Birds in the trees Researchers studied the behavior of birds that were searching for seeds and insects in an Oregon forest. In this forest, $54$% of the trees were Douglas firs, $40$% were ponderosa pines, and $6$% were other types of trees. At a randomly selected time during the day, the researchers observed $156$red-breasted nuthatches: $70$were seen in Douglas firs, $79$in ponderosa pines, and $7$ in other types of trees.^$2$ Do these data suggest that nuthatches prefer particular types of trees when they’re searching for seeds and insects? Carry out a chi-square goodness-of-fit test to help answer this question.

It is given in the question that, In this forest, $54$% of the trees were Douglas firs, $40$% were ponderosa pines, and $6$% were other types of trees. At a randomly selected time during the day, the researchers observed $156$red-breasted nuthatches: $70$were seen in Douglas firs, $79$in ponderosa pines, and $7$in other types of trees.

Given: $n=156$

Let us assume: $\alpha =0.05$

The null hypothesis states that the population proportions are equal to the mentioned proportions:

localid="1650617560505" $$\begin{gathered} H_0:p_1=54\mathrm{\%} \\ =0.54, \end{gathered}$$

$$\begin{gathered} H_0:p_2=40\mathrm{\%} \\ =0.4, \end{gathered}$$

$$\begin{gathered} H_0:p_3=6\mathrm{\%} \\ =0.06 \end{gathered}$$

The alternative hypothesis states the opposite of the null hypothesis:

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

The expected frequency E is the product of the sample size n and the probabilities pi

localid="1650617756516" $$\begin{gathered} E_1=n{p}_1 \\ =156\times 0.54 \\ =84.24 \end{gathered}$$

localid="1650617773816" $$\begin{gathered} E_2=n{p}_2 \\ =156\times 0.4 \\ =62.4 \end{gathered}$$

localid="1650617793473" $$\begin{gathered} E_3=n{p}_3 \\ =156\times 0.06 \\ =9.36 \end{gathered}$$

The squared differences between the actual and predicted frequencies, divided by the expected frequency, make up the chi-square subtotals.

Thevalue of the test statisticis then the sum of the chi-square subtotals: $$\begin{gathered} {\chi }^2=\sum \frac{(O-E{)}^2}{E} \\ =\frac{(70-84.24{)}^2}{84.24}+\frac{(79-62.4{)}^2}{62.4}+\frac{(7-9.36{)}^2}{9.36} \\ =7.4182 \end{gathered}$$

The degree of freedom is the number of categories decreased by 1.

$$\begin{gathered} df=c-1 \\ =3-1 \\ =2 \end{gathered}$$

The $P$-value is the probability of obtaining the value of the test statistic, or a value more extreme. The $P$-value is the number (or interval) in the column title of the chi-square distribution table in the appendix containing the X^$2$-value in row d f=$2$

$0.01<P<0.05$

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

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

There is sufficient evidence to support the claim that nuthatches prefer particular types of trees.