Answer:
(A) The value of t test statistics is 2.767 and P-value is 0.0042.
(B) We conclude that the mean of first group is greater than the mean of second group.
Step-by-step explanation:
We are given the following hypothesis below;
Null Hypothesis,
:
{means that the mean of first group is equal to the mean of second group}
Alternate Hypothesis,
:
{means that the mean of first group is greater than the mean of second group}
The test statistics that would be used here <u>Two-sample t-test statistics</u> as we don't know about population standard deviation;
T.S. =
~ 
where,
= sample mean of first group = 56
= sample mean of second group = 51
= sample standard deviation of first group = 8.2
= sample standard deviation of second group = 6.9
= sample of first group = 30
= sample of second group = 40
Also,
=
= 7.482
So, <em><u>the test statistics</u></em> =
~ 
= 2.767
(A) The value of t test statistics is 2.767.
<u>Also, P-value of the test statistics is given by;</u>
P-value = P(
> 2.767) = 0.0042
(B) <u>Now, at 10% significance level the t table gives critical value of 1.295 at 68 degree of freedom for right-tailed test.</u>
Since our test statistic is more than the critical values of t as 2.767 > 1.295, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis.</em></u>
Therefore, we conclude that the mean of first group is greater than the mean of second group.