Question

A random sample of 200 books purchased at a local bookstore showed that 72 of the books were murder mysteries. Let p be the true proportion of books sold by this store that is murder mystery. Construct a confidence in terval with a 95% degree of confidence.Compute the following:

a.Point estimate

b.Critical value

c.Margin of error

d.Confidence interval

e.Interpretation(confidence statement).

Answer 1

Solution :

Given that,

n = 20

x = 72

Point estimate = = x / n = 72 / 200 = 0.360

1 - = 1 - 0.360 = 0.640

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.360 * 0.640) / 200)

= 0.067

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.360 - 0.067 < p < 0.360 + 0.067

0.293 < p < 0.427

The 95% confidence interval for the population proportion p is : 0.293 to 0.427