Cathy Q. answered • 08/04/21

College Freshman with Perfect on Math SAT

**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:**

- Randomness - The problem states that the data was collected using a random sample of people.
- Normality (Robustness) - The problem states that the population is normally distributed.
- 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):**

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