Let's break down the steps to compute the test statistic based on the image provided.

Problem Summary:

• Sample size $n = 12$
• Sample standard deviation $s = 52$
• Null hypothesis $H_0: \sigma = 69$
• Alternative hypothesis $H_1: \sigma < 69$ (this is a left-tailed test)
• Significance level $\alpha = 0.05$
• We are testing the population standard deviation ($\sigma$)

This is a hypothesis test for the population standard deviation. The test statistic for this kind of problem follows a chi-square ($\chi^2$) distribution and is calculated using the formula:

$\chi^2 = \frac{(n - 1) \cdot s^2}{\sigma_0^2}$

Where:

• $n = 12$ is the sample size.
• $s = 52$ is the sample standard deviation.
• $\sigma_0 = 69$ is the hypothesized population standard deviation.
• $\sigma_0^2 = (69)^2 = 4761$

Step-by-Step Solution:

1. Compute $s^2$ (the sample variance): $s^2 = 52^2 = 2704$

2. Compute the test statistic $\chi^2$: $\chi^2 = \frac{(12 - 1) \cdot 2704}{4761} = \frac{11 \cdot 2704}{4761} = \frac{29744}{4761} \approx 6.245$

So, the test statistic is approximately $\chi^2 \approx 6.245$.

Would you like me to explain how to interpret this test statistic or proceed to find the p-value?

Related Questions:

1. What is the rejection region for this left-tailed chi-square test?
2. How do you find the p-value for a chi-square test?
3. What assumptions must be met for this chi-square test to be valid?
4. How does the sample size affect the outcome of this test?
5. How do you interpret the significance level $\alpha = 0.05$ in this context?

Tip:

When dealing with chi-square tests for standard deviations, always check if the sample size is large enough and the population is normally distributed for accurate results.