Answer:
a) Then, t(s) is in the rejection region. We can not support that the mean annual premium in Pennsylvania is lower than the national mean annual premium
b) The point estimate of the difference is 59 $
Step-by-step explanation:
From sample we get:
sample size n = 25
annual mean x = 1440
standard deviation of the sample s = 150
mean for the mean annual premium μ = 1499
Hypothesis Test:
Null Hypothesis H₀ x = μ
Alternative Hypothesis Hₐ x < μ
Alternative Hypotesis indicates that the test is a one-tail test to the left
Let´s assume confidence Interval 95 % then α = 5 % α = 0.05
As n < 30 and we assume normality we will make a t-student test
degree of freedom df = n - 1 df = 24
With df = 24 and α = 0.05 and from t-student table we get t(c) = - 1.711
To calculate t(s) we apply:
t(s) = ( x - μ ) / s/√n
t(s) = ( 1440 - 1499 ) * 5 / 150
t(s) = - 59/ 30
t(s) = - 1.96
Comparing t(s) and t(c)
|t(s)| > |t(c)|
Then, t(s) is in the rejection region. We can not support that the mean annual premium in Pennsylvania is lower than the national mean annual premium.
b) The point estimate for the difference is : x - μ = 59 $
