Stats: Confidence Interval and Upper Limit > A random sample of n = 18 E-glass fibre test specimens of a certain type yielded a sample average interfacial heard yield of 40 and a standard deviation of 4. Assume the interfacial shear yield stress is normally distributed. Compute the **upper limit** of a 95% confidence interval for the true mean stress. Attempt: $\bar x = 40, n = 18, s=4$. Upper limit : $$ \bar x + t_{\alpha/2,df=17}\frac{4}{\sqrt{18}} = 40 + 2.110 \times \frac{4}{\sqrt{18}} \approx 41.98932708. $$ However, the book uses $2.262$ for the $t_{\alpha/2}$ value, which seems to have df of $9$ rather than $17$ that I've used. Did I make a mistake? Table found @ : <

Seems to me you're right---for the upper end of a two-sided CI. I would round the answer to 41.99.

The wording of the problem (as edited) seems clear, but I wonder why you are asked to compute only the upper limit of a two-sided CI. If this were the _upper bound for a one-sided_ 95% CI, you would put the full 5% in the upper tail of t(df=17) and use 1.740 instead of 2.110.

Below is the printout from Minitab statistical software, which gives both the lower and upper limits of a two-sided 95% CI.


MTB > Onet 18 40 4.

One-Sample T

N Mean StDev SE Mean 95% CI
18 40.000 4.000 0.943 (38.011, 41.989)