Q.27 - Elementary Statistics

Question

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. 

Step-by-Step Solution

Answer

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

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.

$H_0:\mu =468H_a:\mu <468$

Step 3: Test statistic

Calculate the test statistic's value: 

$$\begin{gathered} t=\frac{\overline{x} -{\mu }_0}{s/n} \\ =\frac{455.0-468}{8.6/\sqrt{12}} \\ \approx -0.519 \end{gathered}$$

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
$$\begin{gathered} df=n1 \\ =121 \\ =11 \end{gathered}$$.

$P>0.10$

The null hypothesis is rejected if the P-value is less than the significance level $\alpha$.

$P>0.05$⇒ 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 $2012$.