You are getting ready to take a sample from a large population, and need to determine the appropriate sample size. You are estimating sales at a large chain of department stores. A previous survey from the same population found a standard deviation of $18. What sample size is needed to have a 95% confidence of estimating the population mean within plus/minus $5? Show your work.
95% = mean ± 1.96 SEm
1.96 SEm = 5
SEm = SD/√n
To calculate the required sample size, we can use the formula:
n = (Z * σ / E)²
where:
n = sample size
Z = z-score corresponding to the desired confidence level (95% confidence level corresponds to a Z value of 1.96)
σ = standard deviation of the population
E = margin of error (desired interval width)
In this case, we want the estimate of the population mean to be within plus/minus $5, so the margin of error (E) is $5.
Plugging in the given values, we have:
n = (1.96 * 18 / 5)²
Calculating this equation will provide us with the required sample size.
To determine the appropriate sample size for estimating the population mean within a certain margin of error, we can use the formula for the sample size calculation:
n = (Z * σ / E)^2
Where:
n = sample size
Z = Z-score (corresponding to the desired confidence level)
σ = standard deviation of the population
E = margin of error
In this case, we need to estimate the sample size required to estimate the population mean within plus/minus $5 with a 95% confidence level. The Z-score for a 95% confidence level is approximately 1.96.
Plugging the given values into the formula:
n = (1.96 * 18 / 5)^2
Calculating this:
n = (34.56 / 5)^2
n = 6.912^2
n ≈ 47.7
Rounding up to the nearest whole number, we get:
n = 48
Therefore, a sample size of at least 48 would be needed to estimate the population mean within plus/minus $5 with a 95% confidence level, given the previous survey's standard deviation of $18.