Chapter 10.1, Problem 10.6E

To determine

To Explain: the performance of a chi-square goodness of fit test for the company’s claimed colour distribution and the conclusion.

Answer to Problem 10.6E

Therefore it is not enough evidence to reject the claim that the distribution of colours is as the company claimed.

Explanation of Solution

Given:

12 were blue, 7 orange, 4 yellow, 8 red and 2 brown

  $n=46$

Formula used:

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

Calculation:

Suppose

  $\alpha =0.05$

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

  $H_o:p_1=0.24,p_2=0.2,p_3=0.16,p_4=0.14,p_5=0.13,p_6=0.13$

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

  $H_1:$ At least one of the $p_i$ ’s is different.

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

  $\begin{array}{l}{E}_1=n{p}_1=46\times 0.23=10.58\\ E_2=n{p}_2=46\times 0.23=10.58\\ E_3=n{p}_3=46\times 0.15=6.9\\ E_4=n{p}_4=46\times 0.15=6.9\\ E_5=n{p}_5=46\times 0.12=5.52\\ E_6=n{p}_6=46\times 0.12=5.52\end{array}$

The value of the test-statistic is

  $\begin{array}{c}{x}^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(12-11.04\right)}^2}{11.04}+\frac{{\left(7-9.2\right)}^2}{9.2}\\ +\frac{{\left(13-7.36\right)}^2}{7.36}+\frac{{\left(4-6.44\right)}^2}{6.44}\\ +\frac{{\left(8-5.98\right)}^2}{5.98}+\frac{{\left(2-5.98\right)}^2}{5.98}\\ =9.1872\end{array}$

The degrees of freedom are

  $df=c-1=6-1=5$

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 (or interval) in the column chi-square distribution table in containing the $x^2$ -value in the row $df=c-1=6-1=5$ :

  $0.05<P<0.10$

If the P-value is less than or equal to the significance level, then the Alternate 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 that the distribution of colours is as the company claimed.