Question

A​ university's administrator proposes to do an analysis of the proportion of graduates who have not found employment in their major field one year after graduation. In previous​ years, the percentage averaged​ 15%. He wants the margin of error to be within​ 5% at a​ 99% confidence level. What sample size will​ suffice? Round to the nearest integer.

Answer 1

Answer:

339

Step-by-step explanation:

We have that the formula would come being:

n = p (1-p) * [z (α / 2) / B] ^ 2

let p, percentage averaged (0.15)

B = 5%, 0.05 (bottom margin)

α = 1 - A

A, top margin 99%

α = 1 - 0.99 = 0.01

α / 2 = 0.005

z (0.005) = 2.58 (attached normal distribution table), the value is negative in the table but is taken as positive.

Replacing we have:

n = 0.15 * (1 - 0.15) * [2.58 / 0.05] ^ 2

n = 0.1275 * (2.58 / 0.05) ^ 2

n = 339.4764

Therefore the sample size is 339.

Answer 2

Answer: 400

Step-by-step explanation:

From the question, we see that population proportion (p) = 0.15

Margin of error (m) = 5%

Confidence level = 99%

Level of significance = 100- confidence level = 1%.

The population standard deviation (s) for this data set is

√p(1-p) = √ 0.15 × (1 - 0.15)

√ 0.15 × 0.85 = 0. 1275

Margin of error = critical value × population standard deviation / √n

The critical value (c) for Constructing a 99 % confidence interval for population mean is = 2.58

m = c× √p'(1-p') /√n

We want to make n subject of the formulae, we have that

m = c√p'(1-p') /√n

m×√n = c × √p'(1-p')

√n = c × √p'(1-p') / m

n = c² × {√p'(1-p')}² / m

n = 2.58² × 0.1275/ 0.05²

n = 6.6564 × 0.01625625 / 0.0025

n = 339.47 ~ 400