Null Hypothesis, Alternative hypothesis, Critical Region..

Mar 2011
3
0
I have a test coming up and these questions were given to prepare for the test! Please help! Thank you!(Bow)

(a) A survey on 50 claims for a particular class of customers of a motor insurance company found out that the average cost for car damage is £700 with a standard deviation of 400. The insurance manager believes instead that the average cost is £80 bigger.

(i) Does the sample data suggest that the manager is right? Test at a 5% significance level the hypothesis that the average cost is £780. Specify: null hypothesis, alternative hypothesis, critical region and comment on the result of the test.

(b) For the same type of accidents, the number of claims per vehicle insured is known to follow a Poisson distribution with mean 0.03 per year. Using the normal approximation to the Poisson distribution, compute the probability that a group of 400 insured cars produces :

(i) Less than 10 claims per year.

(ii) More than 20 claims per year.

(c) Assuming the cost per claim is 780 with standard deviation of 400, claims’ frequency and portfolio’s size are as specified above:

(i) Compute the risk premium.

(ii) Compute the value at risk at 99.5% level for the Insurer losses.
 
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Dec 2007
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Zeitgeist
I have a test coming up and these questions were given to prepare for the test! Please help! Thank you!(Bow)

(a) A survey on 50 claims for a particular class of customers of a motor insurance company found out that the average cost for car damage is £700 with a standard deviation of 400. The insurance manager believes instead that the average cost is £80 bigger.

(i) Does the sample data suggest that the manager is right? Test at a 5% significance level the hypothesis that the average cost is £780. Specify: null hypothesis, alternative hypothesis, critical region and comment on the result of the test.

(b) For the same type of accidents, the number of claims per vehicle insured is known to follow a Poisson distribution with mean 0.03 per year. Using the normal approximation to the Poisson distribution, compute the probability that a group of 400 insured cars produces :

(i) Less than 10 claims per year.

(ii) More than 20 claims per year.

(c) Assuming the cost per claim is 780 with standard deviation of 400, claims’ frequency and portfolio’s size are as specified above:

(i) Compute the risk premium.

(ii) Compute the value at risk at 99.5% level for the Insurer losses.
What have you tried? Where are you stuck? Have you asked whoever gave you these questions for solutions or answers?
 
Mar 2011
3
0
For part (a) this is what I did;

H0 : m <= 700
H1 : m > 700

From my test statistic I got 1.41. This is less than 1.96. So you don't reject null hypothesis. Also we can't conclude anything is statistically significant. Therefore, we can't tell if the manger is right or not.

For (b) I don't get what it means by Normal Approximation to the Poisson Distribution but this is what I did. Also I don't know if I should use the 5% significance level from before (If I did I could use the Poisson Table?).

(i)

P(X < 10) = P(0) + P(1) + .. + P(9)

To calculate each value I use: P(X)= (e^-m)*(m^x)/ x!

I don't know if I should be doing it like that or using the poisson table.. It doesn't seem to justify 4 marks..

(ii)

P(X > 20) = 1 - P(X < 20)

From here I use the same method as before but have many more values..? I'm sure there must be a short cut as for 4 marks each.. Seems like a lot of work.

Furthermore, what do I do with the 400? Does it just change my notation on my probabilities? I.e. P(X < 10, 400)?

For part (c)

(i)

RP = average cost per claim * probability of claims. So RP= 780*[the answer I get from (b)(ii)]

However, I don't know if my method of calculating in part (b)(ii) is correct, let alone (b)(i)..

Finally,

(ii) I'm completely lost..
 
Last edited:
Nov 2005
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someplace
For part (a) this is what I did;

H0 : m <= 700
H1 : m > 700
You are told what to use for the null-hypothesis in the question!

for a one tailed test:
H0 : m = 780
H1 : m < 780

for a two tailed test:
H0 : m = 780
H1 : m != 780

CB
 
Last edited:
Mar 2011
3
0
How do I know it's either one tailed or two tailed.. Do I show both one tailed and two tailed testing? Also I was thinking maybe it should be;

H0 : m = 780
H1 : m = 700

or

H0 :m = 780
H1 :m != 780

Does it matter how I phrase the hypotheesis testing? As I could phrase it in different ways?
 
Nov 2005
14,972
5,271
someplace
How do I know it's either one tailed or two tailed.. Do I show both one tailed and two tailed testing? Also I was thinking maybe it should be;

H0 : m = 780
H1 : m = 700

or

H0 :m = 780
H1 :m != 780

Does it matter how I phrase the hypotheesis testing? As I could phrase it in different ways?
First is wrong, the second is right (a typos in my earlier post, now corrected).

CB