Question

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 17 users. Assume normality. Assume normality.
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

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

1. What are the appropriate null and alternative hypotheses?

a. H0: μ = 10 versus Ha: μ ≠ 10
b. H0: μ = 10 versus Ha: μ > 10
c. H0: x = 10 versus Ha: x ≠ 10
d. H0: μ = 10 versus Ha: μ < 10


2. What is the test statistic? Give your answer to four decimal places.
3. What is the critical value for the test? Give your answer to four decimal places.
4. What is the appropriate conclusion?

Answer 1

Answer:

H0 : μ = 10

H1 : μ < 10

Test statistic = - 2.0415

Critical value = -1.4051

There is significant evidence to conclude that the mean time a user stays on site is less than 10 minutes

Step-by-step explanation:

The null and alternative hypothesis :

H0 : μ = 10

H1 : μ < 10

The test statistic :

From the data Given, using calculator :

Sample mean, xbar = 6.3

Sample standard deviation, s = 7.90

Test statistic = (xbar - μ) ÷ (s/sqrt(n))

Number of samples given is 19 and not 17 as stated in the question.

Test statistic = (6.3 - 10) ÷ (7.90 / sqrt(19))

Test statistic = - 3.7 / 1.8123842

Test statistic = - 2.0415

The critical value for the test :

At α = 0.08 ;

Critical value = -1.4051

Since, Test statistic is < Critical value ;

-2.0415 < - 1.0451 ; we reject the Null

There is significant evidence to conclude that the mean time a user stays on site is less than 10 minutes