Answer:
The 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).
Step-by-step explanation:
The (1 - α)% confidence interval for the population standard deviation is:
[tex]CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}[/tex]
The information provided is:
n = 26
s = 4.8 minutes
Confidence level = 90%
Compute the critical values of Chi-square as follows:
[tex]\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.10/2, (26-1)}=\chi^{2}_{0.05, 25}=37.652[/tex]
[tex]\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.10/2, (26-1)}=\chi^{2}_{0.95, 25}=14.611[/tex]
*Use a Chi-square table.
Compute the 90% confidence interval for the population standard deviation waiting time for an oil change as follows:
[tex]CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}[/tex]
[tex]=\sqrt{\frac{(26-1)\times 4.8^{2}}{37.652}}\leq \sigma\leq \sqrt{\frac{(26-1)\times 4.8^{2}}{14.611}}\\\\=3.9113\leq \sigma\leq 6.2787\\\\\approx 3.9 \leq \sigma\leq6.3[/tex]
Thus, the 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).
