Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

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Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

the variability in service time (in hours) spent by mechanics fixing the same automotive problem. A random sample was taken resulting in a sample of size 20 from a substantial file of reported experience. The summary statistics are as follows: n = 20, sample mean = 13.8 hours, sample standard deviation = 3.9 hours. Assume service time follows a normal distribution. Round to two decimal places.

Mathematics

1 answer:

Svet_ta [12.7K] · Answer 1 · 2026-03-10T19:22:28+03:00

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - α)% confidence interval for the variance is defined as:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

We are given:

n = 20

s = 3.9

Confidence level = 95%

Thus, α = 0.05

Calculate the critical Chi-square values:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Refer to a Chi-square table for this.

Now compute the 95% confidence interval for variance like this:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$

Consequently, the 95% confidence interval for the variance is (8.80, 32.45).