Question

10. A survey of 145 randomly selected students at one college showed that only 87 checked their campus email account on a regular basis. Construct and interpret a 90% confidence interval for the percentage of students at that college who do not check their email account on a regular basis. (1pt)

Point Estimate: ______

Margin of Error: ________

Confidence Interval:_________________

Interpretation: ____

Answer 1

Solution :

Given that,

n = 145

x = 87

Point estimate = sample proportion = = x / n = 87/145=0.6

1 -   = 1- 0.6 =0.4

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.6*0.4) /145 )

= 0.0669

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.6- 0.0669< p <0.6+ 0.0669

0.5331< p < 0.6669

The 90% confidence interval for the population proportion p is : 0.5331, 0.6669