91Ó°ÊÓ

Given data:

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:

$\mathrm{Î\pm }=0.05$

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:

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

Where we have,

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

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

Now, the mean is as follows:

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

And the standard deviation is as follows:

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

Now, find the test statistics:

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

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" $2×tcdf(3.892,1E99,85)$ 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,

$P<0.05⇒Reject{H}_0$

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.

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: $2×tcdf(3.892,1E99,85)$ 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,

$P<0.05⇒Reject{H}_0$

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.