Chapter 10.1, Problem 10.7E

To determine

To Explain: a chi-square goodness of fit test for researcher’s prediction and the conclusion.

Answer to Problem 10.7E

  $\begin{array}{c}\text{P<0}\text{.05}\Rightarrow {\text{Fail to reject H}}_{\text{0}}\\ \text{Therefore it is not enough evidence to reject the claim of researcher}\end{array}$

Explanation of Solution

Given:

  $n=423+133=556$

Formula used:

  $x^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$

Calculation:

Assume

  $\alpha =0.05$

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

  $H_0:p_1=\frac{3}{4}=0.75,p_2=\frac{1}{4}=0.25$

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

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

The expected frequencies $E$ are the multiplication of the sample size and the probabilities.

  $\begin{array}{l}{\text{E}}_1=n{p}_1=556\times 0.75=417\\ {\text{E}}_2=n{p}_2=556\times 0.25=139\end{array}$

The value of the test-statistic is

  $\begin{array}{c}{x}^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(423-417\right)}^2}{417}+\frac{{\left(133-139\right)}^2}{139}\\ =0.3453\end{array}$

The degree of freedom is

  $df=c-1=2-1=1$

The P-Value is the probability of getting the value of the test statistic, or a value more extreme. The P-value is the number in the column of the chi-square distribution table containing the $x^2$ -value in the row $df=c-1=2-1=1$ .

P>0.25

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

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

Therefore it is no enough evidence to reject the claim of researcher.