Question

A sample of n = 16 is to be taken from a distribution that can reasonably be assumed
to be Normal with a standard deviation σ of 100. The sample mean comes out to be 110.
1. The standard error of the mean, that is, the standard deviation of the sample mean,
is σx¯ = σ/√
n. What is its numerical value?
2. The 97.5 percentile, 1.96, of the standard Normal distribution is used for a 95% confi-
dence interval. The corresponding margin of error is 1.96 σx¯. What is m.e. here?
3. The lower limit of the interval is ¯x − m.e. What is it here ?
4. The upper limit of the interval is ¯x + m.e. What is it here ?
5. What is the 95% confidence interval ?
(lower limit, upper limit ) = ( , )
6. The 95th percentile, 1.645, of the standard Normal distribution is used for a 90%
confidence interval. The corresponding margin of error m.e. is 1.645 σx¯. What is m.e.
here?
7.The lower limit of the interval is ¯x − m.e. What is it here ?
8. The upper limit of the interval is ¯x + m.e. What is it here ?
9. What is the 90% confidence interval ?
(lower limit, upper limit ) = ( , )
10. Compute the sample size n, needed for the margin of error of the 95% CI to be less
than 2.

Answer 1

Given n = 16, = 100 and = 110

1) The Standard error of the mean =

The Numerical value of = 100/Sqrt(16) = 100/4 = 25

2) Margin of error (ME) for the 95% CI = 1.96 * = 1.96 * 25 = 49

3) The lower limit = - ME = 110 - 49 = 61

4) The Upper Limit = + ME = 110 + 49 = 159

5) The 95% CI = (61 , 159)

6) Margin of error (ME) for the 90% CI = 1.645 * = 1.645 * 25 = 41.125

7) The lower limit = - ME = 110 - 41.125 = 68.875

8) The Upper Limit = + ME = 110 + 41.125 = 151.125

9) The 90% CI = (68.875 , 151.125)

10) Given ME = 2, = 100, = 0.05

The Zcritical at = 0.05 is 1.96

The ME is given by :

Squaring both sides we get: (ME)² = (Z critical)² * ²/n

Therefore n = (Zcritical * /ME)² = (1.96*100/2)²

Therefore n = 9604