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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