48. 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.

To find that the given$t$ statistic is one-sample, paired, or two-sample.

Two sets of birds are compare to the test.
The first group consists of $7$pairings, while the second group consists of $6$pairs.
Because the sample sizes differ, paired data cannot be included in the samples.
As a result, the two-sample is the$t$ statistic

To explain the value of $t$ leads to the quoted conclusion.

Let, the statistic $t=-1.05$

Find the null and alternative hypotheses:
$H_o:{\mu }_1={\mu }_2$
$H_a:{\mu }_1\ne {\mu }_2$
Find the degree of freedom as:
$df=\min \left(n_1-1,n_2-1\right)$

$=\min (7-1,6-1)$

$=5$

Using table $B$, calculate the $p$-value as:

$p-\text{value}=2\times P(t>-1.05\mid )$

$=0.342$

Note that, $p$-value is greater than the significance level.

When the P-value is less than or equal to the significance level, then the null hypothesis will be rejected:
$P>0.05\Rightarrow \text{Fail to reject}{H}_0$

The null hypothesis is not rejected.

Hence, there is insufficient evidence to support the quoted conclusion.