To test the hypothesis that there is no difference between the mean delivery times of the two courier services, we can use a two-sample t-test. Let's denote the mean delivery time for Courier Service A as μA and the mean delivery time for Courier Service B as μB.
The null hypothesis (H0) is that there is no difference between the mean delivery times:
H0: μA - μB = 0
The alternative hypothesis (HA) is that there is a difference between the mean delivery times:
HA: μA - μB ≠ 0 (two-tailed test)
We can use the following formula to calculate the test statistic for the two-sample t-test:
t = (x̄A - x̄B) / √((sA^2 / nA) + (sB^2 / nB))
Where:
x̄A = sample average delivery time for Courier Service A = 42 minutes
x̄B = sample average delivery time for Courier Service B = 38 minutes
sA = population standard deviation for Courier Service A = 14 minutes
sB = population standard deviation for Courier Service B = 10 minutes
nA = number of times Courier Service A was used = 60
nB = number of times Courier Service B was used = 48
Plugging in the values into the formula, we get:
t = (42 - 38) / √((14^2 / 60) + (10^2 / 48))
= 4 / √(1960/60 + 1000/48)
= 4 / √32.6666 + 20.8333
= 4 / √53.4999
≈ 0.5601
Next, we need to find the critical value for a two-tailed test with a 5% significance level. Since there are nA - 1 (60 - 1 = 59) degrees of freedom for Courier Service A and nB - 1 (48 - 1 = 47) degrees of freedom for Courier Service B, we can find the critical value using a t-distribution table or a calculator.
Looking up the critical value for a two-tailed test with 59 and 47 degrees of freedom and a 5% significance level, we find that the critical value is approximately ±2.0106.
As the test statistic (t = 0.5601) does not fall within the critical region (±2.0106), we fail to reject the null hypothesis. There is not enough evidence to support the claim that there is a difference between the mean delivery times of the two courier services at the 5% significance level.
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.) give the critical value.
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