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Q.27

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Elementary Statistics

Found in: Page 393

Book edition 9th

Author(s) Weiss, Neil

Pages 590 pages

ISBN 9780321989505

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Short Answer

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 $468$ in $2012$ . For last year, $12$ randomly selected purse-snatching offenses yielded the following values lost, to the nearest dollar.
Use a t-test to decide, at the $5 %$ significance level, whether last year's mean value lost to purse snatching has decreased from the $2012$ mean. The mean and standard deviation of the data are $455.0$ and $86.8$ , respectively.

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: 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 $2012$ was. Last year, $12$ purse-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 $2012$ mean at the $5 %$ significance level. The data has a mean and standard deviation of $455.0$ and $86.8$ , respectively.

Step 2: Calculation

The data's mean and standard deviation are

455.0

and

86.8

, respectively, according to the information provided. The 5 percent significance level has declined since

2012

, as has the mean value lost to purse theft.From

468

, 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.

H_{0} : μ = 468 H_{a} : μ < 468

Step 3: Test statistic

Calculate the test statistic's value:

t = \frac{x - μ_{0}}{s / n} = \frac{455.0 - 468}{8.6 / \sqrt{12}} \approx - 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 tvalue

0.519

in the row

d f = n 1 = 121 = 11

.

P > 0.10

The null hypothesis is rejected if the P-value is less than the significance level

α

.

P > 0.05

⇒ Fail to reject

H_{0}

As a result, there is insufficient information to support the allegation that the mean value lost to purse snatching in

2013

was lower than in

2012

.

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