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Chapter 9: Q.27 (page 393)

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.

Short Answer

Expert verified

There is insufficient data to support the allegation that the mean value of purses stolen last year was lower than in 2012.

Step by step solution

01

Given information

The facts to come to a conclusion about the previous year's mean.

According to the document, the average amount stolen due to handbag snatching in 2012was. Last year, 12purse-snatching offences were chosen at random and the following values were lost, to the nearest dollar.

Use a t-test to determine whether the mean value lost to purse snatching last year was lower than the 2012mean at the 5%significance level. The data has a mean and standard deviation of 455.0and 86.8, respectively.

02

Calculation

The data's mean and standard deviation are 455.0and 86.8, respectively, according to the information provided. The 5 percent significance level has declined since2012, as has the mean value lost to purse theft.
From468, the average has dropped.
Either the null hypothesis or the alternative hypothesis is the assertion. The null hypothesis states that the population mean is the same as the claimed value. If the claim is the null hypothesis, the alternative hypothesis is the polar opposite of the null hypothesis.
H0:渭=468Ha:渭<468
03

Test statistic

Calculate the test statistic's value:
t=x鈭�渭0s/n=455.0鈭�4688.6/12鈮�鈭�0.519
If the null hypothesis is true, the P-value is the chance of getting the test statistic's value, or a value more high.
The P-value is the number in the Student's T table in the appendix's column title that corresponds to the tvalue0.519 in the row
df=n1=121=11.
P>0.10
The null hypothesis is rejected if the P-value is less than the significance level 伪.
P>0.05鈬� Fail to rejectH0
As a result, there is insufficient information to support the allegation that the mean value lost to purse snatching in 2013was lower than in2012.

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Most popular questions from this chapter

Apparel and Services. According to the document Consumer Expenditures, a publication of the Bureau of Labor Statistics, the average consumer unit spent \( 1736 on apparel and services in 2012 That same year, 25 consumer units in the Northeast had the following annual expenditures, in dollars, on apparel and services.

12791457202016821273
22232233219216111734
26882029216618602444
18441765226715222012
19901751211322021712

At the 5 % significance level, do the data provide sufficient evidence to conclude that the 2012 mean annual expenditure on apparel and services for consumer units in the Northeast differed from the national mean of \)1736? (Note: The sample mean and sample standard deviation of the data are \(1922.76 and \)350.90, respectively.)

Beef Consumption. According to Food Consumption, Prices,\and Expenditures, published by the U.S. Department of Agriculture. the mean consumption of beef per person in 2011 was 57.5 lb. A sample of 40 people taken this year yielded the data, in pounds, on last year's beef consumption given on the Weiss Stats site. Use the technology of your choice to do the following.

a. Obtain a normal probability plot, a boxplot, a histogram, and a stem-and-leaf diagram of the data on beef consumptions.

b. Decide, at the 5% significance level, whether last year's mean beef consumption is less than the 2011 mean of 57.5 lb. Apply the one mean t-test.

c. The sample data contain four potential outliers: 0, 0, 0, and 13.Remove those four observations, repeat the hypothesis test in part (b), and compare your result with that obtained in part (b).

d. Assuming that the four potential outliers are not recording errors, comment on the advisability of removing them from the sample data before performing the hypothesis test.

e. What action would you take regarding this hypothesis test?

9.90 Hotels and Motels. The daily charges, in dollars, for a sample of 15 hotels and motels operating in South Carolina are provided on the WeissStats site. The data were found in the report South Caroline Statistical Abstract, sponsored by the South Carolina Budget and Control Board.
a. Use the one-mean z-test to decide, at the 5%significance level, whether the data provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than \(75. Assume a population standard deviation of \)22.40.

b. Obtain a normal probability plot, boxplot, histogram, and stem-and-leaf diagram of the data.
c. Remove the outliers (if any) from the data and then repeat part (a).
d. Comment on the advisability of using the z-test here.

In each of Exercises 9.41-9.46 ,determine the critical values for a one-mean z-test. For each exercise, draw a graph that illustrates your answer

A right- tailed test with伪=0.01

The following relationship holds between hypothesis tests and confidence intervals for one-mean t-procedures: For a two-tailed hypothesis test at the significance level 伪, the null hypothesis H0:渭=渭0will be rejected in favor of the alternative hypothesis Ha:渭>渭0if and only if 渭0lies outside the 1-伪-level confidence interval for 渭. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean t-interval and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.113

Part (b): Exercise 9.116
See all solutions

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