Question

To build a 95% interval estimate for the population mean textbook expense, from a random sample of 45 IUPUI undergraduate students the following summary measures of expenditure on textbooks were calculated.
	∑x =	21,150
	∑x² =	11,062,900

4	The margin of error for a 95% interval estimate is ________.
A	43.68
B	45.03
C	46.67
D	47.98
5	The 95% interval estimate is ________ , ________.
A	422.02	517.98
B	423.33	516.67
C	424.97	515.03
D	426.32	513.68

6	In the previous question, to obtain a margin of error of ±$20 for a 95% interval estimate the minimum sample size is ________.   For planning value, round the standard deviation obtained above to the nearest integer.
A	282
B	270
C	259
D	246

Answer 1

Solution:

First we need to find sample mean and sample standard deviation s

n = 45

= 21150/45 = 470

Sample variance s² =

= [1/(45 - 1)][11062900- (21150²/45) ]

= 25509.0909091

so

s = 159.715656431

4)

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, d.f = n - 1 = 45 - 1 = 44

    =    =  _0.025,44 = 2.015

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.015 * (159.715656431 / 45)

= 47.98

The margin of error for a 95% interval estimate is

47.98

5)

confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(470 - 47.98)   <   <  (470 + 47.98)

422.02 <   <  517.98

Required 95% confidence interval is

(422.02 , 517.98)

6)

E = 20

Consider , = 159.715656431 = 160

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.025

Using Z table ,

= 1.96

Now, sample size (n) is given by,

=  {(1.96* 160)/ 20 }²

=  245.8624

= 246 ..(round to the next whole number)

Answer : 246