Question

8. An insurance agent theorizes that the fraction of uninsured drivers is higher in states with no-fault auto insurance (e.g. Michigan) than in states with tort auto insurance (e.g. Wisconsin). To test this theory she samples 150 drivers from Michigan and 125 drivers from Wisconsin and finds that 42 and 23 are uninsured in the two samples, respectively. Let Michigan drivers be population 1 and Wisconsin drivers be population 2.

a. Set up the null and alternative hypotheses to test the agent’s theory.

b. Compute an appropriate test statistic.

c. What is the p-value? Conclude at α = 0.10. Explain.

Answer 1

Solution :

Given that,

n₁ = 150

x₁ = 42

Point estimate = sample proportion = ₁ = x₁ / n₁ = 0.28

n₂ = 125

x₂ = 23

Point estimate = sample proportion = ₂ = x₂ / n₂ = 0.184

The value of the pooled proportion is computed as,

= ( x₁ + x₂ ) / ( n₁ + n₂ )

= (42 + 23 ) / (150+125 )

= 0.236

1 - = 0.764

Level of significance = = 0.10
a)

This a right(one)-tailed test.The null and alternative hypothesis is,

Ho: p₁ = p₂

Ha: p₁ > p₂

b)

Test statistics

z = (₁ - ₂ ) / *(1-) ( 1/n₁ + 1/n₂ )

= ( 0.28 - 0.184) / (0.236 * 0.764 ) ( 1/150 + 1/125)

= 1.866

c)

P-value = P(Z>z)

= 1 - P(Z <z )

= 1- P(Z < 1.866)

= 1 - 0.969

= 0.031

The p-value is p = 0.031, and since p = 0.031 < 0.10, it is concluded that the null hypothesis is rejected.

Conclusion :

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p₁ is higher than p₂, at the 0.10 significance level.