Answer:
Step-by-step explanation:
Let x be denoting physics text books and y mathematics text books
Mean 435.00 410.00
SD 100.00 100.00
SEM 15.81 15.81
N 40 40
(SEM is calculated as std deviation/square root of (n))
[tex]\bar x -\bar y = 25[/tex]
Std error for difference = 22.361
Degree of freedom = n1+n2-2=[tex]40+40-2 =78[/tex]
test statistic t = mean difference/std error = [tex]\frac{25}{22.361} =1.180[/tex]
[tex]H_0: \ bar x = \bar y\\H_a: \bar x > \bar Y[/tex]
(Right tailed test at 5% significance level)
p value = 0.2670
Since p value >0.05, our alpha we accept null hypothesis.
the probability of picking samples of these sizes and getting a sample mean so much higher for the physics textbooks (one-sided p-value, to four places)=0.2670