Question

A 95 percent confidence interval for the mean of a population is to be constructedand must be accurate within 0.3 unit. A preliminary sample standard deviation is2.9. The smallest sample size n that provides the desired accuracy with 95 percentconfidence is:________.A. 400 B. 359 C. 389 D. 253

Answer 1

Answer:

B. 359

Step-by-step explanation:

Given that:

The margin of error (e) must be within 0.3 unit, The standard deviation (σ) = 2.9 and the the confidence level = 95% = 0.95

α = 1 =0.95 = 0.05

$\frac{\alpha}{2}=\frac{0.05}{2}=0.025$

The z score of 0.025 corresponds to the z score of 0.0475 (0.5 - 0.025) which can be gotten from the normal probability table.

Therefore: $z_\frac{\alpha}{2}= z_{0.025}=1.96$

The formula for the margin of error is given by:

$e=z_{0.025}\frac{\sigma}{\sqrt{n} } \\Substituting:\\0.3=1.96*\frac{2.9}{\sqrt{n} } \\\sqrt{n}=\frac{1.96*2.9}{0.3}\\\sqrt{n} =18.95\\ n=18.95^2=359$

Therefore the smallest sample size (n) is 359