Chapter 10, Problem R10.4RE

To determine

To describe the shape, center and variability of the sampling distribution ${\overline{x}}_1-{\overline{x}}_2$ .

Answer to Problem R10.4RE

Shape: Approximately Normal.

Center: zero.

Spread: $0.0258$ .

Explanation of Solution

It is given in the question that:

  $\begin{array}{l}{n}_1=49\\ n_2=49\\ {\mu }_1=15\\ {\mu }_2=15\\ {\sigma }_1=0.15\\ {\sigma }_2=0.10\end{array}$

Since both the samples are large the central limit theorem tells us that both samples have a sampling distribution of the sample mean ${\overline{x}}_1\text{ and }{\overline{x}}_2$ that is approximately normal. However, this then also implies that the sampling distribution ${\overline{x}}_1-{\overline{x}}_2$ is also approximately normal.

The mean of the sampling distribution of the difference in sample means is the difference in the population means.

  $\begin{array}{l}{\mu }_{{\overline{x}}_1-{\overline{x}}_2}={\mu }_1-{\mu }_2\\ =15-15\end{array}$

Assuming that the sample of $49$ candles produced by machine $1$ and assuming that the sample of $49$ candles produced by sample $2$ is less than $10\%$ of all candles produced by machine $2$ , the standard deviation of the sampling distribution of the difference in sample mean is given by:

  $\sqrt{\frac{{\sigma }_1{}^2}{n_1}+\frac{{\sigma }_2{}^2}{n_2}}$

Thus, we have,

  $\begin{array}{l}{\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=\sqrt{\frac{{\sigma }_1{}^2}{n_1}+\frac{{\sigma }_2{}^2}{n_2}}\\ =\sqrt{\frac{{0.15}^2}{49}+\frac{{0.10}^2}{49}}\\ =0.0258\end{array}$

Thus we conclude that,

Shape: Approximately Normal.

Center: zero.

Spread: $0.0258$ .