/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 9.124 How Far People Drive. In 2011, t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Expert verified

The significance level is α=0.05

The P-value is 0.026

Step by step solution

01

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.

02

Subpart (a) Step 2: MINITAB output

03

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:

04

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.

05

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:

H0:μ=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:

Ha:μ≠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. α=0.05.

06

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 =13.5vs not =13.5

Variable
NMeanStDevSE Mean95% CITP
Distance50012.9026.0020.268(12.375, 13.429)-2.230.026

From the MINITAB output,

The value of test statistic is -2.23

The P-value is 0.026

07

Definition

If P≤α, 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)<α(=0.05).

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the given exercise, we have provided a sample mean, sample size, and population standard. In each case, use the one-mean z-test to perform the required hypothesis test at the 5% significance level.

x—=20,n=24,σ=4,H0:μ=22,Ha:μ≠22

Dirt Bikes. Dirt bikes are simpler and lighter motorcysles that are designed for off-road events. Specifications for dirt bikes can be found through Motorcycle USA on their website www.motorcycle-usa.com. A random sample of 30 dirt bikes have a mean fuel capacity of 1.91gallons with a standard deviation of 0.74 gallons. At the 10%significance level, do the data provide sufficient evidence to conclude that the mean fuel tank capacity of all dirt bikes is less than 2 gallons?

Purse Snatching. The Federal Bureau of Investigation (FBI) compiles information on robbery and property crimes by type and selected characteristic and publishes its findings in Uniform Crime Reports. According to that document, the mean value lost to purse snatching was 468in 2012. For last year, 12randomly selected purse-snatching offenses yielded the following values lost, to the nearest dollar.

Use a t-test to decide, at the5%significance level, whether last year's mean value lost to purse snatching has decreased from the 2012mean. The mean and standard deviation of the data are 455.0and 86.8, respectively.

In each part, we have given the value obtained for the test statistic, z, in a one-mean z-test. We have also specified whether the test is two tailed, left tailed, or right tailed. Determine the P-value in each case and decide whether, at the 5%significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.

a. z=-1.25; left-tailed test

b. z=2.36; right-tailed test

c. z=1.83; two-tailed test

Data on salaries in the public school system are published annually in Ranking of the States and Estimates of School Statistics by the National Education Association. The mean annual salary of (public) classroom teachers is $55.4thousand. A hypothesis test is to be performed to decide whether the mean annual salary of classroom teachers in Ohio is greater than the national mean.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.