Answer:
We fail to reject the Null ;
(88.323, 130.677)
Step-by-step explanation:
Given the data :
117, 156, 89, 72, 116, 125, 101, 100
H0 : μ ≥ 100
H1: μ < 100
From the data: using calculator :
Mean, x = 109.5
Standard deviation, s = 25.33
The test statistic:
(x - μ) ÷ s/Sqrt(n)
(109.5 - 100) ÷ 25.33/sqrt(8)
Test statistic = 1.06
Since, same size is small, we use t test ;
Using the Pvalue calculator :
T at df = 8 - 1 = 7, α = 0.05
Pvalue = 0.161
Pvalue > α
0.161 > 0.05 ; Hence, we fail to reject H0.
B.)
C. I = mean ± Tcritical *s/√n
Tcritical = 2.365 at 95% and df = 7
109.5 ± 2.365(25.3264/√8)
109.5 ± 2.365(8.9542)
(109.5-21.1768 ; 109.5+21.1768)
(88.323, 130.677)
In conclusion true mean is in the confidence interval, so it is consistent.
