Question

in a random sample of 200 people 153 said that they watched educational television find the 99% confidence interval of the true proportion of people who watched educational television round intermediate answers to at least 5 decimal places and final answer to 3

Answer 1

Solution:

Confidence interval for Population Proportion

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

X = 153

n = 200

P = X/n = 153/200 = 0.765

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.765 ± 2.5758* sqrt(0.765*(1 – 0.765)/200)

Confidence Interval = 0.765 ± 2.5758* 0.029981

Confidence Interval = 0.765 ± 0.077225

Lower limit = 0.765 - 0.077225 = 0.688

Upper limit = 0.765 + 0.077225 = 0.842

Confidence interval = (0.688, 0.842)