Question

A manufacturer of MP3 players conducts a set of comprehensive tests on the electrical functions of its product. All MP3 players must pass all fests pnor to being sold of a random sample of 1000 MP3 players, 40 failed one or more tests. Find a 95% confidence interval for the proportion of MP3 players from the population that pass all tests Click here to view Rage of the standard normal distribution tablo Click here to view page 2 of the standard normal distribution table The 95% confidence interval is (Round to three decimal places as needed)

Answer 1

Confidence interval for Population Proportion is given as below:

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

n = 1000

x = 1000 – 40 = 960

P = x/n = 960/1000 = 0.96

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

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

Confidence Interval = 0.96 ± 1.96* sqrt(0.96*(1 – 0.96)/1000)

Confidence Interval = 0.96 ± 1.96* 0.0062

Confidence Interval = 0.96 ± 0.0121

Lower limit = 0.96 - 0.0121= 0.948

Upper limit = 0.96 + 0.0121= 0.972

0.948 < p < 0.972