclusion?

WE have

Test statistic = t = -2.0421

Critical value = -1.4656

Test statistic < Critical value

So, we reject the null hypothesis

Reject the claim that the mean time is 10 minutes because the test statistic is smaller than the critical point.