Answer:
a) Already in step-by-step explanation
b) t(s) = - 1.91
c) At confidence interval 0.95 we have to reject the null hypothesis H₀
Step-by-step explanation:
We have:
Normal Distribution
Size sample < 30 n = 25
Independent samples
Therefore we can apply t-student analysis
1) Test hypothesis
Null hypothesis H₀ ⇒ μ₀ = 1503
Alternative hypothesis Hₐ ⇒ μ₀ < 1503 (from problem condition)
Now we have to compute t(s)
t(s) = [ ( μ - μ₀ )/(s/√n)
μ = sample mean
s = sandard deviation of sample
Then
t(s) = ( 1440 - 1503 ) / 165 / √25 ⇒ t(s) = - 63 * 5 / 165
t(s) = - 1.91
Now we look in t- student table for t(c) ( 24 degrees of fredom and α = 0,05 and find
t(c) = - 1.711 ( we must remember symmetry in t-studend curve)
So we compare t(s) with t(c)
t(s) = -1.91 t(c) = -1.711
t(s) is smaller than t(c) then our point is in the rejection zone we reject H₀