Please read the question below!

Question

In a random sample of 7 ​people, the mean driving distance to work was 25.9 miles and the standard deviation was 5.3 miles. Assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a ​90% confidence interval for the population mean u . Interpret the results.

Identify the margin of error.

Answer 1

GOAL: We want to construct a 90% confidence interval for the population mean of the driving distance. Let μ = the true mean driving distance.

CHECKING CONDITIONS:

1. Randomness - The problem states that the data was collected using a random sample of people.
2. Normality (Robustness) - The problem states that the population is normally distributed.
3. Independence - It is reasonable to assume that each person's driving distance to work is independent of everyone else's and that the sample size of n = 7 is less than 10% of the population of all people driving to work. That is, n = 7 ≤ 0.10N, so each sample is independent of one another.

CALCULATIONS (manually solving it):

1. margin of error: critical value × standard error ≈ t* × s/√(n)
2. t* (aka Critical Value) — the number of standard deviations away you must be for a given confidence interval level
3. to find t* using a calculator: invT((confidence level + 1)/2) → degrees of freedom (df) = n-1
4. 1.943 × 5.3/√(7) = 3.892
5. interval = statistic ± margin of error
6. = statistic ± (critical value × standard error)
7. = x̄ ± t* × s/√(n) → x̄ - t* × s/√(n) < μ < x̄ + t* × s/√(n)
8. 25.9 ± 3.8926 → 25.9 - 3.8926 < μ < 25.9 + 3.8926 → 22.0074 < μ < 29.7926

(you can also use your calculator to solve this: go to STATS → TESTS → TINTERVAL and plug in your values)

My final answer was 22.0074 < μ < 29.7926, meaning that we can be 90% confident that the true mean driving distance to work lies between 22.0074 miles and 29.7926 miles.