Question

Use the sample data and confidence level given below to complete parts​ (a) through​ (d).

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

n equals 1091n=1091

and

x equals 510x=510

who said​ "yes." Use a

90 %90%

confidence level.

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Click the icon to view a table of z scores.

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

nothing

​(Round to three decimal places as​ needed.)

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

Eequals=nothing

​(Round to three decimal places as​ needed.)

​c) Construct the confidence interval.

nothingless than p less than<p<nothing

​(Round to three decimal places as​ needed.)

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

A. 90​% of sample proportions will fall between the lower bound and the upper bound.

B.There is a 90% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

C.One has 90% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

D.One has 90​% confidence that the sample proportion is equal to the population proportion.

Answer 1

Solution :

Given that,

n = 1091

x = 510

Point estimate = sample proportion = = x / n =510/1091=0.467

1 -   = 1- 0.467 =0.533

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E Z/2 *(( * (1 - )) / n)

= 1.645 *((0.467*0.533) /1091 )

E = 0.025

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.467-0.025< p <0.467+ 0.025

0.442< p < 0.492

The 90% confidence interval for the population proportion p is : 0.442< p < 0.492