Let's work through each part of the question step by step. This is a hypothesis test for the difference in proportions, so we'll follow the procedure for such a test.

Given Information:

• Sample 1 (20th century houses):

  • $n_1 = 144$ (sample size)
  • $x_1 = 51$ (number of houses with pools)
  • Sample proportion: $\hat{p}_1 = \frac{51}{144} = 0.3542$

• Sample 2 (new houses):

  • $n_2 = 77$ (sample size)
  • $x_2 = 22$ (number of houses with pools)
  • Sample proportion: $\hat{p}_2 = \frac{22}{77} = 0.2857$

• Significance level: $\alpha = 0.05$

(a) State the null hypothesis $H_0$ and the alternative hypothesis $H_1$:

• Null hypothesis $H_0$: $p_1 \leq p_2$ (The proportion of 20th-century houses with pools is less than or equal to that of new houses.)
• Alternative hypothesis $H_1$: $p_1 > p_2$ (The proportion of 20th-century houses with pools is greater than that of new houses.)
  This is a one-tailed test.

(b) Determine the type of test statistic to use:

For comparing proportions, we use a z-test for the difference between two proportions.

(c) Find the value of the test statistic:

The formula for the test statistic for two proportions is:

$z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}}$

Where:

• $\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{51 + 22}{144 + 77} = \frac{73}{221} = 0.3303$

Now substitute values into the formula:

$z = \frac{(0.3542 - 0.2857)}{\sqrt{0.3303(1 - 0.3303) \left( \frac{1}{144} + \frac{1}{77} \right)}}$

First, calculate the standard error:

$\text{Standard error} = \sqrt{0.3303(0.6697) \left( \frac{1}{144} + \frac{1}{77} \right)} = \sqrt{0.3303(0.6697) \left( 0.00694 + 0.01299 \right)} = \sqrt{0.3303(0.6697)(0.01993)} = \sqrt{0.004414} \approx 0.06646$

Now calculate the test statistic $z$:

$z = \frac{0.3542 - 0.2857}{0.06646} = \frac{0.0685}{0.06646} \approx 1.0307$

(d) Find the p-value:

For a one-tailed test, we use the standard normal distribution table to find the p-value for $z = 1.0307$.

Using a Z-table or calculator, the p-value associated with $z = 1.0307$ is approximately 0.151.

(e) Conclusion:

Since the p-value (0.151) is greater than the significance level $\alpha = 0.05$, we fail to reject the null hypothesis.

Thus, we cannot conclude that the proportion of Southern California 20th-century houses built with pools is greater than the proportion for new homes.

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Would you like more details on any of the steps, or do you have further questions? Here are 5 related questions for you to explore:

1. How do we interpret p-values in hypothesis testing?
2. What are the assumptions required for conducting a z-test for proportions?
3. How would the conclusion change if the p-value were smaller than 0.05?
4. What is the importance of using a random sample in hypothesis testing?
5. How would a two-tailed test differ from this one-tailed test?

Tip: Always compare the p-value to the significance level to decide whether to reject or fail to reject the null hypothesis.