8.2.3.1.4 - Example: Transportation Costs

According to CNN, in 2011, the average American spent \$16,803 on housing. A suburban community wants to know if their residents spent less than this national average. In a survey of 30 randomly selected residents, they found that they spent an annual average of \$15,800 with a standard deviation of \$2,600.

1. Check assumptions and write hypotheses

Housing costs are quantitative. Because \(n \ge 30\), the sampling distribution can be approximated using the \(t\) distribution.  

This is a left-tailed test because we want to know if residents of this community spent less than the national average.

\(H_{0}:\mu=16803\)
\(H_{a}:\mu<16803\)

2. Calculate the test statistic
Test Statistic: One Group Mean

\(t=\frac{\overline{x}-\mu_0}{\frac{s}{\sqrt{n}}}\)

\(\overline{x}\) = sample mean
\(\mu_{0}\) = hypothesized population mean
\(s\) = sample standard deviation
\(n\) = sample size

\(t=\frac{\overline{x}-\mu_0}{\frac{s}{\sqrt{n}}}=\frac{15800-16803}{\frac{2600}{\sqrt{30}}}=-2.113\)

Our t test statistics is -2.113

3. Determine the p-value

\(df=n-1=30-1=29\)

This is a left-tailed test so we want to know the probability of \(t < -2.113\)

Distribution Plot of Density vs X - T, DF=29

Using Minitab Express we can find that \(p=0.0216634\)

4. Make a decision

\(p\leq .05\), therefore we reject the null hypothesis.

5. State a "real world" conclusion

There is evidence to state that on average residents of this community spent less than the national average on housing in 2011.