Question

In the following problem:
1. Determine the value of the margin of error ( E ) 2. Construct the confidence interval. 3. Write a statement that correctly interprets the confidence interval.
From a Prince Market Research poll in which randomly chosen people were asked if they felt that U.S. nuclear weapons made them feel safer the following data was gathered. Construct a 95% confidence interval for the proportion of all people living in the United States who felt nuclear weapons made them feel safer.
n = 1800 x = 1543 who said “yes”.   

2. A Consumer Reports Research Center randomly surveyed 1000 women to determine what proportion of women purchase books online. Out of 1000 women, 788 reported that they purchase books online.
a. Construct a 90% confidence interval for the true proportion of all women that purchase books online. Write a sentence that correctly interprets the confidence interval.

b. Can we safely conclude that at least 25% of all women purchase books online? Why or why not?

Answer 1

1)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   1543          
Sample Size,   n =    1800          
                  
Sample Proportion ,    p̂ = x/n =    0.857          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0082          
margin of error , E = Z*SE =    1.960   *   0.0082   =   0.0162
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95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.857   -   0.0162   =   0.8411
Interval Upper Limit = p̂ + E =   0.857   +   0.0162   =   0.8734
                  
95%   confidence interval is (   0.841   < p <    0.873   )
------

we are 95% confident that true  proportion of all people living in the United States who felt nuclear weapons made them feel safer lies within confidence interval.

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2)

a)

Level of Significance,   α =    0.10          
Number of Items of Interest,   x =   788          
Sample Size,   n =    1000          
                  
Sample Proportion ,    p̂ = x/n =    0.788          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0129          
margin of error , E = Z*SE =    1.645   *   0.0129   =   0.0213
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.788   -   0.0213   =   0.7667
Interval Upper Limit = p̂ + E =   0.788   +   0.0213   =   0.8093
                  
90%   confidence interval is (   0.767   < p <    0.809   )

b)

Ho :   p ≥ 0.25
H1 :   p <   0.25

since, confidence interval do not contain null hypothesis 0.25 so, reject the null hypothesis

so, we can conclude that percentage of of all women purchase books online is less than 25%