An insurance company selected samples of clients under 18 ye

An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had in the previous year. The results are shown below. What is the point estimate for the difference between the proportions i? Find a 99% confidence interval for the difference between the two proportions. Test if the proportion of clients involved in accidents is higher for under age 18 than over age 18. Use both critical value and p-value approaches. Use a=0.01.

Solution

Here we are given that n1 = 500

x1 = number of accidents = 180

n2 = 600

%x2 = number of accidents = 150

p1^ = x1/n1 = 180 / 600 = 0.36

p2^ = x2/n2 = 150 / 600 = 0.25

q1^ = 1 - p1^ = 1 - 0.36 = 0.64

q2^ = 1 - p2^ = 1 - 0.25 = 0.75

The point estimate for the differnence between the proportions is p1^ - p2^ = 0.36 - 0.25 = 0.11

First we have to check the conditions that,

n1*p1^ = 180

n1*q1^ = 320

n2*p2^ = 150

n2*q2^ = 450

All these four are greator than 5.

99% confidence interval for the difference between the two proportions is,

(p1^ - p2^) - E < p1 - p2 < (p1^ - p2^) + E

where E is the margin of error.

E = Zc * sqrt [ p1^q1^/n1 + p2^q2^/n2 ]

Zc is the critical value for normal distribution.

Zc we can find by using EXCEL.

c = confidence level = 0.99

a = 1 - c = 1 - 0.99 = 0.01

a/2 = 0.01/2 = 0.005

1 - a/2 = 0.995

EXCEL syntax :

=NORMSINV(probability)

probability = 1 - a/2

Zc = 2.58

E = 2.58 * [(0.36*0.64) / 500 + (0.64*0.75) / 600] = 2.58 * sqrt(0.000773)

E = 2.58 * 0.027808 = 0.0716

lower limit = (p1^ - p2^) - E = 0.11 - 0.0716 = 0.0384

upper limit = (p1^ - p2^) + E = 0.11 + 0.0716 = 0.1816

99% confidence interval for the difference between the two proportions is,

(0.0384, 0.1816)

Here we have to test the hypothesis that,

H0 : Proportion of clients involved in accidents is equal for under age 18 than over age 18.

H1 : Proportion of clients involved in accidents is higher for under age 18 than over age 18.

alpha = 0.01

The test statistic is,

Z = (p1^ - p2^) / sqrt[p^q^/n1 + p^q^/n2 ]

where p^ = (x1 + x2) / (n1 + n2)

q^ = 1 - p^

p^ = (180+150) / (500+600) = 0.3

q^ = 1 - 0.3 = 0.7

Z = 0.11 / sqrt [ (0.3*0.7) / 500 + (0.3*0.7) / 600 ] = 3.9641

Now we find critical value and P-value using EXCEL.

syntax :

Critical value :

=NORMSINV(probability)

where probability = 1 - alpha

=NORMSDIST(z)

where z is the test statistic value.

critical value = 2.3263

P-value = 1 - 0.9999 = 0.0000368

P-value < alpha

and Z > critical value

RejectH0 at 0.01 level of significance.

Conclusion : Proportion of clients involved in accidents is higher for under age 18 than over age 18.