The mean annual premium for automobile insurance in the United States is $1,499. Being from Pennsylvania, you believe automobile insurance is 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 $1,440 with a standard deviation of s = $150. 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 (to the nearest dollar)?

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 $

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

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