Question

Find the necessary sample size.

Weights of women in one age group are normally distributed with a standard deviation of 10.43 kg. A researcher wishes to estimate the mean weight of all women in this age group.

Find how large a sample must be drawn in order to be 95% confident (use z=2) that the sample mean will not differ from the population mean by more than 1.30 kg.

Select one:

a. 15

b. 258

c. 350

d. 175

e. 165

Answer 1

Solution

Let X = weight of women.

We are given: X ~ N(μ, 10.43²).

Back-up Theory

Given X ~ N(μ, σ²), 100(1 - α) % Confidence Interval for μ, with σ known is: Xbar ± {(Z_{α /2})σ/√n}   where

Xbar = sample mean, Z_{α /2} = upper (α /2)% point of N(0, 1), σ = population standard deviation and

n = sample size.

Now to work out the solution,

Going by the above theory, 95% CI for μ would be: Xbar ± {(2 x 10.43)/√n} [Strictly Z_{α /2} = 1.96, but given that it can be taken to be 2.]

By the given condition, {(2 x 10.43)/√n} must be 1.3.

=> √n = {(2 x 10.43)/1.3}

          = 16.0462

Or, n = 257.4791.

Thus, the required sample size is 258 Option b ANSWER