Suppose that several insurance companies conduct a survey Th

Suppose that several insurance companies conduct a survey. They randomly surveyed 350 drivers and found that 280 claimed to always buckle up. We are interested in the population proportion of drivers who claim to always buckle up.

n=

P=

The standard deviation for sP=

The z value for a 95% confidence interval is

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) =

Solution

Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=280
Sample Size(n)=350
Sample proportion = x/n =0.8
Confidence Interval = [ 0.8 ±Z a/2 ( Sqrt ( 0.8*0.2) /350)]
= [ 0.8 - 1.96* Sqrt(0) , 0.8 + 1.96* Sqrt(0) ]
= [ 0.758,0.842]

ANS:
Z value = 1.96
[ 0.758,0.842]