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.

Step-by-step explanation:

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:

≅ 43

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%

answer: a and b

step-by-step explanation:

true

step-by-step explanation: