Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.Construct a 95% confidence interval for the population proportion who claim they always buckle up.What is the error bound? (Round to 4 decimal places)

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Asked Nov 16, 2019
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Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.

Construct a 95% confidence interval for the population proportion who claim they always buckle up.

What is the error bound? (Round to 4 decimal places)

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Expert Answer

Step 1

Confidence interval:

The (1-α) % confidence interval for proportion p is given by,

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n pxq p-a p<+ n Here. p

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Step 2

Computing the 99% confidence interval for population proportion:

It was found that, 320 out of 400 drivers claimed that they always buckle up. That is, n=400 and the sample proportion, p-hat = 0.8 (=320/400). The confidence level ...

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(0.8)x (1-0.8) (0.8)x (1-0.8) 0.8-1.96 <p<0.8+1.96. 400 400 0.8-0.0392p<0.8 0.0392 0.761< p<0.839

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