Question

Editors preparing a report on the economy are trying to estimate the percentage of businesses that plan to hire

Answer 1

The number of randomly selected employers that would have to be contacted is 9604

How to solve for the value of n

The confidence level is 95%

1 - 0.95 = 0.05

We have to get the z value at 0.05

0.05/2 = 0.025

Using the standard normal table, the critical value at 0.025 = 1.96

We have to define the probability as P = 50% = 0.5

we have to define q = 1 - 50% = 0.5

The formula that would be used to find the value of n is given as

$n = pq\frac{Zc}{E} ^2$

These variables used are:

pq = variance of population

E = margin of error = 0.01

critical value = Zc = 1.96

$n = 0.5*0.5 (\frac{1.96}{0.01})^2$

= 0.25 * 38416

= 9604

Hence the sample space is given as 9604

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