Question

1) 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 = 1004 and x = 547 who said​ "yes." Use a 99 % confidence level.

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

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

​c) Construct the confidence interval.

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

A. There is a 99​% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

B. 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.

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

D. One has 99​% confidence that the sample proportion is equal to the population proportion.

2). Assume that we want to construct a confidence interval. Do one of the​ following, as​ appropriate: (a) find the critical value t Subscript alpha divided by 2tα/2​, ​(b) find the critical value z Subscript alpha divided by 2zα/2​, or​ (c) state that neither the normal distribution nor the t distribution applies. The confidence level is 99​%, sigmaσequals=3577 thousand​ dollars, and the histogram of 55 player salaries​ (in thousands of​ dollars) of football players on a team is as shown.

3).

Assume that we want to construct a confidence interval. Do one of the​ following, as​ appropriate: (a) find the critical value t Subscript alpha divided by 2tα/2​, ​(b) find the critical value z Subscript alpha divided by 2zα/2​, or​ (c) state that neither the normal distribution nor the t distribution applies. The confidence level is 90​%, sigmaσ is not​ known, and the histogram of 53 player salaries​ (in thousands of​ dollars) of football players on a team is as shown.

Answer 1

a)

Level of Significance,   α =    0.01          
Number of Items of Interest,   x =   547          
Sample Size,   n =    1004          
                  
Sample Proportion ,    p̂ = x/n =    0.545      

b)

z -value =   Zα/2 =    2.576   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0157          
margin of error , E = Z*SE =    2.576   *   0.0157   =   0.0405

c)
                  
99%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.545   -   0.0405   =   0.5043
Interval Upper Limit = p̂ + E =   0.545   +   0.0405   =   0.5853
                  
99%   confidence interval is (   0.5043   < p <    0.5853   )

d)

B. 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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