Suppose we want to calculate an 80% CI for a population mean. We are willing to assume the population is normal, but we don't know its variance. We collect 10 values:

12.3, 5.0, 9.8, 4.5, 6.7, 6.1, 10.1, 11.3, 8.8, and 9.0.

What is the upper bound of the 80% CI? (Use the values from the answers above to avoid any differences due to rounding.)

Question

Suppose we want to calculate an 80% CI for a population mean.  We are willing to assume the population is normal, but we don't know its variance.  We collect 10 values:

12.3, 5.0, 9.8, 4.5, 6.7, 6.1, 10.1, 11.3, 8.8, and 9.0.

What is the upper bound of the 80% CI? (Use the values from the answers above to avoid any differences due to rounding.)

BlackTom AI · Accepted Answer

Since the population variance is unknown and n = 10 is small, use a t-based two-sided confidence interval. For an 80% CI, α = 0.20 and α/2 = 0.10, so use t_{0.10, df=9} = 1.383.

Compute the sample mean and sample standard deviation (s) step by step.

Data sum and mean:
12.3 + 5.0 + 9.8 + 4.5 + 6.7 + 6.1 + 10.1 + 11.3 + 8.8 + 9.0 = 83.6
Sample mean x̄ = 83.6 / 10 = 8.36

Deviations and squared deviations (xi − x̄):
3.94^2 = 15.5236
(−3.36)^2 = 11.2896
1.44^2 = 2.0736
(−3.86)^2 = 14.8996
(−1.66)^2 = 2.7556
(−2.26)^2 = 5.1076
1.74^2 = 3.0276
2.94^2 = 8.6436
0.44^2 = 0.1936
0.64^2 = 0.4096

Sum of squared deviations = 63.924
Sample variance s^2 = 63.924 / (n−1) = 63.924 / 9 = 7.102666...
s = sqrt(7.102666...) ≈ 2.665

Standard error SE = s / sqrt(n) = 2.665 / sqrt(10) ≈ 2.665 / 3.1623 ≈ 0.8429

Margin of error = t * SE = 1.383 * 0.8429 ≈ 1.1657

Upper bound = x̄ + margin = 8.36 + 1.1657 ≈ 9.5257 ≈ 9.53

Why this is correct: the t-distribution is required because σ is unknown and n is small; using the correct t critical value and the computed SE produces the margin shown and gives the upper limit 9.53.

Why the other choices are incorrect:
- 7.19 (and 7.13) are near the lower bound: lower bound = 8.36 − 1.1657 ≈ 7.1943 (≈7.19). 7.13 is a rounding/inaccurate variant.
- 9.59 is larger than the correct upper bound; it would result from an incorrect t or arithmetic error.

Final answer confirmed: 9.53.

类似问题

It would also be appropriate to report this result as mean ± 95% CI

provide information about the unknown values of the true regression coefficients, accounting for sampling error.

If 500 confidence intervals, each having a level of confidence of 90%, were computed for the population proportion, approximately how many of the intervals would be expected NOT to contain the population proportion?

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