Question

5) A simple random sample of​ front-seat occupants involved in car crashes is obtained. Among 2881 occupants not wearing seat​ belts, 27were killed. Among 7900 occupants wearing seat​ belts, 15 were killed. Use a 0.01 significance level to test the claim that seat belts are effective in reducing fatalities. Complete parts​ (a) through​ (c) below.

a) Test the claim using a hypothesis test

  Consider the first sample to be the sample of occupants not wearing seat belts and the second sample to be the sample of occupants wearing seat belts. What are the null and alternative hypotheses for the hypothesis​ test?

a1) Identify the test statistic (Round to two decimal places as needed)

a2) Identify the P Value (Round to three decimal places as needed)

a3) What is the conclusion based on the hypothesis test?

b) Test the claim by constructing an appropriate confidence level. (Round to three decimal places as needed)

b1) What is the conclusion based on the confidence interval?

c) What do the results suggest about the effectiveness of seat belts?

a. The result suggest that the use of seat belts is associated with higher fatality rates than not using seat belts

b. The results suggest that the use of seat belts is associated with the same fatality rates as not using seat belts

c. The result suggests that the use of seat belts is associated with lower fatality rates than not using seat belts

d. The results are inconclusive

Answer 1

For not wearing seat​ belts:

n1 = 2881, x1 = 27

p̂1 = x1/n1 = 0.0094

For wearing seat​ belts:

n2 = 7900, x2 = 15

p̂2 = x2/n2 = 0.0019

α = 0.01

a) Null and Alternative hypothesis:

Ho : p1 = p2

H1 : p1 > p2

Pooled proportion:

p̄ = (x1+x2)/(n1+n2) = (27+15)/(2881+7900) = 0.0039

Test statistic:

z = (p̂1 - p̂2)/√ [p̄*(1-p̄)*(1/n1+1/n2)]

= (0.0094 - 0.0019)/√[0.0039*0.9961*(1/2881+1/7900)] = 5.5119 = 5.51

p-value = 1- NORM.S.DIST(5.5119, 1) = 0.0000

Conclusion:

p-value < α, Reject the null hypothesis

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b) 99% Confidence interval for the difference:  

At α = 0.01, two tailed critical value, z_c = NORM.S.INV(0.01/2) =    2.576

Lower Bound = (p̂1 - p̂2) - z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.0094 - 0.0019) - 2.576*√[(0.0094*0.9906/2881) + (0.0019*0.9981/7900)] = 0.0027

Upper Bound = (p̂1 - p̂2) + z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.0094 - 0.0019) + 2.576*√[(0.0094*0.9906/2881) + (0.0019*0.9981/7900)] = 0.0123

As Both the values are greater than zero. There is enough evidence to conclude that seat belts are effective in reducing fatalities.

c) Answer : c. The result suggests that the use of seat belts is associated with lower fatality rates than not using seat belts.