A study in Orlando, Florida claimed that the mean commute time for all employees working at Disney exceeds 40 minutes. This figure is higher than what was assumed in the past. The plan is to test this claim at a level of significance = 0.05 and a sample size = 100 commuters. For this sample size, it was found the sample mean is 43.5 minutes. Based on previous studies, assume the standard deviation is known to be o = 8 minutes. Conduct a hypothesis test using a rejection region and draw a conclusion.
(please provide complete detailed answer)
To conduct a hypothesis test in this scenario, we need to follow these steps:
Step 1: State the null and alternative hypothesis. The null hypothesis (H0) assumes that the mean commute time for all employees working at Disney is not greater than 40 minutes. The alternative hypothesis (Ha) suggests that the mean commute time for all employees working at Disney is greater than 40 minutes.
H0: µ ≤ 40
Ha: µ > 40
Step 2: Determine the significance level (α) and select the appropriate test. The significance level is given as α = 0.05, which means we will reject the null hypothesis if the probability of observing the sample mean, assuming the null hypothesis is true, is less than 0.05. Since we have a known standard deviation and a large sample size (n > 30), we can use a z-test for population means.
Step 3: Calculate the test statistic. The test statistic for a z-test for population means is given by:
z = (x̄ - µ) / (σ / sqrt(n))
where:
x̄ is the sample mean (43.5 minutes),
µ is the hypothesized population mean (40 minutes),
σ is the population standard deviation (8 minutes), and
n is the sample size (100 commuters).
Plugging in the given values, we have:
z = (43.5 - 40) / (8 / sqrt(100))
= 3.5 / (8 / 10)
= 3.5 / 0.8
= 4.375
Step 4: Determine the critical value(s) or rejection region. Since our alternative hypothesis is one-tailed (greater than 40 minutes), we need to find the critical value for a z-score that corresponds to the given significance level (α = 0.05) in the right tail of the standard normal distribution.
The critical value for α = 0.05 can be found using a Z-table or a statistical calculator, and it is approximately 1.645.
Step 5: Make a decision and draw a conclusion. Compare the test statistic (4.375) with the critical value (1.645).
Since the test statistic falls in the rejection region (test statistic > critical value), we reject the null hypothesis. This means that there is sufficient evidence to support the claim that the mean commute time for all employees working at Disney exceeds 40 minutes based on this sample.
In conclusion, the study suggests that the mean commute time for all employees working at Disney is greater than 40 minutes.