Answer :
By using the concept of mean and standard deviation, it can be calculated that the critical values of t has an absolute value of 2.989
Confidence interval = (-2.989, 2.989)
What is mean and Standard Deviation?
Suppose there is a data set. Mean is used to find the average of all the values of the data set.
Variance is the sum of the square of the deviation from the mean.
On taking the square root of the variance, Standard deviation is obtained.
Biology Psychology
Majors Majors d [tex]d - \bar{d}[/tex] [tex](d - \bar{d})^2[/tex]
87 87 0 9.143 83.5944
78 96 -18 -8.857 78.4464
77 92 -15 -5.857 34.3044
85 90 -5 4.143 17.1644
83 90 -7 2.143 4.5924
88 91 -3 6.143 37.7364
73 89 -16 -6.857 47.0184
Total - -64 - 302.8568
Here number of observations (n) = 7
[tex]\bar{d} = \frac{-64}{7}[/tex]
[tex]\bar{d} = -9.143[/tex]
Standard Deviation (SD) = [tex]\sqrt{\frac{302.8568}{6}}[/tex]
SD = 7.1047
Now,
[tex]H_0 : \mu_d = 0\ against \ H_a : \mu_d \neq 0\\[/tex]
The test statistic used here is t - test
[tex]t = \frac{-9.143 - 0}{\frac{7.1047}{\sqrt{7}}}\\[/tex]
t = 3.4
Degree of freedom (Df) = 7-1 = 6
Critical value of t at the given level of significance at Df of 6 = 2.989
At 0.05 level of significance, 3.4 falls to the right of 2.989 which is the rejection region
So the null hypothesis has been accepted
Absolute value of Critical values of t =2.989
Confidence interval = (-2.989, 2.989)
To learn more about mean and standard deviation, refer to the link -
https://brainly.com/question/28225633
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