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 endered 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 claimof the lawyer? Use 0.05 level or 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 $\alpha = 0.05,$

$df=n-1=35-1=34$ 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:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{4-3.5}{3.5/\sqrt{35}}\approx0.845154$$

Since it is observed that $|t| = 0.845154 \le 2.032244=t_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for $df=34$ degrees of freedom, two-tailed, $t=0.845154$ is $p = 0.403933,$ and since $p=0.403933>0.05=\alpha,$  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.