91Ó°ÊÓ

Consider the given question,

The significance level is$\mathrm{Î\pm }=0.05$.

The null hypothesis is given below,

$H_0:\mathrm{μ}=\$468$

The data does not provide sufficient evidence to conclude that the last year's mean value lost to purse snatching has decreased from the 2012mean.

The alternative hypothesis is given below,

$H_a:\mathrm{μ}>\$468$

The data provide sufficient evidence to conclude that the last year's mean value lost to purse snatching has decreased from the2012mean.

On using the MINITAB procedure,

1. Choose Stat>Basic Statistics>1-Sample t.
2. In Summarized data, enter the sample size 12 and mean 455.
3. In Standard deviation, enter 86.8.
4. In Perform hypothesis test, enter the test mean as 468.
5. Check Options, enter Confidence level as 95.
6. Choose less than in alternative.
7. Click OK in all dialogue boxes.

Hence, from the MINITAB output, the value of test statistics is -0.52 and the P-value is 0.307.

If $P≤\mathrm{Î\pm }$, then reject the null hypothesis.

The P-value is 0.307which is greater than the level of significance that is $P\left(=0.307\right)>\mathrm{Î\pm }\left(=0.05\right)$.

Therefore, the null hypothesis is not rejected at 5% level.

It can be concluded that the test results are not statistically significant at 5% level of significance.

On interpreting, we can say that the data does not provide sufficient evidence to conclude that the last year's mean value lost to purse snatching has decreased from the2012 mean.