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.