Many states are carefully considering steps that would help them collect sales taxes on items purchased through the Internet. How many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the Internet? Assume that we want to be 90% confident that the sample percentage is within four percentage points of the true population percentage for all sales transaction.
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To determine the sample size (n) required to estimate the percentage of sales transactions that occurred over the internet with a 90% confidence level and a 4-percentage point margin of error, we can use the following formula:
n = (Z^2 * p * (1-p)) / E^2
Where:
n = sample size
Z = Z-score corresponding to the desired confidence level (90%) -> Z = 1.645 for a 90% confidence level
p = estimated proportion of sales transactions over the internet (we'll assume 50% for a conservative estimate)
E = margin of error as a decimal (4% -> 0.04)
Plugging in the values:
n = (1.645^2 * 0.5 * (1-0.5)) / 0.04^2
n = (2.705 * 0.5 * 0.5) / 0.0016
n = 0.67625 / 0.0016
n ≈ 422.65625
We would round up the sample size to the nearest whole number because you cannot have a fraction of a transaction surveyed.
Therefore, approximately 423 randomly selected sales transactions should be surveyed to determine the percentage of transactions that occurred over the internet with a 90% confidence level and a 4-percentage point margin of error.
To determine the sample size required to estimate the percentage of sales transactions that transpired over the Internet with a 90% confidence level and a margin of error of four percentage points, you can use the following formula:
n = (Z^2 * p * (1-p)) / E^2
Where:
n = sample size
Z = Z-score corresponding to the desired confidence level (in this case, 90% confidence level corresponds to a Z-score of 1.645)
p = estimated proportion (if unknown, you can assume 0.5 for a conservative estimate, which yields the largest necessary sample size)
E = margin of error (in this case, four percentage points, which is 0.04)
Plugging in the values into the formula, we get:
n = (1.645^2 * 0.5 * (1-0.5)) / 0.04^2
Simplifying the equation:
n = (2.706 * 0.5 * 0.5) / 0.0016
n = 0.6765 / 0.0016
n ≈ 422
Therefore, based on these assumptions, you would need to survey approximately 422 randomly selected sales transactions to determine the percentage of transactions that transpired over the Internet with a 90% confidence level and a margin of error of four percentage points. Keep in mind that this number may vary if you have more accurate estimates of the proportion or employ other sampling techniques.