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

Explanation:

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

`\bar X=(42.16+57.84)/(2) \\\\=50`

From standard normal tables, we have:

P( -1.96 < Z < 1.96 ) = 0.95

Therefore the margin of error here is computed as:

`MOE=z*(\sigma)/(√(n) )`

`1.96*(\sigma)/(√(n) ) =57.84-50\\\\1.96*(\sigma)/(√(25) ) =57.84-50\\\\ \sigma=(7.84 * 5)/(1.96) \\\\=20`

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

`1.96 * (20)/(√(n) ) =(12)/(2) \\\\1.96 * (20)/(√(n) )=6\\\\n=(1.96*(20)/(6) )^2\\\\=42.7`

≅ 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

`z*(20)/(√(25) ) =6\\\\z=(6 * 5)/(20) \\\\=1.5`

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%