Question

In a study of 11,000 car crashes, it was found that 5720 of them occurred within 5 miles of home. Use a .05 significance level to test the claim that more than 50% of car crashes occur within 5 miles of home. Fill in the requested information. Claim: ??0: ????: Test you’ll use on your calculator: Values for each required entry on calculator: p-value: Alpha: Decision: Reason for decision: Conclusion (complete sentence):

Answer 1

We want to test hypothesis: Ho : p 0.5 Ha: p>0.5 This is an upper tailed test Null Hypothesis, Ho p 0.5 is also correct) (Alternative Hypothesis, also called H1 5720 n 11000 Sample proportion, p - 0.52, n 11000 and claimed proportion. P = 0.5 Significance level, α-0.05 now, calculating statistics P × (1-P) 0.5 × (1-0.5) 11000 Standard Deviation of p, ơp ะ 0.00477 (0.52)-0.5 0.00477 p-0.5 ะ 4.195235 ~ 4.20 Test statistic,20-420 Since this is upper tailed test, p-value- P(Z > Zobserved) - P(Z > 4.20) P(Z<-4.20) 0 p-value:0.0000 Note that exact p-value for above test is: 0.0000136294 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 0.5 Conclusion: Sufficient evidence to support the claim that more than 50% of car crashes occur within 5 miles of home