An insurance company, based on past experience, estimates the mean damage for a

Question

An insurance company, based on past experience, estimates the mean damage for a natural disaster in its area is $ 5,000. After introducing several plans to prevent loss, it randomly samples 200 policyholders and finds the mean amount per claim was $ 4,800 with a standard deviation of $ 1,300.

Use the .05 significance level.

1. What is the population in this case? What is the sample? How big is the sample?
2. State your null and alternative hypotheses.
3. What kind of tests do you plan to use? Why?
4. Calculate the required test statistics and find the p-value associated with it.
5. Do you reject Ho or accept Ha ? Why?
6. Does it appear the prevention plans were effective in reducing the mean amount of a claim?What should the insurance company do?

Explanation / Answer

1. H0 and H1;
2. the critical variable(s);
3. the test statistic;
4. the decision to reject or to fail to reject;
5. the p-value, if requested

Solution:

Given m = 5000, n = 200, X(bar) = 4800, S = 1300 and a = 0.05

The hypothesis to be tested is

Ho: The prevention plan does not make any change is the mean amount (ie, m = 5000)

H1: The prevention plans were effective in reducing the mean amount (ie, m < 5000)

The test statistic is

Z = [X(bar) - m]/(S/Ön)

= [4800-5000]/( 1300/Ö200)

= - 2.1757

From the standard normal table for 0.05 level of significance we get the critical value
Za = -1.645 (since left tailed test)

The decision rule is Reject Ho if Z < - Za

Since Z < - Za we reject the null hypothesis with 95% confidence. Hence we conclude that the prevention plans were effective in reducing the mean amount.

P-value = P(Z < - 2.1757)

= 0.5 – P(-2.1757 < Z < 0)
= 0.5 – P(0 < Z < 2.1757), since normal curve is symmetric.

= 0.5 – 0.4854 = 0.0146

P-value = 0.0146

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