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.

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.

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step-by-step explanation:

sorry no step by step

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer