Question

A poll of 2514 U.S. adults found that 41% regularly used Facebook as a news source.

Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places.

Margin of Error (as a percentage):

Confidence Interval:

Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 95% level of confidence. Round all answers to 2 decimal places.

Margin of Error (as a percentage):

Confidence Interval:

Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 99% level of confidence. Round all answers to 2 decimal places

Margin of Error (as a per

Answer 1

n = number of U.S. adults in a poll = 2514

Sample proportion = = proportion of US adults who regularly used Facebook as a news source = 41% = 0.41

a)

Confidence level = c = 0.90

z critical value for (1 + c)/2 = (1 + 0.90)/2 = 0.95 is

zc = 1.645 (From statistical table of z values, average of 1.64 and 1.65, (1.64+1.65)/2 = 1.645)

Margin of error (E) :

E = 1.645 * 0.0098092

E = 0.0161    (Round to 4 decimal)

E = 1.61%

Margin of error = 1.61%

90% Confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source is

41% - 1.61% < p < 41% + 1.61%

39.39% < p < 42.61%

90% Confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source is

(39.39% , 42.61%)

b)

Confidence level = c = 0.95

z critical value for (1 + c)/2 = (1 + 0.95)/2 = 0.975 is

zc = 1.96 (From statistical table of z values)

Margin of error (E) :

E = 1.96 * 0.0098092

E = 0.0192 (Round to 4 decimal)

E = 1.92%

Margin of error = 1.92%

95% Confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source is

41% - 1.92% < p < 41% + 1.92%

39.08% < p < 42.92%

95% Confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source is

(39.08% , 42.92%).

c)

Confidence level = c = 0.99

z critical value for (1 + c)/2 = (1 + 0.99)/2 = 0.995 is

zc = 2.58 (From statistical table of z values, average of 1.64 and 1.65, (1.64+1.65)/2 = 1.645)

Margin of error (E) :

E = 2.58 * 0.0098092

E = 0.0253 (Round to 4 decimal)

E = 2.53%

Margin of error = 2.53%

99% Confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source is

41% - 2.53% < p < 41% + 2.53%

38.47% < p < 43.53%

99% Confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source is

(38.47% , 43.53%)