I took a large sample of households in a city, and based on that, estimate the standard deviation of the income for all households in the city is $800. In order to make a desired conclusion about the income for all the households in the city, I want the sample mean for another sample to have a standard deviation of no more than $100. How many households must I have in this new sample?
A. 100
B. 8
C. 64
Please help :)
64
Compute a 99% confidence interval for the mean age of the youngest child in all employees’ households
To determine how many households you need in the new sample, we can use the formula for the standard error of the mean (SE):
SE = standard deviation / √n
where SE represents the standard deviation of the sample mean, standard deviation represents the known standard deviation of the population (which is $800 in this case), and n represents the sample size.
In this scenario, you want the standard deviation of the new sample mean to be no more than $100. Therefore, we can set up the following inequality:
100 ≥ 800 / √n
To solve for n, we can rearrange the equation as follows:
√n ≥ 800 / 100
Simplifying, we get:
√n ≥ 8
To eliminate the square root, we square both sides of the inequality:
n ≥ 8^2
n ≥ 64
Therefore, the minimum number of households you need in the new sample is 64.
So, the answer is option C: 64.