Question

A simple random sample ct tront-seat occupants involved in car crashes is obtained. Among 2765 occupants not wearng seat beits, 33 were kiled. Among 7806 occupans wearnng seat belts, 10 were killed. Use a 005 significance level to test the claim that seat bets are etective in reducing fatalties. Complete parts (a) through (c) below dentity the test stat stic. Round to twa decimal places as needed) dentity the P-value P-value- (Rouns to three decimal places as needed) hat is the conclusion based on the hypothesis test? The P value is the signifcance level c a 005, so ▼ | the null hypothesis. There | relect fall to reject ▼| suncert evidence to support the Caim that re ratary rate is nigher for those not wearing seat belts. Is not ls less than greater than b. Test the ciaim by construcring an approprite confidence interval The appropriate corrounce ntervalisD-(p.P2)다 Round to three decimal places as needed) at is he conclusion based on the conndence intervar? Bccouse the confidence interval imts o, it appears that the two tatallty rates are Secouse the confidence interval limits include valucs, it appears that the tatality rate is for thosc not wearing seat bcts Include does not include equal not equal only posltive positlve and negative only negative the same lower higher e. what do the resuts suggest about the efecaiveness of seat bets? o A. The results suggest that ne use or seat belts is associated with lower tility rates than not using seat bens ○ B. The results suggest that Te use of seat belts is associated with higher tatality rates than not using seat belts. O C. The resuits supgest that he use of seat belts is associated with the same fata ity rates as not using seat belts

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The following null and alternative hypotheses need to be tested Ha pi P2 This corresponds to a left-tailed test, for which a z-test for two population proportions needs to be conducted (2) Rejection Region Based on the information provided, the significance level is α 0.05, and the critical value for a left tailed test is zc--1.64 The rejection region for this left-tailed test is R (3)_Test Statistics The z-statistic is computed as follows <1.64 P1 P2 0.0119 -0.0013 0.0041 (1-0.0041) (1/2765 1/7806) (4) Decision about the null hypothesis Since it is observed that z = 7.563 > ze =-1.64, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p 1, and since p 10.05, it is concluded that the null hypothesis is not rejected (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion pi is less than p2, at the 0.05 significance level Confidence Interval The 95% confidence interval for pi-P2 is: 0.007 < pi-P2 < 0.015

p value greater than 0.05

Fail to reject

not enough evidence

confidence interval does not include zero

equal

because interval includes only positive

same

c) C option

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