Question

Confidence Interval:

1. A new Pew Research Center survey of 4,594 U.S. adults finds that 78% of the public thinks it likely that social media platforms actively censor political views that those companies find objectionable.

1. What is the value of n?
2. What is the value of pˆp^
3. Compute the margin of error, E, that corresponds to a 95% confidence level.
4. Compute the 95% confidence interval estimate of the population proportion p.
5. Based on the results, can we safely conclude that more than three-quarters of U.S. adults feel it likely that social media platforms actively censor political views that those companies find objectionable?
6. Assuming that you are a newspaper reporter, write a brief statement that accurately describes the results and includes all of the relevant information.

Answer 1

a)

Sample Size,   n =    4594

b)

Sample Proportion ,    p̂ =    0.780

c)

Level of Significance,   α =    0.05
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]  
              
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0061      
margin of error , E = Z*SE =    1.960   *   0.0061   =   0.0120

d)

95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.780   -   0.0120   =   0.7680
Interval Upper Limit = p̂ + E =   0.780   +   0.0120   =   0.7920
                  
so, confidence interval is (   0.7680   < p <    0.7920   )  

e)

Yes, because confidence interval is greater than 0.75

f)

we can be 95% confident that proportion of  U.S. adults feel it likely that social media platforms actively censor political views that those companies find objectionable lie within confidence interval.