Question

Q1

A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll,

n=911 and

x=582 who said "yes." Use a

99% confidence level.

**a)** Find the best point estimate of the
population proportion p.

(Round to three decimal places as needed.)

**b)** Identify the value of the margin of error
E.

(Round to three decimal places as needed.)

**c)** Construct the confidence interval.

less than p less than

<p<

(Round to three decimal places as needed.)

**d)** Write a statement that correctly interprets
the confidence interval. Choose the correct answer below.

**A.**

99

99% of sample proportions will fall between the lower bound and the upper bound.

**B.**

There is a

99

99% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

**C.**

One has

99

99% confidence that the sample proportion is equal to the population proportion.

**D.**

One has

99

99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

Answer #1

a)

Number of Items of Interest,x =582

Sample Size,n = 911

Sample Proportion , p̂ = x/n = 0.639

best point estimate of the population proportion p = 0.639

b)

z -value =Zα/2 = 2.576 [excel formula =NORMSINV(α/2)]

Standard Error , SE = √[p̂(1-p̂)/n] = 0.0159

margin of error , E = Z*SE = 2.576*0.0159=0.041

c)

99%Confidence Interval is

Interval Lower Limit = p̂ - E = 0.639-0.0410=0.5979

Interval Upper Limit = p̂ + E =0.639+0.0410=0.6799

**99% confidence interval is (0.598< p <
0.680)**

d)

One has

99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

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