Supermercado Central: estimating mean spending from receipts
A regional supermarket chain wants to estimate the mean amount spent per customer. A random sample of 64 receipts from one week gives a sample mean of €38.50 and a sample standard deviation of €12.00. The finance director needs a 95% confidence interval for the population mean and is considering a larger survey that would give a 99% confidence interval with a margin of error of €1.50. A board member claims that because €37 lies inside the 95% interval, the chain can state that the true mean is €37 with 95% probability. The chain also wants to know whether the sample gives evidence that the true mean is below €37, because a promotion is planned if average spending is lower than that.
1. Construct and interpret the 95% confidence interval for the population mean spending. State any assumption needed for the method.
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With n = 64, x̄ = 38.50 and s = 12.00, the estimated standard error is s/√n = 12/8 = 1.50. For 95% confidence the standard normal critical value is z = 1.96, so the margin of error is 1.96 × 1.50 = 2.94. The interval is 38.50 ± 2.94 = (35.56, 41.44) euros. Interpretation: if the chain repeated this sampling procedure many times, about 95% of the intervals produced would contain the true population mean. It is not correct to say that this particular fixed interval has a 95% probability of containing the mean, because the mean is a fixed unknown number, not a random variable. The calculation uses the large-sample normal approximation; with n = 64 this is reasonable unless the population is extremely skewed or the sample is not random.
2. How many receipts would be needed for a 99% confidence interval with a margin of error of €1.50, using the same preliminary standard deviation? Show the rounding decision.
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For 99% confidence, z = 2.576. The required sample size satisfies E = z s/√n, so n = (z s/E)^2 = (2.576 × 12 / 1.50)^2 = (20.608)^2 ≈ 424.7. The sample size must be an integer and the margin must not exceed €1.50, so round up: n = 425 receipts. Rounding down to 424 would give a slightly larger margin than requested. This calculation assumes the preliminary s = 12 is a reliable estimate of the population standard deviation; if the true standard deviation is larger, the achieved margin will be wider than €1.50.
3. Assess the board member's statement that the true mean is €37 with 95% probability. Does the sample provide evidence at the 5% level that the true mean is below €37?
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The statement misinterprets the confidence interval. The true mean μ is fixed; the 95% refers to the long-run success rate of the interval-building procedure, not to a probability attached to μ after the sample is taken. Also, €37 lies inside (35.56, 41.44), so the data are compatible with μ = 37, but the interval also contains many other plausible values. It does not single out €37. For a formal one-sided test of H0: μ = 37 against H1: μ < 37, the test statistic is z = (38.50 - 37)/1.50 = 1.0. The sample mean is above 37, so there is no evidence for μ < 37. The one-sided p-value is about 0.1587, far above 0.05. The chain should not trigger the promotion on this sample alone; if anything, the sample points slightly above €37.