Chapter 10, Problem 10.60RE

To determine

To Explain: a goodness of fit test to examine that this student’s sample reflects the population at his high school.

Answer to Problem 10.60RE

There is no enough evidence to help the claim that the distribution in the sample is different from the distribution in school roster.

Explanation of Solution

Given:

  $n=54+66+56+30=206$

Formula used:

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ E=np\end{array}$

Calculation:

The null hypothesis statement is that the population proportions are equal to the given proportions:

  $H_0:p_1=29\%=0.29,p_2=27\%=0.27,p_3=25\%=0.25,p_4=19\%=0.19$

The alternative hypothesis statement is the opposite of the null hypothesis:

  $H_1:{\text{At least one the P}}_i\text{'}\text{s is different}\text{.}$

The expected frequencies $\text{E}$ are

  $\begin{array}{l}{\text{E}}_1=n{p}_1=206\times 0.29=59.74\\ {\text{E}}_2=n{p}_2=206\times 0.27=55.62\\ {\text{E}}_3=n{p}_3=206\times 0.25=51.5\\ {\text{E}}_4=n{p}_4=206\times 0.19=39.14\end{array}$

The value of the test-statistic is

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(54-59.74\right)}^2}{59.74}+\frac{{\left(66-55.62\right)}^2}{55.62}\\ +\frac{{\left(56-51.5\right)}^2}{51.5}+\frac{{\left(30-39.14\right)}^2}{39.14}\\ =5.01625\end{array}$

The degree of freedom is:

  $df=c-1=4-1=3$

The P-value is the probability of getting the value of the test statistic, or a value more extreme and $df=c-1=4-1=3$ :

  $0.15<P<0.20$

If the P-value is equal or less than to the significance level, then the alternative hypothesis is accepted null hypothesis is rejected:

  $P>0.05\Rightarrow {\text{Fail to reject H}}_0$

There is no enough evidence to help the claim that the distribution in the sample is different from the distribution in school roster.