Question

In: Statistics and Probability

Based on information from a large insurance company, 67% of all damage liability claims are made...

Based on information from a large insurance company, 67% of all damage liability claims are made by single people under the age of 25. A random sample of 52 claims showed that 43 were made by single people under the age of 25. Does this indicate that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company? Give the test statistic and your conclusion.

a) z = 2.407; reject Ho at the 5% significance level

b) z = 2.407; fail to reject Ho at the 5% significance level

c) z = 1.907; reject Ho at the 5% significance level

d) z = -1.907; fail to reject Ho at the 5% significance level

e) z = -2.407; reject Ho at the 5% significance level

Solutions

Expert Solution

H0: p = 0.67

Ha: p > 0.67

Sample proportion = 43 / 52 = 0.827

Test statistics

z = - p / sqrt( p (1 - p) / n)

= 0.827 - 0.67 / sqrt ( 0.67 * 0.33 / 52)

= 2.407

This is test statistics value.

p-value = P( Z > z)

= P( Z > 2.407)

= 0.008

Since p-value < 0.05 level, we have sufficient evidence to reject H0.

Conclusion -

z = 2.407 , Reject H0 at 5% significance level.


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