In: Statistics and Probability
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
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.