Suppose that a random sample of n kernels of popcorn is taken and x of these kernels pop. The buyer wants to be 95% sure that the point estimate for p will be within 0.01.If no preliminary study is done to estimate p, how large a sample should the buyer use? If a preliminary study is done and p hat is 0.86. How large a sample is needed if we use the results of the preliminary study?
To determine the sample size needed in each case, we can use the formula for sample size calculation in estimating a proportion.
1. Case without preliminary study:
When no preliminary study is done, we don't have any estimate of p. In this case, we assume the worst case scenario which is p = 0.5 (maximum variability). The formula to calculate the sample size is:
n = (Z^2 * p * (1-p)) / E^2
Where:
- Z is the Z-score corresponding to the desired confidence level (95% confidence level corresponds to Z = 1.96)
- p is the estimated proportion of kernels that pop (0.5 in this case as we assume maximum variability)
- E is the desired margin of error (0.01)
Plugging in the values, we get:
n = (1.96^2 * 0.5 * (1-0.5)) / 0.01^2
Simplifying the equation, we get:
n = (3.8416 * 0.25) / 0.0001
n = 9,604
Therefore, the buyer should use a sample size of at least 9,604 kernels to be 95% sure that the point estimate for p will be within 0.01.
2. Case with preliminary study:
When a preliminary study is done, we have an estimate of p (denoted as p-hat). In this case, we can use p-hat to reduce the required sample size. The formula to calculate the sample size becomes:
n = (Z^2 * p-hat * (1-p-hat)) / E^2
Plugging in the values, we get:
n = (1.96^2 * 0.86 * (1-0.86)) / 0.01^2
Simplifying the equation, we get:
n = (3.8416 * 0.86 * 0.14) / 0.0001
n = 439.3624 / 0.0001
n ≈ 4,393.62
Therefore, if we use the results of the preliminary study with p-hat = 0.86, the buyer should use a sample size of at least 4,394 kernels to be 95% sure that the point estimate for p will be within 0.01.