Question

A random sample of 120 observations results in 90 successes. [You may find it useful to reference the z table.]

a. Construct the 90% confidence interval for the population proportion of successes. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)  

Confidence interval _____ to _____

b. Construct the 90% confidence interval for the population proportion of failures. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)

Confidence interval _____ to _____

Answer 1

Solution :

Given that,

a)

n = 120

x = 90

Point estimate = sample proportion = = x / n = 0.75

1 - = 0.25

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z 0.05 = 1.645

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

= 1.645 * (((0.75 * 0.25) / 120)

= 0.0650

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.75 - 0.0650 < p < 0.75 + 0.0650

0.6815 < p < 0.8185

Confidence interval 0.6815 to 0.8185.

b)

n = 120

x = 10

Point estimate = sample proportion = = x / n = 0.083

1 - = 0.917

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z 0.05 = 1.645

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

= 1.645 * (((0.083 * 0.917) / 120)

= 0.0414

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.083 - 0.0414 < p < 0.083 + 0.0414

0.0416 < p < 0.1244

Confidence interval 0.0416 to 0.1244.