Question

How Far People Drive. In 2011, the average car in the United States was driven 13.5 thousand miles, as reported by the Federal Highway Administration in Highway Statistics. On the WeissStats site, we provide last year's distance driven, in thousands of miles, by each of 500 randomly selected cars. Use the technology of your choice to do the following.

a. Obtain a normal probability plot and histogram of the data.

b. Based on your results from part (a), can you reasonably apply the one-mean t-test to the data? Explain your reasoning.

c. At the 5 % significance level, do the data provide sufficient evidence to conclude that the mean distance driven last year differs from that in 2011?

Short answer

The significance level is $\alpha =0.05$

The P-value is $0.026$

Step-by-step solution

Subpart (a) Step 1: 

MINITAB is used to create the normal probability plot.

MINITAB procedure:

Step 1: Choose Graph > Probability Plot.

Step 2: Choose Single, and then click OK.

Step 3: In Graph variables, enter the column ofDISTANCE.

Step 4: Click OK.

Subpart (a) Step 2: MINITAB output

Subpart (a) Step 3: 

MINITAB is used to create the histogram.

MINITAB procedure:

Step 1: Choose Graph > Histogram.

Step 2: After choosing Simple, click OK.

Step 3: In Graph variables, enter the corresponding column of DISTANCE.

Step 4: Click OK.

MINITAB output:

Subpart (b) Step 1:

Conditions to use of the one mean t-test procedure are given below:

Small Sample size:

- The population is divided into samples at random.

- Population follows normal distribution or the sample size is larger.

- The standard deviation is unknown.

Explanation:

The sample is drawn from the entire population, and the sample size is rather huge. Furthermore, the data has a considerable number of outliers, indicating that there is some caution in the data. As a result, the single mean t-test technique is appropriate. When outliers are eliminated, the single mean t-test technique becomes more accurate.

Subpart (c) Step 1:

Examine the data to see if there is enough information to establish that the average distance traveled last year varies from the average distance driven in 2011.

State the null and alternative hypothesis:

Null hypothesis:

$H_0:\mu =13.5$

That is, the data does not provide sufficient evidence to conclude that the mean distance driven last year differs from that in 2011 .

Another possibility:

$H_a:\mu \ne 13.5$

That is, the facts are adequate to determine that the average distance traveled last year varies from the average distance driven in 2011.

Decide a significance level

The significance level is, in this case. $\alpha =0.05$.

Subpart (c) Step 2:

MINITAB may be used to calculate the test statistic and P-value.

Procedure for MINITAB:

Step 1: Select Stat > Basic Statistics > 1-Sample t from the drop-down menu.

Step 2: In Samples in Column, enter the column of DISTANCE.

Step 3: In Perform hypothesis test, enter the test mean as 13.5.

Step 4: Go to Options and choose Confidence Level 95.

Step 5: In the alternative, choose not equal.

Step 6: In all dialogue boxes, click OK.

MINITAB output:

One-Sample T: DISTANCE

Test of mu $=13.5$vs not $=13.5$

Variable	N	Mean	StDev	SE Mean	95% CI	T	P
Distance	500	12.902	6.002	0.268	(12.375, 13.429)	-2.23	0.026

From the MINITAB output,

The value of test statistic is $-2.23$

The P-value is $0.026$

Definition

If $P\le \alpha$, then reject the null hypothesis.

Here, the P-value is 0.026 which is less than the level of significance. That is,

$P(=0.026)<\alpha (=0.05)\text{.}$

Therefore, the null hypothesis is rejected at 5% level.

Thus, it can be conclude that the test results are statistically significant at 5 % level of significance.

Interpretation:

As a result, the statistics are adequate to determine that the average distance driven last year varies from that of 2011.