Question

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)

Step 1

**Confidence interval:**

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

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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