Question

An insurance company selected a random sample of 500 clients under 18 years of age and found that 180 of them had had an accident in the previous year. A random sample of 600 clients aged 18 and older was also selected and 150 of them had had an accident in the past year. We want to conduct a hypothesis test to determine if the accident proportions differ between the two age groups.
a. What is the pooled proportion?

b. The p-value for this test is...

c. If we want to create a 95% confidence interval for the difference in accident rates between younger and older drivers, what is the LOWER bound of the interval? Round to 4 decimal places.

d. If we want to create a 95% confidence interval for the difference in accident rates between younger and older drivers, what is the UPPER bound of the interval? Round to 4 decimal places.

Answer 1

Answer:

a) The pooled proportion is p=0.3.

b) P-value = 0.000078

c) Lower bound = 0.0556

d) Upper bound = 0.1644

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the accident proportions differ between the two age groups .

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

The significance level is 0.05.

The sample 1, of size n1=500 has a proportion of p1=0.36.

$p_1=X_1/n_1=180/500=0.36$

The sample 2, of size n2=600 has a proportion of p2=0.25.

$p_1=X_1/n_1=150/600=0.25$.

The difference between proportions is (p1-p2)=0.11.

$p_d=p_1-p_2=0.36-0.25=0.11$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{180+150}{500+600}=\dfrac{330}{1100}=0.3$

The standard error for the difference between proportions can now be calculated as:

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.3*0.7}{500}+\dfrac{0.3*0.7}{600}}\\\\\\s_{p1-p2}=\sqrt{0.00042+0.00035}=\sqrt{0.00077}=0.0277$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.11-0}{0.0277}=\dfrac{0.11}{0.0277}=3.964$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=2\cdot P(t>3.964)=0.000078$

As the P-value (0.000078) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that the accident proportions differ between the two age groups.

If we want to calculate the bounds of a 95% confidence interval, we start by calculating the margin of error.

For a 95% CI, the critical value for z is z=1.96.

Then, the margin of error is:

$MOE=z \cdot s_{p1-p2}=1.96\cdot 0.0277=0.0544$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.11-0.0544=0.05561\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.11+0.0544=0.16439$

The  95% confidence interval for the population mean is (0.0556, 0.1644).