AP Statistics 4.6 Difference of Means

Two students at a large high school, Peter and Rania, wanted to estimate $\mu$ , the mean number of soft drinks that a student at their school consumes in a week. A complete roster of the names and genders for the 2,000 students at their school was available. Peter selected a simple random sample of 100 students. Rania, knowing that 60 percent of the students at the school are female, selected a simple random sample of 60 females and an independent simple random sample of 40 males. Both asked all of the students in their samples how many soft drinks they typically consume in a week.

Peter and Rania conducted their studies as described. Peter used the sample mean $\bar{X}$ as a point estimator for $\mu$ . Rania used $\bar{X}_{\text {overall }}=(0.6) \bar{X}_{\text {female }}+(0.4) \bar{X}_{\text {male }}$ as a point estimator for $\mu$ , where $\bar{X}_{\text {female }}$ is the mean of the sample of 60 females and $\bar{X}_{\text {male }}$ is the mean of the sample of 40 males.

Summary statistics for Peter's data are shown in the table below.

Table for Question AP Statistics 4.6 A Calculate the Mean and Standard Deviation of a Sampling Distribution for the Difference Between Two Sample Means Questions question 1 — AP Statistics

Summary statistics for Rania's data are shown in the table below.

A dotplot of Peter's sample data is given below.

Figure for Question AP Statistics 4.6 A Calculate the Mean and Standard Deviation of a Sampling Distribution for the Difference Between Two Sample Means Questions question 1 — AP Statistics

Comparative dotplots of Rania's sample data are given below.

Based on the summary statistics, calculate the estimated standard deviation of the sampling distribution of Rania's point estimator $\bar{X}_{\text {overall }}=(0.6) \bar{X}_{\text {female }}+(0.4) \bar{X}_{\text {male }}$ .

Part (c):
The variance of Rania's estimator is $(0.6)^{2} \operatorname{Var}\left(\bar{X}_{f}\right)+(0.4)^{2} \operatorname{Var}\left(\bar{X}_{m}\right)$ , where $\operatorname{Var}\left(\bar{X}_{f}\right)=\frac{\sigma_{f}^{2}}{n_{f}}$ represents the variance of the point estimator for females and $\operatorname{Var}\left(\bar{X}_{m}\right)=\frac{\sigma_{m}^{2}}{n_{m}}$ represents the variance of the point estimator for males.

The estimated standard deviation is the square root of the variance. Using the respective sample standard deviations $s_{f}$ and $s_{m}$ for the population parameters, Rania's estimate is calculated as:

\sqrt{(0.6)^{2} \frac{s_{f}^{2}}{n_{f}}+(0.4)^{2} \frac{s_{m}^{2}}{n_{m}}}=\sqrt{(0.6)^{2} \frac{(1.80)^{2}}{60}+(0.4)^{2} \frac{(2.22)^{2}}{40}} \approx \sqrt{0.01944+0.01972} \approx 0.198 .

AP Statistics 4.6 A Calculate the Mean and Standard Deviation of a Sampling Distribution for the Difference Between Two Sample Means Questions question 1