The mean length of a work week for the population of workers was reported to be
39.2 hours. Suppose that we would like to take a current sample of workers to see
whether the mean length of a work week has changed from the previously
reported 39.2 hours.
a. State the hypotheses that will help us determine whether a change occurred in
the mean length of a work week.
b. Suppose a current sample of 112 workers provided a sample mean of 38.5
hours. Use a population standard deviation σ = 4.8 hours. What is the p-
value?
c. At =0.05, can the null hypothesis be rejected? What is your conclusion
regarding the mean length of a work week?
d. Now suppose population standard deviation is not known and from the above
sample you calculate standard deviation to be 6 hours. Test the hypothesis
that the mean length of a work week has changed from the previously
reported 39.2 hours.
Ho: mean1 = mean2
Ha: mean1 ≠ mean2
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
a. The null hypothesis (H0) states that there is no change in the mean length of a work week, and the alternative hypothesis (Ha) states that there is a change in the mean length of a work week.
H0: μ = 39.2 (Mean length of a work week is 39.2 hours)
Ha: μ ≠ 39.2 (Mean length of a work week is not equal to 39.2 hours)
b. To determine the p-value, we can use the t-test.
The formula for the t-test statistic is:
t = (x̄ - μ) / (σ / √n)
where x̄ is the sample mean, μ is the population mean (assumed to be 39.2 in this case), σ is the population standard deviation, and n is the sample size.
Given:
x̄ = 38.5
μ = 39.2
σ = 4.8
n = 112
Substituting the values into the formula:
t = (38.5 - 39.2) / (4.8 / √112)
Calculating the t-value:
t ≈ -1.458
Next, we can compute the p-value using the t-distribution table or a statistical software. The p-value represents the probability of observing a sample mean as extreme as the one obtained if the null hypothesis is true.
c. Since we have a two-tailed test (we're checking for any difference, whether greater or smaller), we need to compare the p-value with the significance level (α) to determine if the null hypothesis can be rejected.
At α = 0.05, the critical value for a two-tailed test is approximately ±1.96.
If the absolute value of the t-value is greater than the critical value, then we reject the null hypothesis.
In this case, the absolute value of the t-value (1.458) is less than 1.96. Therefore, we fail to reject the null hypothesis.
Conclusion: There is not enough evidence to conclude that the mean length of a work week has changed from the previously reported value of 39.2 hours.
d. When the population standard deviation is not known, we need to use the t-distribution instead of the z-distribution.
The formula for the t-test statistic is the same as before:
t = (x̄ - μ) / (s / √n)
where s is the sample standard deviation, calculated from the sample. In this case, s = 6.
Given the same null and alternative hypotheses as before (H0: μ = 39.2, Ha: μ ≠ 39.2), we can follow the same steps as in part b to calculate the t-value and p-value using the t-distribution with n-1 degrees of freedom (n = 112 - sample size).
Once the t-value and p-value are calculated, we can compare the p-value with the significance level to determine if the null hypothesis can be rejected, following the same procedure as in part c.