Question

Perform a goodness-of-fit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity.

Short answer

The$p-$value is 0, we reject the null hypothesis. There is evidence to conclude that the local results do not follow the distribution of the overall student population.

Step-by-step solution

Given Information

The distribution of the U.S. overall student population based on ethnicity.

Explanation

The right column contains the result for $1000$locals. We have to use the percentages from the overall student population to find the expected values. The table with data is given by

$\begin{array}{|ccc|}\hline \text{Observed}& \text{Expected}& \text{How we calculated Exp}\\ 113& 54& 0.054·1000\\ 94& 145& 0.145·1000\\ 136& 159& 0.159·1000\\ 10& 12& 0.12·1000\\ 604& 616& 0.616·1000\\ 43& 14& 0.014·1000\\ \hline\end{array}$

Explanation

To test these hypothesis:

$H_0$:The local results follow the distribution of the overall student population $H_1$: The local results do not follow the distribution of the overall student population

Since we are given $6$different ethnicities, the number of degrees of freedom is $6-1=5$.

We are using the ${\chi }^2$distribution.

Test statistic is given by:

${\chi }^2=\frac{(113-54{)}^2}{54}+\frac{(94-145{)}^2}{145}+\frac{(136-159{)}^2}{159}++\frac{(10-12{)}^2}{12}+\frac{(604-616{)}^2}{616}+\frac{(43-14{)}^2}{14}$

$=64.46+18+3.33+0.333+0.234+60.1$

role="math" localid="1648727321937" $=146.457$

Explanation

Using the applet, we can find that $p$-value is $0$:

Taking $\alpha =0.05$, we can see that

$0.05>0=p$

This means that we reject null hypothesis. There is evidence to conclude that the local results do not follow the distribution of the overall student population.