Question

A survey in the N.Y. Times Almanac finds the mean commute time (one way) is $25.4$ minutes for the $15$largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for $25$randomly selected commuters is $22.1$minutes with a standard deviation of $5.3$minutes. At the $\alpha =$ $0.10$ level, is the Austin, TX commute significantly less than the mean commute time for the $15$largest US cities?

Short answer

There is enough data to infer that commuting times in Austin, TX are much shorter than the average of the largest US cities by$25.4$ minutes.

Step-by-step solution

Given information

$\text{M}=22.1\text{minutes},\text{S}=5.3\text{minutes},\text{n}=25$

Explanation

The null hypothesis asserts that the average commute time in Austin, TX is greater than or equal to the average time for the largest U cities of$25.4$ minutes.

$H_0:\mu \ge 25.4$

According to the alternative hypothesis, the average commute time in Austin, TX is shorter than the $25.4$minutes of average travel time for the largest US cities.

The alternative hypothesis states that the mean commute time in Austin, TX is less than the mean commute time for the largest US cities, $25.4$minutes.

$H_a:\mu <25.4$

The degree of freedom is:

$$\begin{gathered} df=n-1 \\ =25-1 \\ =24 \end{gathered}$$

The test statistic is:

$$\begin{gathered} t=\frac{M-\mu }{S/\sqrt{n}} \\ =\frac{22.1-25.4}{5.3/\sqrt{25}} \\ =\frac{-3.30}{1.06} \\ =-3.11 \end{gathered}$$

The p-value is $0.0024$

Since $p-value<0.10$, reject the null hypothesis. We can conclude that the mean commute time in Austin, TX is less than the mean commute time for the largest US cities, $25.4$minutes.