To solve this hypothesis testing problem, we'll address each part systematically.

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(a) Compute the test statistic

The formula for the test statistic in a chi-square test for variance is:

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

where:

• $n = 22$ (sample size)
• $s = 61.3$ (sample standard deviation)
• $\sigma_0 = 70$ (hypothesized population standard deviation)

Substitute values into the formula:

1. Compute $s^2$ (sample variance):
   $s^2 = 61.3^2 = 3757.69$

2. Compute the test statistic:
   $\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{(22-1)(3757.69)}{70^2}$ $\chi^2 = \frac{21 \times 3757.69}{4900} = \frac{78811.49}{4900} \approx 16.10$

So, the test statistic is: $\chi^2 \approx 16.10$

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(b) Determine the P-value

For a one-tailed chi-square test, the P-value is calculated from the chi-square distribution with $df = n - 1$. Here:

• $df = 22 - 1 = 21$
• Test statistic: $\chi^2 = 16.10$

The P-value corresponds to the area in the left tail of the chi-square distribution (since $H_1$ is $\sigma < 70$). Using technology or a chi-square table:

$P(\chi^2 < 16.10, df = 21)$

Using statistical software or a calculator, the P-value is approximately:

$P \approx 0.372$

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(c) Decision at $\alpha = 0.05$

• The null hypothesis $H_0: \sigma = 70$ is rejected if the P-value $< \alpha$.
• Here, $P = 0.372$ is much larger than $\alpha = 0.05$.

Thus, the researcher does not reject the null hypothesis.

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Final Summary:

(a) Test statistic: $\chi^2 = 16.10$
(b) P-value: $P \approx 0.372$
(c) Decision: Do not reject $H_0$.

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Would you like more details or have additional questions?

Expansion Questions:

1. How is the chi-square test used in other variance testing scenarios?
2. How does the chi-square distribution shape change with degrees of freedom?
3. What are the assumptions underlying a chi-square test?
4. How would the conclusion change if $\alpha = 0.10$?
5. How can technology (e.g., software like Excel or Python) help in hypothesis testing?

Tip: Always verify that your sample is from a normally distributed population before using chi-square tests for variance.