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 19 users. Assume normality.
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
Conduct the appropriate hypothesis test for Company B using a 0.08 level of significance.
a) What is the critical value for the test? Give your answer to four decimals.
b) What is the appropriate conclusion?
A. Reject the claim that the mean time is 10 minutes because the test statistic is larger than the critical point.
B. Fail to reject the claim that the mean time is 10 minutes because the test statistic is larger than the critical point.
C. Reject the claim that the mean time is 10 minutes because the test statistic is smaller than the critical point.
D. Fail to reject the claim that the mean time is 10 minutes because the test statistic is smaller than the critical point.

Answer 1

Answer:

a) Critical value = -1.4052

Since we are checking if the mean time is less than 10 minutes, the rejection area would be

z < -1.4052

b) Option C is correct.

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

That is, the mean time is significantly less than 10 minutes.

Step-by-Step Explanation:

a) Using z-distribution, the critical value is obtained from the confidence level at which the test is going to be performed. Since the hypothesis test tests only in one direction (checking if the claim is less than 10 minutes significantly)

P(z < Critical value) = 0.08

From the z-tables, critical value = -1.4052

Since we are checking if the mean time is less than 10 minutes, the rejection area would be

z < -1.4052

b) We first give the null and alternative hypothesis

The null hypothesis is that there isn't significant evidence to suggest that the mean time is less than 10 minutes.

And the alternative hypothesis is is that there is significant evidence to suggest that the mean time is less than 10 minutes.

To now perform this hypothesis test, we need to obtain the test statistic

Test statistic = (x - μ)/σₓ

x = sample mean = (Σx/N)

The data is

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

Σx = 119.7

N = Sample size = 19

x = sample mean = (119.7/19) = 6.3

μ = standard to be compared against = 10 minutes

σₓ = standard error = (σ/√N)

where N = Sample size = 19

σ = √[Σ(x - xbar)²/N]

x = each variable

xbar = mean = 6.3

N = Sample size = 19

Σ(x - xbar)² = 1122.74

σ = (√1122.74/19) = 7.687

σₓ = (7.687/√19) = 1.7635

Test statistic = (x - μ)/σₓ

Test statistic = (6.3 - 10)/1.7635

= -2.098 = -2.10

z = -2.10 and is in the rejection region, (z < -1.4052), hence, we reject the null hypothesis and the claim and say that the mean time is significantly less than 10 minutes.

The test statistic is less than the critical point, hence, we reject the null hypothesis and the claim and conclude that the mean time is less than 10 minutes.

Hope this Helps!!!