Chapter 10.1, Problem 10.29E

To determine

To find: a chi-square goodness of fit test to help answer this question.

Answer to Problem 10.29E

There is enough evidence to help the claim that nuthatches prefer particular types of trees.

Explanation of Solution

Given:

  $n=156$

Formula used:

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

Calculation:

Assume

  $\alpha =0.05$

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

  $H_0:p_1=54\%=0.54,p_2=40\%=0.4,p_3=6\%=0.06$

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

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

The expected frequencies $E$

  $\begin{array}{c}{E}_1=n{p}_1=156\times 0.54=84.24\\ E_2=n{p}_2=156\times 0.4=62.4\\ E_3=n{p}_3=156\times 0.06=9.36\end{array}$

The value of the test- statistic is

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(70-84.24\right)}^2}{84.24}+\frac{{\left(79-62.4\right)}^2}{62.4}+\frac{{\left(7-9.36\right)}^2}{9.36}\\ =7.4182\end{array}$

The degrees of freedom is

  $df=c-1=3-1=2$

The P-Value is the probability of getting the value of the test statistic, or a value more extreme and $df=2:$

  $0.01<P<0.05$

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

  $P<0.05\Rightarrow \text{Reject }{H}_0$

There is enough evidence to help the claim that nuthatches prefer particular types of trees.