Question

In: Statistics and Probability

Suppose that insurance companies did a survey. They randomly surveyed 440 drivers and found that 330...

Suppose that insurance companies did a survey. They randomly surveyed 440 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.

NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

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

(i) State the confidence interval. (Round your answers to four decimal places.)

  ,  



(ii) Sketch the graph.

(iii) Calculate the error bound. (Round your answer to four decimal places.)

Solutions

Expert Solution

Solution:

(i) State the confidence interval.

Answer: The formula for finding the 95% confidence interval is:

Where:

is the sample proportion of drivers who always buckle up.

is the critical value at 0.05 significance level

is the sample size

Therefore, the confidence interval is:

Therefore, the 95% confidence interval estimate for the population proportion who claim they always buckle up is

(ii) Sketch the graph.

(iii) Calculate the error bound.

The error bound is:

orchestra answered 2 years ago
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