A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period.
Single Policyholders Married Policyholders
n1 = 500 n2 = 800
Number making claims = 95 Number making claims = 80
(a) Use a level of significance a = 0.05. Test to determine whether the claim rates differ between single and married male policyholders. Explain your determination.
(b) Provide a 95% confidence interval for the difference between the proportions for the two populations.

Answer:

a

The  determination is  

    There is sufficient to conclude that proportion of the single policy

b

  The 95% confidence interval is  [tex] 0.0498    < p_1 - p_2 < 0.1302 [/tex]  

holders is difference from the proportion of married policy holders

Step-by-step explanation:

From the question we are told that

The sample size for single policy holders is [tex]n_1 = 500[/tex]

The number of single police holders making claim is k = 95

The sample size for married policy holders [tex]n_2 = 800[/tex]

The number of double police holders making claim is u = 80

The level of significance is [tex]\alpha = 0.05[/tex]

Generally the critical value of [tex]\frac{\alpha }{2}[/tex] obtained from the

normal distribution table is

[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]

The null hypothesis is [tex]H_o : p_1 = p_2[/tex]

The alternative hypothesis is [tex]H_a : p_1 \ne p_2[/tex]

The sample proportion of single police holders making claim is

[tex]\r p_ 1 = \frac{k}{n_1}[/tex]

=> [tex]\r p_ 1 = \frac{95}{500}[/tex]

=> [tex]\r p_ 1 = 0.19[/tex]

The sample proportion of married police holders making claim is

[tex]\r p_ 2 = \frac{u}{n_2}[/tex]

=> [tex]\r p_ 2 = \frac{80}{800}[/tex]

=> [tex]\r p_ 2 = 0.1[/tex]

Generally the pooled sample proportion is mathematically represented as

[tex]\r p = \frac{k + u}{n_1 + n_2}[/tex]

=> [tex]\r p = \frac{95 + 80}{500 + 800}[/tex]

=> [tex]\r p = 0.1346 [/tex]

Generally the test statistics is mathematically represented as

[tex]z = \frac{\r p_1 - \r p_2 }{ \sqrt{ \r p (1 - \r p )* [\frac{1}{n_1} + \frac{1}{n_2} ]} }[/tex]

=> [tex]z = \frac{0.19 - 0.1 }{ \sqrt{0.1346 (1 - 0.1346 )* [\frac{1}{500} + \frac{1}{800} ]} }[/tex]

=>[tex]z = 4.6256 [/tex]

Generally the p-value is mathematically represented as

[tex]p-value = 2* P(Z >4.6256 )[/tex]

From the z table [tex]P(Z >4.6256 ) = 0 [/tex]

[tex]p-value = 2* 0 = 0[/tex]

From the calculation we see that the p-value < [tex]\alpha[/tex] so the null hypothesis is rejected

Hence there is sufficient to conclude that proportion of the single policy holders is difference from the proportion of married policy holders

Generally the standard error is mathematically represented as

[tex]SE = \sqrt{ \frac{\r p_1 (1 -\r p_1)}{n_1} + \frac{\r p_2 (1 -\r p_2)}{n_2} }[/tex]

=> [tex]SE = \sqrt{ \frac{0.19 (1 -0.19)}{500} + \frac{ 0.1 (1 - 0.1)}{800} }[/tex]

=> [tex]SE =0.0205012[/tex]

Generally the margin of error is mathematically represented as

[tex]E = Z_{\frac{\alpha }{2} } * SE[/tex]

=> [tex]E = 1.96 *0.0205012 [/tex]

=> [tex]E = 0.0402035[/tex]

Generally the 95% confidence level is mathematically represented as

[tex]\r p_1 - \r p_2 - E < p_1 - p_2 < \r p_1 - \r p_2 + E[/tex]

=> [tex]0.19  - 0.10 -0.0402035    < p_1 - p_2 < 0.19  - 0.10 + 0.0402035  [/tex]

=>  [tex]0.09 -0.0402035    < p_1 - p_2 < 0.09 + 0.0402035  [/tex]

=> [tex] 0.0498 < p_1 - p_2 < 0.1302 [/tex]

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