# 9.2 - Confidence Intervals for a Population Mean

9.2 - Confidence Intervals for a Population Mean## Example 9.3

Over the three-day period from April 1 to April 3, 2015, a national poll surveyed 1500 American households to gauge their levels of discretionary spending. The question asked was how much the respondent spent the day before; not counting the purchase of a home, motor vehicle, or normal household bills. For these sampled households, the average amount spent was \(\bar x\) = \$95 with a standard deviation of *s* = \$185.

How close will the sample average come to the population mean?

Let's follow the same reasoning as developed in section 9.2 for proportions. We have:

\[\text{Sample average} = \text{population mean} + \text{random error}\]

The **Normal Approximation** tells us that the distribution of these random errors over all possible samples follows the normal curve with a standard deviation of \(\frac{\sigma}{\sqrt{n}}\). Notice how the formula for the standard deviation of the average depends on the true population standard deviation \(\sigma\). When the population standard deviation is unknown, like in this example, we can still get a good approximation by plugging in the sample standard deviation (*s*). We call the resulting estimate the Standard Error of the Mean (SEM).

**Standard Error of the Mean** (SEM) = estimated standard deviation of the sample average =

\[\frac{\text{standard deviation of the sample}}{\sqrt{n}} = \frac{s}{\sqrt{n}}\]

In the example, we have *s* = \$185 so the Standard Error of the Mean =

\[\frac{\text{\$185}}{\sqrt{1500}} = \$4.78\]

**Recap:** the estimated daily amount of discretionary spending amongst American households at the beginning of April 2015 was \$95 with a standard error of \$4.78

The Normal Approximation tells us, for example, that for 95% of all large samples, the sample average will be within two SEM of the true population average.

Thus, a 95% confidence interval for the true daily discretionary spending would be \$95 ± 2(\$4.78) or\$95 ± \$9.56.

Of course, other levels of confidence are possible. When the sample size is large, *s* will be a good estimate of \(\sigma\) and you can use multiplier numbers from the normal curve. When the sample size is smaller (say *n* < 30), then s will be fairly different from \(\sigma\) for some samples - and that means that we need a bigger multiplier number to account for that. (see the optional material on "t-multipliers" in chapter 21).

## Confidence Intervals for a population mean (n \(\ge\)) 30

For large random samples, a confidence interval for a population mean is given by

\[\text{sample mean} \pm z^* \frac{s}{\sqrt{n}}\]

where z* is a multiplier number that comes from the normal curve and determines the level of confidence (see Table 9.1 in section 9.2).

## Example 9.4

The equatorial radius of the planet Jupiter is measured 40 times independently by a process that is practically free of bias. These measurements average \(\bar x\) = 71492 kilometers with a standard deviation of s = 28 kilometers. Find a 90% confidence interval for the equatorial radius of Jupiter.

**Note!**Note that the equatorial radius of the planet is a fixed number (Jupiter is not changing in size). But measurements are random quantities that might come out different when repeated independently. If the measurement process is unbiased, then repeating the process many times and taking the average gives a better estimate of the true value.

**Solution:**since

*s*= 28 km, the SEM = \(\frac{28}{\sqrt{40}}=4.4 km\). With

*n*= 40, using the multiplier number from the normal curve for 90% confidence (z*=1.645) will work pretty well so our confidence interval would be: 71492 km ± 1.645(4.4 km) or 71492 km ± 7.3 km

## Example 9.5

How much credit card debt do students typically have when they graduate from Penn State University? A sample of 15 recent Penn State graduates is obtained. Each of these recent graduates is asked to indicate the amount of credit card debt they had at the time of graduation. It turns out that the sample mean was \(\bar x\) = \$2430 with a sample standard deviation of *s* = \$2300. Would it be appropriate to use the method above to find a 99% confidence interval for the average credit card debt for all recent Penn State graduates?

**Solution**: No, with

*n*= 15, using s as an estimate of \(\sigma\) would add quite a bit of extra variability; so it would not be appropriate to use the normal curve multiplier associated with 99% confidence (z* = 2.576). Also, we can tell from the large value of

*s*relative to the sample average that the data here are quite skewed and so the normal curve would not be a good approximation to the sampling distribution regardless.