Question

1. In a study of habits of undergraduate students, a researcher sampled 53 students and found that the mean number of classes missed over the course of a school year was 18. Assume the population standard deviation, σ = 5.7.
   1. (5 points) Calculate the 95% confidence interval for the mean number of classes missed during the school year by undergraduate students.

2. (2 points) If the researcher had sampled 75 students, would the margin of error be larger or smaller?

3. (2 points) If the researcher used a 90% confidence level instead of 95%, would the margin of error be larger or smaller?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 18

Population standard deviation = =5.7

a) Sample size = n =53

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z0.025 = 1.960

Z/2 = 1.960  

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

= 1.960 * (5.7 / 53 )

= 1.53

At 95 % confidence interval estimate of the population mean is,

- E < < + E

18 - 1.53 <   < 18+ 1.53

16.47 <   < 19.53

( 16.47 ,19.53 )

b)

Sample size = n = 75

Z/2 = 1.960  

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

= 1.960 * (5.7 / 75 )

= 1.29

  If the researcher had sampled 75 students, would the margin of error be smaller.

c)

Point estimate = sample mean = = 18

Population standard deviation = =5.7

Sample size = n =53

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z0.05 = 1.645

Z/2 = 1.645  

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

= 1.645 * (5.7 / 53 )

= 1.29

If the researcher used a 90% confidence level instead of 95%, would the margin of error be smaller.