Suppose that insurance companies did a survey. They randomly surveyed 430 drivers and found that 330 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.
a) x = ? b) n = ? c) p' = ? d)
Which distribution should you use for this problem? (Round your answer to four decimal places.)
P- (_) (_,_)
e) Construct a 95% confidence interval for the population proportion who claim they always buckle up.
i. state the confidence interval
ii. sketch graph
iii. calculate error bound

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Qwtiger Qwtiger · Answer 1 · 2022-12-04T12:05:48+01:00

The 95% confidence interval for the population proportion who claim they always buckle up is 0.0204.

a) x: 330

b) n: Number of drivers who are surveyed = 430

c) p': proportion of drivers in a sample who claimed to always buckle up = 0.7674

d) normal distribution should be used for this problem.

e) 95% confidence interval for the population proportion who claim they always buckle up.

The error margin is:

1.96 × sqrt(p(1-p) ÷ n)

1.96 × sqrt(0.7674 × 0.2326 ÷ 430) = 0.0204

(0.7674 - 0.0204,0.7674 + 0.0204) = (0.747, 0.7878)

The confidence interval is (0.747, 0.7878).

Learn more about the normal distribution at

https://brainly.com/question/29509087?referrer=searchResults

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