In a order to compare the means of two populations, independent random samples of 410 observations are selected from each population, with Sample 1 the results found in the table to the right. Complete parts a through e below. X1 = 5,319 S1= 143 a. Use a 95% confidence interval to estimate the difference between the population means (H - H2) Interpret the contidence interval. The contidence interval IS (Round to one decimal place as needed.) Sample 2 X2 = 5,285 S2 = 198 Aa. Use a 95% confidence interval to estimate the difference between the population means (A1 - M2) Interpret the contidence interval. The contidence interval Is (Round to one decimal place as needed.) b. Test the null hypothesis Ho versus alternative hypothesis Ha (H What is the test statistic? H2) + Give the significance level of the test, and interpret the result. Use a = 0.05. Z=

To estimate the difference between the population means (μ1 - μ2) using a 95% confidence interval, we can use the formula:

\text{{Confidence Interval}} = (\bar{x_1} - \bar{x_2}) \pm Z \cdot \sqrt{\frac{{s_1^2}}{{n_1}} + \frac{{s_2^2}}{{n_2}}}

where:
- μ1 and μ2 are the population means
- \bar{x_1} and \bar{x_2} are the sample means
- s_1 and s_2 are the sample standard deviations
- n1 and n2 are the sample sizes
- Z is the Z-score for a 95% confidence interval, which is approximately 1.96

a. Using the given values:
\bar{x_1} = 5,319, s_1 = 143, n_1 = 410
\bar{x_2} = 5,285, s_2 = 198, n_2 = 410
Z = 1.96

Substituting these values into the formula:

\text{{Confidence Interval}} = (5,319 - 5,285) \pm 1.96 \cdot \sqrt{\frac{{143^2}}{{410}} + \frac{{198^2}}{{410}}}

Simplifying the expression within the square root:

\sqrt{\frac{{143^2}}{{410}} + \frac{{198^2}}{{410}}} = \sqrt{\frac{{20,449}}{{410}}} + \sqrt{\frac{{39,204}}{{410}}}

\sqrt{\frac{{20,449 + 39,204}}{{410}}} = \sqrt{\frac{{59,653}}{{410}}} = \sqrt{\frac{{146}}{1}} \approx 12.083

Therefore, the confidence interval for the difference between the population means (μ1 - μ2) is:

(5,319 - 5,285) \pm 1.96 \cdot 12.083

34 \pm 23.733

Answer:
The confidence interval for the difference in population means is (10.3, 57.7). This means that we can be 95% confident that the true difference between the population means lies between 10.3 and 57.7.

b. To test the null hypothesis (H0: μ1 = μ2) against the alternative hypothesis (Ha: μ1 ≠ μ2), we can use a two-sample t-test. Since the sample sizes are large (both are 410), we can approximate the test statistic using a Z-test.

The test statistic for a two-sample Z-test is given by:

Z = \frac{{(\bar{x_1} - \bar{x_2}) - (\mu_1 - \mu_2)}}{{\sqrt{\frac{{s_1^2}}{{n_1}} + \frac{{s_2^2}}{{n_2}}}}}

where:
- μ1 and μ2 are the population means
- \bar{x_1} and \bar{x_2} are the sample means
- s_1 and s_2 are the sample standard deviations
- n1 and n2 are the sample sizes

Using the given values:
\bar{x_1} = 5,319, s_1 = 143, n_1 = 410
\bar{x_2} = 5,285, s_2 = 198, n_2 = 410

Substituting these values into the formula:

Z = \frac{{(5,319 - 5,285) - 0}}{{\sqrt{\frac{{143^2}}{{410}} + \frac{{198^2}}{{410}}}}}

Simplifying the expression within the square root:

\sqrt{\frac{{143^2}}{{410}} + \frac{{198^2}}{{410}}} = \sqrt{\frac{{20,449}}{{410}}} + \sqrt{\frac{{39,204}}{{410}}}

\sqrt{\frac{{20,449 + 39,204}}{{410}}} = \sqrt{\frac{{59,653}}{{410}}} = \sqrt{\frac{{146}}{1}} \approx 12.083

So the test statistic Z is:

Z = \frac{{34}}{{12.083}} \approx 2.810

Answer:
The test statistic is Z = 2.810. The significance level of the test is α = 0.05. Since the test statistic Z is greater than the critical value Zα/2 (approximately ±1.96 for a 95% confidence level), we reject the null hypothesis. This means there is sufficient evidence to suggest that the population means are not equal.