Question

QUESTION 2

1. Fifty students are enrolled in an Economics class. After the first examination, a random sample of 5 papers

   was selected. The grades were 65, 73, 80, 72, and 95

   a.	Calculate the estimate of the standard error of the mean.
   b.	What assumption must be made before we can determine an interval for the mean grade of all the students in the class? Explain why.
   c.	Assume the assumption of Part b is met. Provide a 90% confidence interval for the mean grade of all the students in the class.
   d.	If there were 200 students in the class, what would be the 90% confidence interval for the mean grade of all the students in the class?

Answer 1

Answer:

Given,

65 , 73 , 80 , 72 , 95

Mean = x/n

= (65+73+80+72+95)/5

= 77

Standard deviation = sqrt((x-xbar)^2/(n-1))

On solving we get

= 11.38

a)

standard error of the mean = s/sqrt(n)

= 11.38/sqrt(5)

= 5.089

b)

Here the sample is small i.e., (n < 30) and σ is estimated from s.

So that we must assume the distribution of all the grades is normal.

c)

Here degree of freedom = n-1= 5-1=4

t(alpha/2,df) = t(0.05,4) = 2.13

90% CI = xbar +/- t*s/sqrt(n)

= 77 +/- 2.13*5.089

= 77 +/- 10.84

= (66.16 , 87.84)

d)

Here degree of freedom = n-1= 200-1=199

t(alpha/2,df) = t(0.05,199) = 1.653

90% CI = xbar +/- t*s/sqrt(n)

= 77 +/- 1.653*11.38/sqrt(200)

= 77 +/- 1.33

= (75.67 , 78.33)