Question

Results of a survey conducted in 15 large cities within the U.S. finds that the average commute time one way to/from work is 25.4 minutes. An executive of chamber of commerce believes that the commute time in her city is less than the amount reported by the survey. She randomly selects 25 commuters from her city and finds the average is 22.1 minutes and the standard deviation is 5.3 minutes. At ???? = 0.10, is her claim supported?

Answer 1

Answer:

We conclude that commute time in her city is less than the amount reported by the survey which is 25.4 minutes.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 25.4 minutes

Sample mean, $\bar{x}$ = 22.1 minutes

Sample size, n = 15

Alpha, α = 0.10

Sample standard deviation, s = 5.3 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 25.4\text{ minutes}\\H_A: \mu < 25.4\text{ minutes}$

We use one-tailed t(left) test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{22.1 - 25.4}{\frac{5.3}{\sqrt{15}} } = -2.411$

Now, $t_{critical} \text{ at 0.10 level of significance, 14 degree of freedom } = -1.345$

Since,                    

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, there is enough evidence to conclude that commute time in her city is less than the amount reported by the survey which is 25.4 minutes.