A client used two courier services on a random daily basis over a period of 3 months for similar destinations. Delivery times where noted by a clerk. During this time Courier Service A was used 60 times with a sample average delivery time calculated as 42 minutes. During this same time Courier Service B was used 48 times and a sample average delivery time is calculated as 38 minutes. Suppose that the population standard deviation for Courier Service A is 14 minutes and for Courier Service B is 10 minutes. Test the hypothesis that there is no difference between the mean delivery times of the two courier services ( on a 5% significance level.) Calculate the test statistic.

Answer 1

To test the hypothesis that there is no difference between the mean delivery times of the two courier services, we will perform a two-sample t-test.

The null hypothesis (H0) is that there is no difference between the mean delivery times of the two courier services.
The alternative hypothesis (H1) is that there is a difference between the mean delivery times of the two courier services.

Let's denote the sample mean delivery time for Courier Service A as X̄A = 42 minutes, and the sample mean delivery time for Courier Service B as X̄B = 38 minutes.

The population standard deviation for Courier Service A is σA = 14 minutes.
The population standard deviation for Courier Service B is σB = 10 minutes.

The number of observations for Courier Service A is nA = 60, and for Courier Service B is nB = 48.

The degrees of freedom for this two-sample t-test is calculated as:
df = nA + nB - 2 = 60 + 48 - 2 = 106.

The test statistic for the two-sample t-test is calculated as:
t = (X̄A - X̄B) / √[(σA^2 / nA) + (σB^2 / nB)].

Substituting the given values into the formula, we get:
t = (42 - 38) / √[(14^2 / 60) + (10^2 / 48)]
= 4 / √[(196 / 60) + (100 / 48)]
= 4 / √[3.2667 + 2.0833]
= 4 / √5.35
≈ 1.756.

Therefore, the test statistic for this hypothesis test is approximately 1.756.