Question

z-distribution

USING TI-84 plus only

Number of young men.

Suppose that scores of men aged 21 to 25 years on the quantitative part of the national assessment of educational Progress (NAEP) test follow a Normal Distribution with standard devviation 60. You want to estimate the mean score with +/- 10 with 90% confidence. How large an SRS of scores must you choose.

Answer 1

Given that, population standard deviation = 60

margin of error (E) = 10

A 90% confidence level has significance level of 0.10 and critical value is,

We want to find, the sample size (n),

Therefore, required sample size is 98

If you want sample size is to nearest whole number then, n = 97

Note: Using TI-83 plus calculator command : "in norm(0.05)" we get critical value as shown in below image:

Answer 2

SOLUTION :

Let the sample size be n.

SE of sample means distribution = 60/sqrt(n) 

Confidence interval (90%) has following interval of z-values :

= - 1.645 to + 1.645

So,

E = z * SE 

10 = 1.645 * 60/sqrt(n)

=> n = (1.645*60/10)^2 = 97.4169 = 98 

Sample size will be 98 (ANSWER).