Certainly! Let's go through each part of the problem step-by-step.
### a. State the null and alternative hypotheses.
The null and alternative hypotheses are formulated as follows:
- Null Hypothesis [tex]\( H_0: \mu \leq 15 \)[/tex] days
- Alternative Hypothesis [tex]\( H_1: \mu > 15 \)[/tex] days
### b. Using a critical value, test the null hypothesis at the 5% level of significance.
1. Calculate the Z value for the sample mean:
[tex]\[ Z = \frac{16.2 - 15}{\frac{5.6}{\sqrt{49}}} = \frac{1.2}{0.8} = 1.5 \][/tex]
2. Determine the critical value for a one-tailed test at the 5% significance level:
[tex]\[ \text{Critical value} \approx 1.645 \][/tex]
3. Compare the calculated Z value to the critical value:
[tex]\[ 1.5 < 1.645 \][/tex]
Since the Z value (1.5) is less than the critical value (1.645), we fail to reject the null hypothesis at the 5% significance level.
### c. Using a p-value, test the hypothesis at the 5% level of significance.
4. Determine the p-value for the calculated Z value:
[tex]\[ \text{p-value} \approx 0.0668 \][/tex]
5. Compare the p-value to the significance level:
[tex]\[ \text{Significance level} = 0.05 \][/tex]
Since the p-value (0.0668) is greater than the significance level (0.05), we fail to reject the null hypothesis at the 5% level of significance.
### d. Compute the probability of a Type II error if the true average delivery time is 17 days after purchase.
6. Calculate the Z value for the true mean of 17 days:
[tex]\[ Z = \frac{16.2 - 17}{\frac{5.6}{\sqrt{49}}} = \frac{-0.8}{0.8} = -1.0 \][/tex]
7. Find the probability corresponding to this Z value:
[tex]\[ \text{Probability} \approx 0.1587 \][/tex]
The probability of a Type II error, denoted as [tex]\( \beta \)[/tex], is approximately 0.1587 when the true average delivery time is 17 days.
In summary:
- We fail to reject the null hypothesis both by using the critical value and the p-value at the 5% significance level.
- The probability of a Type II error when the true average delivery time is 17 days is approximately 0.1587.
