A sample of n = 278 observations has sample mean 8.83 and sample standard deviation 41.40. Construct a 95% confidence interval for the population mean.

Step-by-step solution

1. 1

   Choose the right interval formula. The population standard deviation $\sigma$ is unknown and has been estimated from the sample, so the correct construction is the $t$-interval

   $\bar x\pm t_{\alpha/2,\,n-1}\cdot\frac{s}{\sqrt n}$

   with $n-1=277$ degrees of freedom. With $n=278$ the sample is large enough that the central limit theorem makes the interval valid even if the underlying data are not normal — which matters here, because $s=41.40$ is nearly five times the mean, a strong hint of a heavily skewed variable.

2. 2

   Compute the standard error of the mean. This measures how much $\bar x$ itself would bounce around from sample to sample:

   $SE=\frac{s}{\sqrt n}=\frac{41.40}{\sqrt{278}}=\frac{41.40}{16.6733}=2.4830$

   The $\sqrt n$ in the denominator is the reason a large sample helps: the raw spread of $41.40$ is cut by a factor of nearly $17$.

3. 3

   Look up the critical value for $95\%$ confidence. For $95\%$ confidence, $\alpha=0.05$ and each tail carries $0.025$. With $df=277$,

   $t_{0.025,\,277}=1.9686$

   This is only a whisker above the normal value $z_{0.025}=1.9600$, because a $t$ distribution with hundreds of degrees of freedom is almost indistinguishable from the standard normal. Using $z$ here is a defensible shortcut, but $t$ is the technically correct choice whenever $\sigma$ is estimated.

4. 4

   Multiply to get the margin of error.

   $ME=t\cdot SE=1.9686\times2.4830=4.8880$

   Round only at the very end; rounding $SE$ to $2.48$ first would shift the endpoints in the third decimal place.

5. 5

   Assemble the interval.

   $\bar x\pm ME=8.83\pm4.888\;\Longrightarrow\;(8.83-4.888,\;8.83+4.888)=(3.942,\;13.718)$

   Reported to two decimals, the $95\%$ confidence interval for $\mu$ is $(3.94,\;13.72)$. For comparison, the $z$-based interval would be $(3.96,\;13.70)$ — the same to within two hundredths, as expected at this sample size.

6. 6

   Interpret it correctly. The statement is about the procedure, not about this one interval: if the sampling were repeated many times, $95\%$ of the intervals built this way would contain the true mean $\mu$. It is not correct to say that $95\%$ of the individual data values fall in $(3.94,13.72)$ — the data spread is governed by $s=41.40$, roughly seventeen times wider than this interval, which describes only the uncertainty in the mean.