Question

1. A popular professor wants to construct a 90% confidence interval with a margin of error = 0.03 on the proportion of students that make a thorough “cheat sheet” for her statistics tests.
   1. (4 points) Through experience, the professor estimates that about 42% of students make thorough “cheat sheets.” Using this estimate for the sample proportion, how many students should the professor sample?

2. (4 points) How many students should the professor sample if no estimate is available?

Answer 1

Solution :

Given that,

= 0.42

1 - = 1 - 0.42 = 0.58

margin of error = E =0.03

At 90% confidence level z

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.645 / 0.03)2 * 0.42 * 0.58

=732.43

Sample size = 732

( b)

Solution :

Given that,

= 0.5( use 0.5)

1 - = 1 - 0.5 = 0.5

margin of error = E =0.03

At 90% confidence level z

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.645 / 0.03)2 * 0.5 * 0.5

=751.67

Sample size = 752