Sample Statistics Homework Solution – Central Limit Theorem

Working on some Sample Statistics Problems

Sample Stats Question 1: (a) A 144-person random sample is drawn from the local population of grade-school pupils. Each child calculates the amount of hours they spend watching television each week. What can be said about the sampling distribution at this point?

(b) Assume that a standard deviation, $$\sigma$$, of 8 hours describes the TV estimates for the local population of schoolchildren. At this point, what can be said about the sampling distribution?

(c) Assume that a mean, $$\mu$$ , of 21 hours describes the TV estimates for the local population of schoolchildren. Now what can be said about the sampling distribution?

(d) Roughly speaking, the sample means in the sampling distribution should deviate, on average, about ___ hours from the mean of the sampling distribution and from the mean of the population.

(e) About 95 percent of the sample means in this sampling distribution should be between ___ hours and ___ hours.

Solution: (a) We cannot say anything about the individual scores, but the sampling distribution of sampling means is at least approximately normal, due to Central Limit Theorem.

(b) The sampling distribution of sampling means is at least approximately normal, due to Central Limit Theorem, and the standard error is $$\frac{\sigma }{\sqrt{n}}=\frac{8}{\sqrt{144}}=0.75$$.

(c) The sampling distribution of sampling means is at least approximately normal, due to Central Limit Theorem, with a mean of 21 and a standard error of $$\frac{\sigma }{\sqrt{n}}=\frac{8}{\sqrt{144}}=0.75$$.

(d) 0.75

(e) 19.53, 22.47

Sample Stats Question 2: The standard range for the body mass index (BMI), a commonly recognized indicator of body fat, is 18.5 to 25. The null hypothesized value for the population mean is 21.75, which is the mid-range BMI performance, test this hypothesis at the .01 level of significance given a random sample of 30 weight-watcher participants who show a mean BMI 5 22.2 and a standard deviation of 3.1.

Solution: Our interest is in testing the following null and alternative hypotheses

\begin{align}{{H}_{0}}:\mu {=} {21.75}\, \\ {{H}_{A}}:\mu {\ne} {21.75} \\ \end{align}

Considering that the population standard deviation$\sigma$ is unknown, we have to use a t-test using the following formula:

$t =\frac{\bar{X}-\mu }{s / \sqrt{n}}$

This corresponds to a two-tailed t-test. The t-statistics is given as follows:

$t=\frac{\bar{X}-\mu }{s /\sqrt{n}}=\frac{{22.2}-21.75}{3.1/\sqrt{30}}={0.7951}$

The critical value for $$\alpha = 0.01$$ and for $$df = n- 1 = 30 -1 = 29$$ degrees of freedom for this two-tailed test is $$t_{c} = 2.756$$. The rejection region is given by

$R=\left\{ t:\,\,\,|t|>{2.756} \right\}$

Since $$|t| = 0.7951 {<} t_c = 2.756$$, then we fail to reject the null hypothesis H0.

Hence, we don’t have enough evidence to support the claim that the population mean is different from to 21.75.