Question

Year-to-year comparison Rather than comparing the two groups in each year, we could compare the behavior of each group in the first and second years. The study report says: "Our main prediction was that females receiving additional food in the nestling period should not change laying date the next year, whereas controls, which (in our area) breed too late in their first year, were expected to advance their laying date in the second year."

Comparing days behind the caterpillar peak in Years 1 and 2 gave $t=0.63$ for the control group and $t=-2.63$ for the supplemented group.

(a) What type of $t$ statistic (one-sample, paired, or two-sample) are these? Justify your answer.

(b) What are the degrees of freedom for each $t$ ?

(c) Explain why these t-values do not agree with the prediction.

Short answer

a)Paired t statistic

b) Control group:$t=0.63$and$df=5$

Supplemented group: $t=-2.63$and $df=6$

c) The allegation that the birds in the control group did not move their laying date in the second year is unsupported by evidence. There is enough data to suggest that the supplemented group's birds shift their laying schedule.

Step-by-step solution

Part(a) Step 1: Given Information

Need to find what type of statistics.

Part(a) Step 2: Explanation

There will be two samples because the test compares one group of birds from two separate years.

Because each sample contains the same birds, the samples have paired data.

As a result, the paired t statistic is the $t$-statistic.

Part(b) Step 1: Given Information

Given for the control group

$t=0.63$

Given for the supplemented group

$t=-2.63$

Part(b) Step 2: Explanation

Degrees of freedom for the control group:

$$\begin{gathered} df=n_1-1 \\ =6-1 \\ =5 \end{gathered}$$

Degrees of freedom for the supplemented group:

$$\begin{gathered} df=n_1-1 \\ =7-1 \\ =6 \end{gathered}$$

Part(c) Step 1: Given Information

$t=0.63$

$t=-2.63$

Part(c) Step 2: Explanation

The $P$-value is the chance of getting the test statistic's result, or a number that is more severe. The $P$-value is the number (or interval) in Table B's column title that corresponds to the t-value in row$df=5$:

$P>0.25$

The null hypothesis is rejected if the P-value is less than or equal to the significance level:

$P>0.05\Rightarrow \text{Fail to reject}{H}_0$

The allegation that the birds in the control group did not move their laying date in the second year is unsupported by evidence.

The $P$-value is the chance of getting the test statistic's result, or a number that is more severe. The P-value is the number (or interval) in Table B's column title that corresponds to the t-value in row localid="1650517845586" $df=6$:

$0.01<P<0.02$

The null hypothesis is rejected if the P-value is less than or equal to the significance level:

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

There is enough data to suggest that the supplemented birds modify their laying date.