Question

You obtain 75.7% successes in a sample of size n₁=247 from the first population. You obtain 80.5% successes in a sample of size n₂=328 from the second population. Suppose that all assumptions are met, so that the difference in sample proportions follows a normal distribution. What is the test statistic for this sample?

Answer 1

The test statistic for the difference in sample proportions is approximately -2.21.

Step-by-step explanation:

In hypothesis testing for the difference in proportions, the test statistic is calculated using the formula:

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

where
`\( \hat{p}_1 \) and \( \hat{p}_2 \) are the sample proportions, \( p \)` is the pooled sample proportion, \
`\( n_1 \) and \( n_2 \)`are the sample sizes. The numerator
`(\( \hat{p}_1 - \hat{p}_2 \))` represents the difference in sample proportions, and the denominator is the standard error of the difference.

For this scenario, substituting the given values:

`\[ z = \frac{(0.757 - 0.805)}{\sqrt{0.781(1-0.781)\left((1)/(247) + (1)/(328)\right)}} \]`

After calculation, the test statistic is approximately -2.21. This negative value indicates that the sample proportion from the first population is lower than the sample proportion from the second population. In hypothesis testing, we would compare this test statistic to critical values or use it to calculate a p-value to determine the statistical significance of the observed difference in proportions.

In this case, a negative value suggests a lower success rate in the first population compared to the second, but further analysis is needed to assess if this difference is statistically significant.