# Sample Homework Solution – T test

## Sample Homework Solutions

Question 1: Assume that a paper and pencil measure of fear has yielded a mean score of 35 for all incoming college freshmen over the years. We want to see whether the scores of a random group of 20 incoming freshmen, with a mean of 30 and a standard deviation of 10, can be viewed as coming from this population. Test at the .05 level of significance.

Solution: We are interested in testing the following null and alternative hypotheses

\begin{align}{{H}_{0}}:\mu {=} {35}\, \\ {{H}_{A}}:\mu {\ne} {35} \\ \end{align}

Since the population standard deviation $$\sigma$$ is not known, we must use a t-test using the following term:

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

This corresponds to a two-tailed t-test. The t-statistics is given by the following formula:

$t=\frac{\bar{X}-\mu }{s /\sqrt{n}}=\frac{{30}-35}{10/\sqrt{20}}={-2.2361}$

The critical value for $$\alpha = 0.05$$ and for $$df = n- 1 = 20 -1 = 19$$ degrees of freedom for this two-tailed test is $$t_{c} = 2.093$$. The rejection region is given by

$R=\left\{ t:\,\,\,|t|>{2.093} \right\}$

Since $$|t| = 2.2361 {>} t_c = 2.093$$, then we reject the null hypothesis H0.

Hence, we have enough evidence to claim that the population mean is different from to 35.

Question 2: According to the California Educational Code ( http://www.cde.ca.gov/ls/fa/sf/peguidemidhi.asp) , students in grades 7–12 should receive 400 minutes of physical education every 10 school days. The average time spent by 48 students was 385 minutes, with a standard deviation of 53 minutes. Test the hypothesis that the sampled population follows the criteria at the.05 degree of significance.

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

\begin{align}{{H}_{0}}:\mu {=} {400}\, \\ {{H}_{A}}:\mu {\ne} {400} \\ \end{align}

Considering that the population standard deviation $$\sigma$$ is not known, we have to use a t-test with the following term:

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

This corresponds to a two-tailed t-test. The t-statistics is computed by the following formula:

$t=\frac{\bar{X}-\mu }{s /\sqrt{n}}=\frac{{385}-400}{53/\sqrt{48}}={-1.9608}$

The critical value for $$\alpha = 0.05$$ and for $$df = n- 1 = 48 -1 = 47$$ degrees of freedom for this two-tailed test is $$t_{c} = 2.012$$. The rejection region is given by

$R=\left\{ t:\,\,\,|t|>{2.012} \right\}$

Given that $$|t| = 1.9608 {<} t_c = 2.012$$, then we fail to reject the null hypothesis H0.

Therefore, we don’t have enough evidence to reject the claim the sampled population satisfies the requirement.

Question 3: According to a 2009 survey based on the United States Census ( http://www.census.gov/prod/2011pubs/acs-15.pdf ), the daily one-way commute time of U.S. workers averages 25 minutes with (we’ll assume) a standard deviation of 13 minutes. An investigator wishes to determine whether the national average describes the mean commute time for all workers in the Chicago area. Commute times are obtained for a random sample of 169 workers from this area, and the mean time is found to be 22.5 minutes. Test the null hypothesis at the .05 level of significance.

Solution: We want to test the following null and alternative hypotheses

\begin{align}{{H}_{0}}:\mu {=} {25}\, \\ {{H}_{A}}:\mu {\ne} {25} \\ \end{align}

Considering that the population standard deviation is given, with $$\sigma = 13$$ we use the normal distribution. The z-statistic is calculated as

$z =\frac{\bar{X}-\mu }{\sigma / \sqrt{n}}$

This corresponds to a two-tailed z-test. The z-statistics is computed by the following calculation:

$z =\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}=\frac{{22.5}-25}{13 /\sqrt{169}}={-2.5}$

The critical value for $$\alpha = 0.05$$ for this two-tailed test found to be $$z_{c} = {1.96}$$. The rejection region is corresponds to

$R=\left\{ z:\,\,\,|z|>{1.96} \right\}$

Since $$|z| = 2.5 {>} z_c = 1.96$$, then we reject the null hypothesis H0.

Thus, we have enough evidence to support the claim that the population mean is different from 25.