The mean annual premium for automobile insurance in the united states is $1503 (insure. com website, march 6, 2014). being from pennsylvania you believe automobile insurance s cheaper there and wish to develop statistical support for your opinion. a sample of 25 automobile insurance policies from the state of pennsylvania showed a mean annual premium of $1440 with a standard deviation of s = $165.

a. develop a hypothesis test that can be used to determine whether the mean annual premium in pennsylvania is lower than the national mean annual premium.

b. what is a point estimate of the difference between the mean annual premium in pennsylvania and the national mean?

c at a = .05, test for a significant difference. what is your conclusion?

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

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.