Chapter 10.1, Problem 10.10E

(a)

To determine

To Calculate: the chi-square statistic of observed counts for a bag with 60 candies that result in a P-value between 0.01 and 0.05.

Answer to Problem 10.10E

  ${\chi }^2=12.5$

Explanation of Solution

Given:

  $0.01<P<0.05$

Formula used:

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

  $E=np$

Calculation:

  $0.01<P<0.05$

The previous part solved that the P-value is less than 0.05 if the ${\chi }^2$ -statistic is less than 9.49.

  ${\chi }^2>9.49$

The previous part solved the P-value is less than 0.01 if the ${\chi }^2$ -statistic is less than 13.28 ${\chi }^2>13.28$

Therefore require to find a set of observations such that the chi-square statistic is between 9.49 and 13.28.

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

  $\begin{array}{c}{\text{E}}_1=n{p}_1=60\times 0.20=12\\ {\text{E}}_2=n{p}_2=60\times 0.20=12\\ E_3=n{p}_3=60\times 0.20=12\\ E_4=n{p}_4=60\times 0.20=12\\ E_5=n{p}_5=60\times 0.20=12\end{array}$

For example, Assume the observe value be for the flowers be: 7 for 3 flowers each and 17 for 3 flowers each.

7, 7, 7, 17, 17, 17

The value of the test-statistic is

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(7-12\right)}^2}{12}+\frac{{\left(7-12\right)}^2}{12}+\frac{{\left(7-12\right)}^2}{12}\\ +\frac{{\left(17-12\right)}^2}{12}+\frac{{\left(17-12\right)}^2}{12}+\frac{{\left(17-12\right)}^2}{12}\\ =12.5\end{array}$

(b)

To determine

To Explain: the conclusion drawn in this case.

Answer to Problem 10.10E

  $\begin{array}{c}P<0.05\Rightarrow {\text{Reject H}}_0\\ P>0.01\Rightarrow {\text{Fail to reject H}}_0\end{array}$

Explanation of Solution

Given:

  $0.01<P<0.05$

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

  $\begin{array}{c}P<0.05\Rightarrow {\text{Reject H}}_0\\ P>0.01\Rightarrow {\text{Fail to reject H}}_0\end{array}$

Therefore it would reject the flavour bend contains 20 percent of each flavour at the $\alpha =0.05$ significance level, while would fail to reject the claim that the flavour bend contains 20 percent of each flavour at the $\alpha =0.01$ significance level.