Question

Did the treatment have an effect? The investigators expected the control group to adjust their breeding date the next year, whereas the well-fed supplemented group had no reason to change. The report continues: "But in the following year, food-supplemented females were more out of synchrony with the caterpillar peak than the controls." Here are the data (days behind caterpillar peak):

Carry out an appropriate test and show that it leads to the quoted conclusion.

Short answer

There is sufficient evidence to support the claim that the food-supplemented females were more out of synchrony with the caterpillar peak than the controls..

Step-by-step solution

Given Information

The mean is the sum of all values divided by the number of values:

$$\begin{gathered} {\overline{x}}_1=4 \\ {\overline{x}}_2=11.3 \end{gathered}$$

$n$is the number of values in the data set?

The standard deviation is the square root of the sum of squared deviations from the mean divided by $n-1$:

$s_1=3.1093$

$s_2=3.9256$

Explanation

Determine the hypothesis:

$H_0:{\mu }_1={\mu }_2$

$H_a:{\mu }_1<{\mu }_2$

Determine the test statistic:

localid="1650516568809" $$\begin{gathered} t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{4-11.3}{\sqrt{\frac{3.{1093}^2}{6}+\frac{3.{9256}^2}{7}}} \\ \approx -3.739 \end{gathered}$$

Determine the degrees of freedom:

localid="1650516589517" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (7-1,6-1) \\ =5 \end{gathered}$$

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 Table B containing the t-value in the row$\cap df=5$:

$0.005<P$$<0.01$

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

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