The average total dollar purchase at a convenience store is less than that at a supermarket. Despite smaller purchases, convenience stores can still be profitable because of the size of operation, the volume of business and the mark-up. A researcher is interested in estimating the average purchase amount for convenience stores in suburban Melbourne. To do so, she randomly samples 24 purchases from several convenience stores in suburban Melbourne and tabulates the amounts to the nearest dollar and the sample average is 5.625. Use the following data to construct a 90% confidence interval for the population's average amount of purchases. Assume that the population standard deviation is 3.23 dollars and the population is normally distributed.

( Upper and Lower Limit)

Question

The average total dollar purchase at a convenience store is less than that at a supermarket. Despite smaller purchases, convenience stores can still be profitable because of the size of operation, the volume of business and the mark-up. A researcher is interested in estimating the average purchase amount for convenience stores in suburban Melbourne. To do so, she randomly samples 24 purchases from several convenience stores in suburban Melbourne and tabulates the amounts to the nearest dollar and the sample average is 5.625. Use the following data to construct a 90% confidence interval for the population's average amount of purchases. Assume that the population standard deviation is 3.23 dollars and the population is normally distributed.

( Upper and Lower Limit)

BlackTom AI · Accepted Answer

We are asked to construct a 90% confidence interval for the population mean given: n = 24, sample mean = 5.625, and population standard deviation σ = 3.23 (known), with the population assumed normally distributed.

First, identify the appropriate formula for a known-σ z-interval: 
CI = x̄ ± z_(α/2) * (σ / sqrt(n)).

Compute the standard error: σ / sqrt(n) = 3.23 / sqrt(24).
- sqrt(24) ≈ 4.899
- 3.23 / 4.899 ≈ 0.660 (rounded to three decimals).

Find the critical z-value for a 90% CI. Since it is two-tailed, α = 0.10, so α/2 = 0.05. The z-score corresponding to 0.95 cumulative probability is z_(0.95) ≈ 1.645.

Calculate the margin of error: z_(0.95) * (σ / sqrt(n)) ≈ 1.645 * 0.660 ≈ 1.086 (rounded to three decimals).

Form the confidence interval by adding and subtracting the margin of error from the sample mean:
x̄ ± ME = 5.625 ± 1.086.
- Lower limit = 5.625 − 1.086 ≈ 4.539 → when rounded to two decimals, 4.54; to two decimals this could be reported as 4.54 or, if rounding differently, 4.54–4.55.
- Upper limit = 5.625 + 1.086 ≈ 6.711 → rounded to two decimals gives 6.71 (or 6.70 if following a slightly different rounding convention).

Thus, the 90% confidence interval for the population mean purchase amount is approximately (4.54, 6.71) dollars when rounded to two decimals, noting that small differences in rounding can yield (4.55, 6.70) in some sources. The essential steps were computing the standard error with the known σ, selecting the correct z critical value for a 90% interval, multiplying to obtain the margin of error, and applying it to the sample mean to form the interval.

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