In: Statistics and Probability
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.
a) The sample mean is computed as the mid point of the given confidence interval. It is computed as:
From standard normal tables, we have:
P( -1.96 < Z < 1.96 ) = 0.95
Therefore the margin of error here is computed as:
Now for confidence interval width as 12, and above standard deviation the minimum sample size is computed as:
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
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%