: Human Resources took a survey and found that the average commute time one way is 25.4 minutes. However one of the executives feels that the commute is less. He randomly selects 25 commuters and finds that the average is 22.1 minutes with a standard deviation of 5.3 minutes. At Ą=0.10, is he correct?
I don't know what "Ą" means.
Z = (mean1 -mean2)/Standard Error (SE) of the mean
SE = SD/√(n-1)
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
To determine whether the executive is correct about the average commute time being less than 25.4 minutes, we need to perform a hypothesis test. This will help us evaluate whether the observed sample supports or contradicts his claim.
Here's how to conduct the hypothesis test:
1. Null Hypothesis (H₀): The average commute time is not less than 25.4 minutes. Therefore, H₀: µ ≥ 25.4.
2. Alternative Hypothesis (H₁): The average commute time is less than 25.4 minutes. Therefore, H₁: µ < 25.4.
3. Select a significance level (α): In this case, the significance level is given as 0.10, which means we want to be 90% confident in our conclusion.
4. Calculate the test statistic: We'll use the one-sample t-test, since the population standard deviation (σ) is unknown. The formula for the test statistic is:
t = (x̄ - µ₀) / (s / √n)
where x̄ is the sample mean (22.1 minutes), µ₀ is the hypothesized population mean (25.4 minutes), s is the sample standard deviation (5.3 minutes), and n is the sample size (25 commuters).
5. Find the critical value: Since the alternative hypothesis is less than (<), we'll use a one-tailed t-distribution to find the critical value. At α = 0.10, with 24 degrees of freedom (n - 1), the critical t-value is -1.317 (obtained from a t-table or statistical software).
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; otherwise, we fail to reject it.
Now, let's calculate the test statistic and make the conclusion:
t = (22.1 - 25.4) / (5.3 / √25)
≈ -3.45
Since -3.45 is smaller than -1.317 (the critical value), we can reject the null hypothesis. This means there is evidence to support the executive's claim that the average commute time is less than 25.4 minutes at a significance level of 0.10.
In conclusion, based on the sample data, the executive's belief that the average commute time is less than 25.4 minutes seems to be correct.