A food store owner wants to determine the type of products customers prefer to buy. The owner surveyed 40 customers on a day when the store had 335 customers, and found that 40% of those surveyed preferred to buy organic products over non-organic products. Assuming a 95% confidence level, which statement holds true? A. As the sample size is too small, the margin of error is 0.127. B. As the sample size is too small, the margin of error cannot be trusted. C. As the sample size is appropriately large, the margin of error is 0.127. D. As the sample size is appropriately large, the margin of error is 0.152.
Question
A food store owner wants to determine the type of products customers prefer to buy. The owner surveyed 40 customers on a day when the store had 335 customers, and found that 40% of those surveyed preferred to buy organic products over non-organic products. Assuming a 95% confidence level, which statement holds true? A. As the sample size is too small, the margin of error is 0.127. B. As the sample size is too small, the margin of error cannot be trusted. C. As the sample size is appropriately large, the margin of error is 0.127. D. As the sample size is appropriately large, the margin of error is 0.152.
Solution
To answer this question, we first need to understand the concept of margin of error and confidence level in statistics.
The margin of error is a statistic expressing the amount of random sampling error in a survey's results. It asserts a likelihood (not a certainty) that the result from a sample is close to the number one would get if the whole population had been queried.
The confidence level tells you how sure you can be about your result. It is expressed as a percentage and represents how often the true percentage of the population who would pick an answer lies within the confidence interval.
In this case, the sample size is 40 out of a population of 335. This is a relatively small sample size, which can lead to a larger margin of error.
The margin of error can be calculated using the formula:
Margin of Error = Z * sqrt[(p(1-p))/n]
where: Z is the z-score, which is a measure of how many standard deviations an element is from the mean. For a 95% confidence level, the z-score is approximately 1.96. p is the sample proportion, in this case, 40% or 0.4. n is the sample size, in this case, 40.
Substituting these values into the formula, we get:
Margin of Error = 1.96 * sqrt[(0.4(1-0.4))/40] = 0.152
So, the correct answer is D. As the sample size is appropriately large, the margin of error is 0.152.
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