First, it's important to note the question text mentions the data came from a normally distributed population; otherwise, you can't assume a normal sampling distribution because of your low sample size (n=11) - see Central Limit Theorem for more information.
There are three pieces of information you need to construct a confidence interval:
- Estimate → in this case, your x̄ (mean), which is given as 37
- Critical Value → Zcritical. Note the population standard deviation (σ) is not known and we are told that the population distribution is normal (this information is important because the sample size is not large enough to assume a normal sampling distribution without it), so we would use the t-distribution and tcritical. To find tcritical we need to calculate degrees of freedom (df), which is (n-1) = (11-1) = 10. From a t-table, using the df, find the tcritical. With 10 df, tcritical is 1.812 (look for two-tailed prob. of 0.10 for a 90% confidence interval)
- Standard Error → (σ/√n). Recall the population standard deviation (σ) is not known, so we estimate it with the sample standard deviation, s, which is given as 14. The formula becomes (s/√n)
Now, we have everything we need to calculate a 90% Confidence Interval:
x̄ +/- tcritical * (s/√n) → 37 +/- 1.812 * (14/√11) = 37 +/- 7.649.
Rounding down to one decimal place, as required in the question, our 90% Confidence Interval is (29.4,44.6)
