A survey found 195 of 250 randomly selected Internet users have high-speed Internet access at home. Construct a 90% confidence interval for the proportion of all Internet users who have high-speed Internet access at home. a. (1.19, 1.38)
b. (0.728,0.832)
c. (0.731, 0.829)
d. (0.737, 0.823)

d. (0.737, 0.823)

The 90% confidence interval is = (0.737, 0.823)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 195/250 = 0.78

Number of samples n = 250

Confidence interval = 90%

z value(at 90% confidence) = 1.645

Substituting the values we have;

0.78 +/- 1.645√(0.78(1-0.78)/250)

0.78 +/- 1.645√(0.0006864)

0.78 +/- 1.645(0.026199236630)

0.78 +/- 0.043097744256

0.78 +/- 0.043

(0.737, 0.823)

The 90% confidence interval is = (0.737, 0.823)

question is incomplete.