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.
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.
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.
The null hypothesis (H0) is that there is no difference between the mean delivery times of the two courier services, while the alternative hypothesis (H1) is that there is a difference.
Step 1: Specify the hypothesis:
H0: μA - μB = 0 (There is no difference between the mean delivery times)
H1: μA - μB ≠ 0 (There is a difference between the mean delivery times)
Step 2: Determine the significance level:
The significance level, denoted as α, is given as 0.05 or 5%.
Step 3: Conduct the t-test and calculate the test statistic:
The formula for the t-test statistic for independent samples is:
t = ((X̄A - X̄B) - D) / √((sA^2 / nA) + (sB^2 / nB))
Where:
X̄A = sample mean for Courier Service A
X̄B = sample mean for Courier Service B
sA = sample standard deviation for Courier Service A
sB = sample standard deviation for Courier Service B
nA = sample size for Courier Service A
nB = sample size for Courier Service B
D = hypothesized difference (in this case, 0)
Plugging in the values:
X̄A = 42 minutes
X̄B = 38 minutes
sA = 14 minutes
sB = 10 minutes
nA = 60
nB = 48
D = 0
t = ((42 - 38) - 0) / √((14^2 / 60) + (10^2 / 48))
Step 4: Determine the critical value:
Since this is a two-tailed test and α = 0.05, we divide the significance level by 2 to get α/2 = 0.025. Looking up the critical value of t with (nA + nB - 2) degrees of freedom and α/2 as the tail probability, we find the critical value.
Step 5: Calculate the degrees of freedom:
The degrees of freedom for this two-sample t-test is calculated as (nA + nB - 2).
Step 6: Compare the test statistic with the critical value:
If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is a significant difference between the mean delivery times of the two courier services. Otherwise, we fail to reject the null hypothesis.
I will now calculate the test statistic, critical value, and degrees of freedom.
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 are the steps to calculate the test statistic and critical value:
1. Define the null and alternative hypotheses:
- Null hypothesis (H₀): The mean delivery times of Courier Service A and Courier Service B are the same.
- Alternative hypothesis (H₁): The mean delivery times of Courier Service A and Courier Service B are different.
2. Calculate the test statistic using the formula:
t = (x₁ - x₂) / sqrt((s₁² / n₁) + (s₂² / n₂))
where:
- x₁ and x₂ are the sample means of Courier Service A and Courier Service B, respectively.
- s₁ and s₂ are the sample standard deviations of Courier Service A and Courier Service B, respectively.
- n₁ and n₂ are the sample sizes of Courier Service A and Courier Service B, respectively.
In this case,
x₁ = 42 minutes, x₂ = 38 minutes,
s₁ = 14 minutes, s₂ = 10 minutes,
n₁ = 60 times, n₂ = 48 times.
Plugging in these values:
t = (42 - 38) / sqrt((14² / 60) + (10² / 48))
3. Determine the degrees of freedom (df) using the formula:
df = (s₁² / n₁ + s₂² / n₂)² / ((s₁² / n₁)² / (n₁ - 1) + (s₂² / n₂)² / (n₂ - 1))
In this case,
df = (14² / 60 + 10² / 48)² / ((14² / 60)² / (60 - 1) + (10² / 48)² / (48 - 1))
4. Calculate the critical value at a 5% significance level (α = 0.05) for a two-tailed test with the degrees of freedom (df) obtained in step 3. This critical value will determine the rejection region.
You can use a t-distribution table or a statistical software to find the critical value. The critical value depends on the significance level and degrees of freedom.
5. Compare the test statistic from step 2 with the critical value from step 4:
- If the test statistic falls within the non-rejection region, we fail to reject the null hypothesis.
- If the test statistic falls within the rejection region, we reject the null hypothesis.
Thus, to answer your question, the critical value needs to be calculated using the degrees of freedom obtained in step 3 and a t-distribution table or statistical software.