Question

Do birds learn to time their breeding? Blue titmice eat caterpillars. The birds would like lots of caterpillars around when they have young to feed, but they must breed much earlier. Do the birds learn from one year’s experience when to time their breeding next year? Researchers randomly assigned $7$pairs of birds to have the natural caterpillar supply supplemented while feeding their young and another $6$pairs to serve as a control group relying on natural food supply. The next year, they measured how many days after the caterpillar peak the birds produced their nestlings. The investigators expected the control group to adjust their breeding date the following year, whereas the well-fed supplemented group had no reason to change. Here are the data (days after caterpillar peak):

a. Do the data provide convincing evidence that birds like these that have to rely on the natural food supply produce their nestlings closer to the caterpillar peak, on average, than birds like these that have the caterpillar supply supplemented?

b. Interpret the P-value from part (a) in the context of this study.

Short answer

Part a. There is convincing evidence that bird like these that have to rely on the natural food supply produce their nestlings closer to the caterpillar peak on average than the birds like these that have the caterpillar supply supplemented.

Part b. There is$0.575\%$ chance of obtaining similar results or more extreme when that bird like these that have to rely on the natural food supply produce their nestlings an equal distance from the caterpillar peak on average than the birds like these that have the caterpillar supply supplemented.

Step-by-step solution

Part a. Step 1. Given information

Given data:

Part a. Step 2. Explanation

In this study the researcher wants to know that that bird like these that have to rely on the natural food supply produce their nestlings closer to the caterpillar peak on average than the birds like these that have the caterpillar supply supplemented or not.

It is given that:

$\alpha =0.05$

The given claim that: Mean is higher for the fish oil.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

$$\begin{gathered} H_0:{\mu }_1={\mu }_2 \\ {H}_a:{\mu }_1<{\mu }_2 \end{gathered}$$

Where we have,

${\mu }_1=$the true mean number of days after the caterpillar peak the birds produced their nestlings.

${\mu }_2=$the true mean number of days after the caterpillar peak the well-fed produced their nestlings.

Now, the mean is as follows:

$$\begin{gathered} {\overline{x}}_1=\frac{{\displaystyle \sum _{i-1}^n}{x}_i}{n} \\ =\frac{4.6+2.3+7.7+6+4.6+(-1.2)}{6} \\ =4 \\ {\overline{x}}_2=\frac{{\displaystyle \underset{i-1}{\overset{n}{\sum x_i}}}}{n} \\ =\frac{15.5+11.3+5.4+16.5+11.3+11.4+7.7}{7} \\ =11.3 \end{gathered}$$

And the standard deviation is as follows:

$$\begin{gathered} s_1=\sqrt{\frac{\sum _{}{(x-\overline{x})}^2}{n-1}} \\ =\sqrt{\frac{{(4.6-4)}^2+{(2.3-4)}^2+...+{(-1.2-4)}^2}{6-1}} \\ =3.1093 \\ {s}_2=\sqrt{\frac{\sum _{}{(x-\overline{x})}^2}{n-1}} \\ =\sqrt{\frac{{(15.5-11.3)}^2+{(11.3-11.3)}^2+...+{(7.7-11.3)}^2}{7-1}} \\ =3.9256 \end{gathered}$$

Now, find the test statistics:

$$\begin{gathered} t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{{\displaystyle \frac{s_1^2}{n_1}}+{\displaystyle \frac{s_2^2}{n_2}}}} \\ =\frac{4-11.3-0}{\sqrt{{\displaystyle \frac{3.{1093}^2}{6}}+{\displaystyle \frac{3.{9256}^2}{7}}}} \\ =3.892 \end{gathered}$$

Now, the degree of freedom will be:

$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(6-1,7-1) \\ =5 \end{gathered}$$

So the P-value will be:

$0.005<P<0.01$

By the calculator we calculate the P-value that will be: role="math" localid="1664258954074" $2\times tcdf(3.892,1E99,85)$ which results in the P-value of $0.00575$.

And we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, then,

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

Thus, we conclude that there is convincing evidence that bird like these that have to rely on the natural food supply produce their nestlings closer to the caterpillar peak on average than the birds like these that have the caterpillar supply supplemented.

Part b. Step 1. Explanation

From part (a), we have that,

So the P-value will be:

$0.005<P<0.01$

By the calculator we calculate the P-value that will be: $2\times tcdf(3.892,1E99,85)$ which results in the P-value of $0.00575$.

And we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, then,

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

Thus, we conclude that there is convincing evidence that bird like these that have to rely on the natural food supply produce their nestlings closer to the caterpillar peak on average than the birds like these that have the caterpillar supply supplemented.

Thus we can say that there is$0.575\%$ chance of obtaining similar results or more extreme when that bird like these that have to rely on the natural food supply produce their nestlings an equal distance from the caterpillar peak on average than the birds like these that have the caterpillar supply supplemented.