Assume that previous fast-food research has shown that 80 percent of the consumers like curly French fries. The researcher wishes to have a standard error 6 percent or less and be 95 percent confident of an estimate to be made about curly French fry consumption from a survey. What sample size should be used for a simple random sample?

Question

Assume that previous fast-food research has shown that 80 percent of the consumers like curly French fries. The researcher wishes to have a standard error 6 percent or less and be 95 percent confident of an estimate to be made about curly French fry consumption from a survey. What sample size should be used for a simple random sample?

Answer 1

Formula to find sample size:

n = [(z-value)^2 * p * q]/E^2
... where n = sample size, z-value is found using a z-table for 95% confidence (which is 1.96), p = .80, q = 1 - p, ^2 means squared, * means to multiply, and E = .06.

Plug values into the formula and calculate n.

I hope this will help get you started.

Answer 2

To determine the sample size required for a simple random sample, we can use the formula:

n = (Z * σ / E)²

Where:
n = sample size
Z = Z-score corresponding to the desired confidence level
σ = standard deviation (which we will estimate using the assumed proportion)
E = desired margin of error

In this case, the researcher wants to be 95% confident with a standard error of 6% or less. Therefore, the desired margin of error (E) is 6%.

First, we need to calculate the estimated standard deviation (σ). Since there is no information given about the population standard deviation, we can use the assumed proportion. The assumed proportion is 80%, which corresponds to a probability of success (p) of 0.8.

The formula to calculate the estimated standard deviation for a proportion is:

σ = sqrt(p * (1 - p))

σ = sqrt(0.8 * (1 - 0.8))
= sqrt(0.8 * 0.2)
= sqrt(0.16)
= 0.4

Next, we need to find the Z-score corresponding to the desired confidence level. For a 95% confidence level, the Z-score is approximately 1.96 (which corresponds to the 97.5th percentile of the standard normal distribution). You can find this value in a Z-table or use statistical software.

Finally, we can substitute the values into the formula:

n = (1.96 * 0.4 / 0.06)²
= (0.784 / 0.06)²
= 13.0667²
= 170.6042

Rounding up to the nearest whole number, the required sample size is 171.

Therefore, the researcher should use a sample size of 171 for a simple random sample.