To test the hypothesis that the community's mean household size differs from the countrywide average, we will follow the hypothesis testing process.

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Step 1: State the hypotheses

• Null Hypothesis ($H_0$): The community's mean household size is equal to the countrywide average. $H_0: \mu = 2.23$
• Alternative Hypothesis ($H_a$): The community's mean household size is different from the countrywide average. $H_a: \mu \neq 2.23$

This is a two-tailed test.

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Step 2: Determine the test statistic

The test statistic for this situation is a t-statistic because we are using the sample standard deviation. The formula is: $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$ Where:

• $\bar{x} = 2.61$ (sample mean)
• $\mu_0 = 2.23$ (population mean under the null hypothesis)
• $s = 2.20$ (sample standard deviation)
• $n = 81$ (sample size)

Plugging in the values: $t = \frac{2.61 - 2.23}{2.20 / \sqrt{81}}$ $t = \frac{0.38}{2.20 / 9}$ $t = \frac{0.38}{0.2444} \approx 1.554$

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Step 3: Find the critical value

For a two-tailed test with $\alpha = 0.05$ and $df = n - 1 = 80$, we find the critical t-value from the t-distribution table. Using degrees of freedom (80), the critical values are approximately: $t_{\text{critical}} = \pm 2.00$

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Step 4: Decision Rule

• If $|t| > 2.00$, reject the null hypothesis ($H_0$).
• If $|t| \leq 2.00$, fail to reject the null hypothesis ($H_0$).

Here, $|t| = 1.554$, which is less than $2.00$. Therefore, we fail to reject the null hypothesis.

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Step 5: p-value

Using a t-distribution table or statistical software, the p-value for $t = 1.554$ with $df = 80$ is approximately: $p = 0.124$

Since $p > \alpha$, we fail to reject the null hypothesis.

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Conclusion

At the 0.05 significance level, there is not enough evidence to conclude that the mean household size in the community differs from the countrywide average of 2.23.

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Summary of Results:

1. Null Hypothesis ($H_0$): $\mu = 2.23$
2. Alternative Hypothesis ($H_a$): $\mu \neq 2.23$
3. Critical t-value: $\pm 2.00$
4. Test statistic: $t = 1.554$
5. p-value: $p = 0.124$

The planner concludes that the average household size in the community is not significantly different from the countrywide average.

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Would you like additional details or further clarification on this analysis?
Here are 5 related questions to deepen your understanding:

1. How is the test statistic affected by the sample size?
2. What is the difference between a one-tailed and two-tailed test?
3. Why do we use the t-distribution instead of the z-distribution in this case?
4. How would the conclusion change if the alpha level was 0.10?
5. What assumptions are made when conducting a t-test?

Tip: Always ensure the sample data satisfies the assumptions of normality and independence when performing a t-test.