Question

Did the randomization produce similar groups? First, compare the two groups of birds in the first year. The only difference should be the chance effect of the random assignment. The study report says: "In the experimental year, the degree of synchronization did not differ between food-supplemented and control females." For this comparison, the report gives $t=-1.05.$

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

(b) Explain how this value of $t$ leads to the quoted conclusion.

Short answer

a) Two-sample t statistic

b) There is not sufficient evidence to support the claim of a difference.

Step-by-step solution

Part(a) Step 1: Given information

Need to find what type of t statistics.

Part(a) Step 2: Explanation

The test compares two groups of birds.

7 pairs are in the first group and 6 pairs in the second group.

The samples cannot contain paired data, because the sample sizes differ.

Thus the t-statistic is then the two-sample $t$ statistic.

Part(b) Step 1: Given Information

$t=-1.05$

Part(b) Step 2: Explanation

Determine the hypothesis:

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

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

Determine the degrees of freedom:

localid="1650516349894" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (7-1,6-1) \\ =5 \end{gathered}$$

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 $df=5$:

localid="1650516387942" $0.30=2\times 0.15<P<2\times 0.20=0.40$

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

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