Question

In: Statistics and Probability

Suppose that several insurance companies conduct a survey. They randomly surveyed 300 drivers and found that...

Suppose that several insurance companies conduct a survey. They randomly surveyed 300 drivers and found that 240 claimed to always buckle up. We are interested in the population proportion of drivers who claim to always buckle up.



(.20)  n=   

(.20)  p′=   

(.20) The standard deviation for   p=   

(.20) The   z   value for a   95%   confidence interval is   


(.20) Construct a   95%   confidence interval for the population proportion that claim to always buckle Fill in the blanks to clarify the following diagram.

LL   (lower limit)   =       UL   (upper limit)   =   

limit) =

Solutions

Expert Solution

Suppose,

x: Number of drivers who claim to always buckle up = 240

n: Sample size = 400

The standard deviation for p is,

For a 95% confidence interval,

The z value is,

or

The 95% confidence interval for the population proportion that claims to always buckle up is

orchestra answered 2 years ago
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