Q.7 - The Practice of Statistics for AP

Question

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. 

Step-by-Step Solution

Answer

From the given information, We cannot reject the null hypothesis because the chi-square value is relatively insignificant, thus we conclude that there is no significant difference between the observed and expected pea count. In other words, after performing the chi-square goodness of fit test, we can conclude that Mendel's prediction is correct.

Step 1: Given Information

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.

Step 2: Explanation

Given: $n=156$

Let us assume: $\alpha =0.05$

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

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

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

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

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

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

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

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

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

Step 3: Explanation

The value of the test statistic is 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.

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

Step 4: Explanation

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.