SOLUTION: In preparing a report on the economy, we need to estimate the percentage of businesses that plan to hire additional employees in the next 60 days.
a) How many randomly selec
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Question 1197972: In preparing a report on the economy, we need to estimate the percentage of businesses that plan to hire additional employees in the next 60 days.
a) How many randomly selected employers must we contact in order to create an estimate in which we are 99% confident with a margin of error of 6%?
b) Suppose we want to reduce the margin of error to 4%. What sample size will suffice?
c) Why might it not be worth the effort to try to get an interval with a margin of error of 1%?
a) A sample size of ? is needed.
Part 2
b) A sample size of ? is needed.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
Part (a)
At 99% confidence, the z critical value is roughly z = 2.576
Use a table like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to get that value. Look at the bottom row labeled "Z" and above the 99% confidence level.
You can also use a stats calculator or spreadsheet to determine this z critical value.
The desired margin of error is 6%, which means we want E = 0.06 or smaller.
We're not told the value of phat, which is the sample proportion of businesses that plan to hire additional employees in the next 60 days.
Use phat = 0.5 as a conservative estimate. This is the default value of phat if none is stated.
n = min sample size
n = phat*(1-phat)*(z/E)^2
n = 0.5*(1-0.5)*(2.576/0.06)^2
n = 460.817778 approximately
n = 461 always round UP to the nearest whole number
Here's why we round up to the nearest whole number.
Let's try n = 460 in the margin of error formula
E = z*sqrt(phat*(1-phat)/n)
E = 2.576*sqrt(0.5*(1-0.5)/460)
E = 0.060053
We're slightly over the 6% target.
Now try n = 461
E = z*sqrt(phat*(1-phat)/n)
E = 2.576*sqrt(0.5*(1-0.5)/461)
E = 0.059988
Now the error is either 6% or less, which meets the goal we're after.
This is why we round up to the nearest whole number for min sample size problems. This is to clear the hurdle needed.
Answer: 461 employers
===============================================================
Part (b)
Repeat the same set of steps as part (a), but this time we use E = 0.04
Keep everything else the same.
n = phat*(1-phat)*(z/E)^2
n = 0.5*(1-0.5)*(2.576/0.04)^2
n = 1036.84
n = 1037
Answer: 1037 employers
===============================================================
Part (c)
Now use E = 0.01
n = phat*(1-phat)*(z/E)^2
n = 0.5*(1-0.5)*(2.576/0.01)^2
n = 16589.44
n = 16590
We need to sample a lot more employers at this point (more than ten times as much compared to the result of part (b)), so it's more practical to go with the 6% or 4% margin of error instead.
Answer: Minimum sample size gets way too large
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