Question

A random sample of n = 25 observations is taken from a N(µ, σ ) population. A 95% confidence interval for µ was calculated to be (42.16, 57.84). The researcher feels that this interval is too wide. You want to reduce the interval to a width at most 12 units.

a) For a confidence level of 95%, calculate the smallest sample size needed.

b) For a sample size fixed at n = 25, calculate the largest confidence level 100(1 − α)% needed.

Answer 1

a) The sample mean is computed as the mid point of the given confidence interval. It is computed as:

42.1657.84 X -50 2

From standard normal tables, we have:

P( -1.96 < Z < 1.96 ) = 0.95

Therefore the margin of error here is computed as:

MOE 2 n

1.96 57.84 50 n

57.84 50 1.96 25

7.84 5 20 1.96

Now for confidence interval width as 12, and above standard deviation the minimum sample size is computed as:

20 1.96 /n 12 2

20 6 /n 1.96

20 n (1.96 =42.7 6

Therefore 43 is the minimum sample size required here.

b) Here for n = 25, we need to find the critical z value first. It is computed as

20 6 V25

6 5 - 1.5 20

We now have to find the probability now:

P( -1.5 < Z < 1.5)

= 2*P(0 < Z < 1.5)

From standard normal tables, we have:

P(Z < 1.5) = 0.9332

Therefore P( 0 < Z < 1.5) = 0.9332 - 0.5 = 0.4332

Therefore the required probability here is:

= 2*P(0 < Z < 1.5) = 2*0.4332 = 0.8664

Therefore the largest confidence interval here is given as 86.64%