Question

1. Adam wants to know what percentage of YU students want to stay in Canada after Graduation. Previous studies suggest 85%. He has taken a sample of 25 students.
   1. What is the probability that he finds a percent less than 80%?

2. Assuming that he found 90% would like to stay in Canada, construct the 95% confidence interval for the true population’s proportion.

Answer 1

Solution :

a) Given ,

p = 0.85 (population proportion)

1 - p = 0.15

n = 25 (sample size)

Let be the sample proportion.

The sampling distribution of is approximately normal with

mean = =  p = 0.85

SD =   =     

=  \sqrt{0.85(1 - 0.85)/25 }

=  0.07141428428

Find P(less than 0.80)

= P( < 0.80)

= P((\hat p-\mu_{\hat p})/\sigma_{\hat p} <(0.52-\mu_{\hat p})/\sigma_{\hat p} )

=  P(Z <(0.80 - 0.85)/0.07141428428 )

= P(Z < -0.70)

= 0.2420 ...use z table

P( less than 0.52) = 0.2420

b) Given,

n = 25 ....... Sample size

   = 0.90

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

= 1.96

Now , the margin of error is given by

E = *  

= 1.96 * [ 0.90 *(1 - 0.90 )/25]

= 0.1176

Now the confidence interval is given by

( - E)   ( + E)

( 0.90 - 0.1176 )   ( 0.90 + 0.1176 )

0.782   1.018

Required 95% Confidence Interval is ( 0.782 , 1.018 )