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.
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.
To test the hypothesis that there is no difference between the mean delivery times of the two courier services, we can conduct a two-sample t-test.
The null hypothesis (H0) is that there is no difference between the means of the two courier services, while the alternative hypothesis (Ha) is that there is a difference between the means of the two courier services.
H0: μA - μB = 0 (There is no difference between the means)
Ha: μA - μB ≠ 0 (There is a difference between the means)
The test statistic for comparing two sample means is given by:
t = (x̄A - x̄B) / √[(sA^2 / nA) + (sB^2 / nB)]
where x̄A and x̄B are the sample means, sA and sB are the sample standard deviations, and nA and nB are the sample sizes for Courier Service A and Courier Service B, respectively.
Given information:
x̄A = 42 minutes
x̄B = 38 minutes
sA = 14 minutes
sB = 10 minutes
nA = 60 times
nB = 48 times
Substituting the values into the formula, we get:
t = (42 - 38) / √[(14^2/60) + (10^2/48)]
Calculating the values within the square root:
t = (4) / √[(1.96) + (4.17)]
Now, let's calculate the square root within the square root:
t = (4) / √[6.13]
Calculating the square root:
t = (4) / 2.474
Finally, calculating the test statistic:
t ≈ 1.616
Therefore, the test statistic is approximately 1.616.
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. Here's how to calculate the test statistic:
Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): There is no difference between the mean delivery times of Courier Service A and Courier Service B.
- Alternative hypothesis (Ha): There is a difference between the mean delivery times of Courier Service A and Courier Service B.
Step 2: Determine the significance level:
The given significance level is 5% or 0.05.
Step 3: Determine the test statistic:
We will calculate the test statistic using the formula:
t = ((X1 - X2) - D) / sqrt((s1^2 / n1) + (s2^2 / n2))
Where:
- X1 and X2 are the sample means
- D is the hypothesized difference between the population means (0 in this case)
- s1 and s2 are the sample standard deviations
- n1 and n2 are the sample sizes
In this case:
X1 = 42 (sample mean for Courier Service A)
X2 = 38 (sample mean for Courier Service B)
D = 0 (hypothesized difference between the population means)
s1 = 14 (population standard deviation for Courier Service A)
s2 = 10 (population standard deviation for Courier Service B)
n1 = 60 (number of samples for Courier Service A)
n2 = 48 (number of samples for Courier Service B)
Calculating the test statistic:
t = ((42 - 38) - 0) / sqrt((14^2 / 60) + (10^2 / 48))
= 4 / sqrt(3.453 + 2.083)
= 4 / sqrt(5.536)
= 4 / 2.354
= 1.700
The calculated test statistic is 1.700.
Step 4: Determine the critical value:
Since the significance level is 5%, we need to find the critical value for a two-sample t-test at a 5% level of significance. This can be done using a t-table or a statistical software. The degrees of freedom for this test is given by df = (n1 - 1) + (n2 - 1) = 60 - 1 + 48 - 1 = 106.
For a two-tailed test at a 5% level of significance with 106 degrees of freedom, the critical value is approximately ±1.984.
Step 5: Make a decision:
If the absolute value of the test statistic is greater than the critical value (|t| > 1.984), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, |t| = |1.700| = 1.700, which is less than 1.984. Therefore, we fail to reject the null hypothesis.
Conclusion:
Based on the calculated test statistic, we do not have enough evidence to conclude that there is a significant difference between the mean delivery times of Courier Service A and Courier Service B at a 5% significance level.