Question

Based on information rom a large insurance company, 67% of all damage liability claims are made by single people under the age of 25 A random sample of 51 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) O z-2.130: reject Ho at the 5% significance level b) O z--2. 130; fail to reject Ho at the 5% significance level c) O z 2630: fail to reject Ho at the 5% significance level d) O e) O z--2.630: reject Ho at the 5% significance level z 2.630: reject Ho at the 5% significance level

Answer 1

We want to test hypothesis: Ho p 0.67 Ha: p>0.67 This is an upper tailed test Null Hypothesis, Ho p0.67 is also correct) (Alternative Hypothesis, also called H1 r43 Sample proportion, p= 0.843137, n = 51 n 51 and claimed proportion. P = 0.67 Significance level, a = 0.05 now, calculating statistics P × (1-P) 0.67 × (1-0.67) 51 Standard Deviation of p, ơp 0.0658 (43/51)-0.67 -- 0.0658 p-0.67 «observed 2.629549 ~ 2.63 Test statistic,20-2.63 Since this is upper tailed test, p-value = P(Z > zobserved) = P(Z > 2.63) P(Z<-2.63) 0.0043 p-value:0.0043 Note that exact p-value for above test is: 0.0042749123 Rejection Criteria: Reject H0 ifp-value <a Decision : Since p-value < α, we reject) the null hypothesis. Hence at 5% significance level, we have sufficient evidence to conclude that p is greater than 0.67 Option D