Grocery-spending survey: confidence intervals and sample size
A consumer-research team in Madurai selects a simple random sample of 49 households from a district list of 1,200 households to estimate average monthly grocery spending. The sample mean is ₹8,400 and the sample standard deviation is ₹1,400. The population standard deviation is not known. The team's report must give a 95% confidence interval for the district mean and, separately, a 95% confidence interval for the district's total monthly grocery spending. A junior analyst has already typed: 'There is a 95% probability that the population mean lies between the two limits,' and the team leader is uneasy about both the wording and the use of z = 1.96 with a sample of 49. The team also needs to plan a follow-up survey in a neighbouring district and wants the 95% margin of error to be no more than ₹250, using ₹1,400 as a planning value for the standard deviation. The team must decide the required sample size and explain the assumptions behind the interval.
1. Construct the 95% confidence interval for the mean monthly grocery spending. Justify the use of t rather than z.
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The population standard deviation σ is unknown, so use the sample standard deviation s = 1400 with the t distribution. With n = 49, df = 48. The 95% two-tailed critical value is t(0.025, 48) ≈ 2.011.
Standard error = s/√n = 1400/√49 = 1400/7 = 200. Margin of error = 2.011 × 200 = 402.2.
Interval: 8400 ± 402.2, i.e. approximately (₹7,997.8, ₹8,802.2).
The t procedure is appropriate because σ is estimated by s. If the population were strongly skewed or had extreme outliers, the interval would need a larger sample or a different method; with n = 49 the central limit theorem gives reasonable protection for a mean, but the team should still inspect the data for outliers.
2. Construct the corresponding 95% confidence interval for the total monthly grocery spending of the 1,200 households. State any assumption needed.
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The district total is T = 1200μ. Multiply each endpoint of the mean interval by 1200:
Lower = 1200 × 7997.8 = 9,597,360. Upper = 1200 × 8802.2 = 10,562,640.
So the 95% interval for the total is approximately ₹95.97 lakh to ₹1.056 crore.
This interval treats the 1,200 households as the population of interest and assumes that the sample came from that list by simple random sampling. The sampling fraction is 49/1200 ≈ 4.1%, so a finite population correction would be small and is reasonably ignored; if the team wants extra precision, it can multiply the standard error by √((1200 − 49)/(1200 − 1)). The interval does not allow for non-response or under-reporting, which are practical risks in a spending survey.
3. Find the sample size needed for a 95% margin of error of at most ₹250. Then criticise the junior analyst's sentence about probability.
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For planning, use the normal critical value z = 1.96 and the planning value σ = 1400:
n = (zσ/E)^2 = (1.96 × 1400 / 250)^2 = (10.976)^2 = 120.47.
Round up, so n = 121 households are needed. If a t critical value were used iteratively, the answer would still round to about 121; the normal value is standard for sample-size planning.
The junior analyst's sentence is not strictly correct. Once the interval (7997.8, 8802.2) has been computed, the population mean is either inside it or not; the 95% refers to the long-run procedure that produces intervals capturing the true mean in 95% of repeated samples. A better sentence is: 'We are 95% confident that the population mean lies between ₹7,997.8 and ₹8,802.2.' The analyst also used z = 1.96 despite unknown σ; t is the correct multiplier for the realised sample, although z is acceptable for sample-size planning.