Question

Suppose that insurance companies did a survey. They randomly surveyed 450 drivers and found that 340...

Suppose that insurance companies did a survey. They randomly surveyed 450 drivers and found that 340 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.)

A. (i) Enter an exact number as an integer, fraction, or decimal.

x =

(ii) Enter an exact number as an integer, fraction, or decimal.

n =

(iii) Round your answer to four decimal places.

p' =

(rounded to four decimal places)

B.

Which distribution should you use for this problem? (Round your answer to four decimal places.)

P' ~___ (________,_________)

C. 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.

a/2=

C.L.=

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

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