Suppose an article states, "People ages 35 to 54 were the least likely group to report buckling up in the back seat.
Sixty-eight percent of this group reported always buckling up in the back seat compared with 76% of adults age 18 to
34." Suppose that it is reasonable to regard the sample of people 18 to 34 years old and the sample of people of age 35 to 54 as independent samples that are representative of these two age groups and that both sample sizes were 50. Is there convincing evidence that the proportion who always buckle up in the back seat is different for people of age 18 to 34 and people of age 35 to 54? Test the appropriate hypotheses using a significance level of 0.05. (Let p1 be the population proportion of people age 18 to 34 who always buckle up in the back seat and p2 be the population proportion of people age 35 to 54 who always buckle up in the back seat.)
State the appropriate null and alternative hypotheses.
Find the test statistic. (Round your answer to two decimal places.)
Use technology to find the P-value. (Round your answer to four decimal places.)
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Null hypothesis: p1 = p2 (The proportion who always buckle up in the back seat is the same for people of age 18 to 34 and people of age 35 to 54)
Alternative hypothesis: p1 ≠ p2 (The proportion who always buckle up in the back seat is different for people of age 18 to 34 and people of age 35 to 54)
To find the test statistic, we need to calculate the test statistic The formula for the test statistic for comparing two proportions is given by:
z = ((p1 - p2) - 0) / √((p1*(1-p1))/n1 + (p2*(1-p2))/n2)
where p1 and p2 are the sample proportions and n1 and n2 are the sample sizes.
Given that the sample sizes for both age groups are 50, we can calculate the test statistic as follows:
p1 = 0.76
p2 = 0.68
n1 = 50
n2 = 50
z = ((0.76 - 0.68) - 0)/√((0.76*(1-0.76))/50 + (0.68*(1-0.68))/50)
Using technology to find the P-value, we can calculate that the P-value is 0.3601.
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