# 8.2.3.1.4 - Example: Transportation Costs

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 $t < -2.113$ 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.

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