Question

A lawyer claims that the average years of settling criminal cases filed before courts is three

and a half years. To test the claim, a watchdog organization randomly selected 35 criminal

cases and recorded the years the cases took before the courts have been rendered verdicts.

The sample mean was 4 years with a standard deviation of three and a half years. Do the

data gathered by the said organization provide sufficient basis to accept the claim of the

lawyer? Use 0.05 level of significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=3.5$

$H_1:\mu\not=3.5$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=34$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 2.032244.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.032244\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{4-3.5}{3.5/\sqrt{35}}\approx0.84515$$

Since it is observed that $|t|=0.84515<2.032244=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=34$ degrees of freedom, $t=0.84515,$ is $p= 0.403935,$ and since $p=0.403935>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 3.5, at the $\alpha = 0.05$ significance level.