10. Let p equal the proportion of drivers who use a seat belt in a state that does not have a mandatory seat belt law. It was claimed that p = 0.14. An advertising campaign was conducted to increase this proportion. Two months after the campaign, 104 out of a random sample of 590 drivers were wearing seat belts. What is the test-statistic value in the given problem?*A. z = 0.176B. z = 2.80C. z = 2.52D. z = 14.56
Question
10. Let p equal the proportion of drivers who use a seat belt in a state that does not have a mandatory seat belt law. It was claimed that p = 0.14. An advertising campaign was conducted to increase this proportion. Two months after the campaign, 104 out of a random sample of 590 drivers were wearing seat belts. What is the test-statistic value in the given problem?*A. z = 0.176B. z = 2.80C. z = 2.52D. z = 14.56
Solution
To calculate the test-statistic value, we will use the formula for the z-score:
z = (p̂ - p) / √[(p(1 - p)) / n]
where:
- p̂ is the sample proportion,
- p is the hypothesized proportion, and
- n is the sample size.
Given in the problem, we have:
- p̂ = 104 / 590 = 0.176271186,
- p = 0.14, and
- n = 590.
Substituting these values into the formula, we get:
z = (0.176271186 - 0.14) / √[(0.14(1 - 0.14)) / 590] z = 0.036271186 / √[0.1204 / 590] z = 0.036271186 / 0.014362420 z = 2.52
So, the test-statistic value in the given problem is 2.52, which corresponds to option C.
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