Explanation:
We have bar x = 934
S = 616
n = 12
A.
Null hypothesis: h0: u = 1250
Alternative hypothesis: h1: u< 1250
B. To get the t test statistic
T = 934-1250/616/√12
T = -316/616/3.4641
= -316/117.824
= -1.777
C. O value = p(t<-1.777)
This gives 0.0516
D. At 10% significance
O.0516<0.1 so we reject the null hypothesis and conclude that this deduction is < than 1250
At 5% level of significance
O.o516>0.05. we do not reject the null hypothesis. We conclude there is insufficient evidence of claim being less than 1250