Chapter 7: Q. 7.45 (page 301)

NBA Champs Repeat parts (b) and (c) of Exercise 7.41 for samples of size 5. For part (b), use your answer to Exercise 7.15(b).

Expert verified

For the samples of size 5, the mean of the variable $\bar{x}$ is $μ_{\bar{x}} = 78.6$ .

Step by step solution

We have to determine the sample mean for exercise 7.41 for samples of size 4. The given sample in exercise 7.41 as:

In the table below, the samples of size 5 and their respective means are obtained:

Sample	Height	Mean Height ( $\bar{x}$ )
B,W,J,C,H	83,76,80,74,80	$\frac{83 + 76 + 80 + 74 + 80}{5} = 78.6$

The variable $\bar{x}$ has the following mean

$μ_{\bar{x}} = \frac{78.6}{1} μ_{\bar{x}} = 78.6$

So when the sample size is 5, the variable $\bar{x}$ has a mean $μ_{\bar{x}} = 78.6$ .

We know that mean of the sample mean is equal to the population mean irrespective of the sample size.

Here, for the given data the population mean is $μ = 78.6$ .

So the mean of the sample mean of sample size 5 is

$μ_{\bar{x}} = μ = 78.6$

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