Question

What proportion of JMU students like online courses? To estimate this, a random sample of 200 students was taken and found that 50 of them like online courses. (1) Are the assumptions for constructing a confidence interval for the population proportion satisfied? Explain.(2) Construct a 99% confidence interval for the population proportion of JMU students who like online courses.(3) What is the margin of error in (2) ?______________(4) Interpret the 99% confidence interval.

Answer 1

Solution :

Given that,

n = 200

x = 50

Point estimate = sample proportion = $\hat p$ = x / n = 50/200=0.25

1 - $\hat p$   = 1- 0.25 =0.75

At 99% confidence level the z is ,

$\alpha$  = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z$\alpha$/2 = Z0.005 = 2.576 ( Using z table )

  Margin of error = E = Z$\alpha$ / 2    * ($\sqrt$((($\hat p$ * (1 - $\hat p$ )) / n)

= 2.576* ($\sqrt$((0.25*0.75) / 200)

E = 0.079

A 99% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.25- 0.079< p < 0.25+0.079

0.171< p < 0.329

(0.171 , 0.329)

lower bound 0.171

upper bound 0.329