Question #333068

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.


1
Expert's answer
2022-04-27T00:34:15-0400

The following null and alternative hypotheses need to be tested:

H0:μ=3.5H_0: \mu=3.5

H1:μ3.5H_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 α=0.05,\alpha = 0.05,

df=n1=351=34df=n-1=35-1=34 degrees of freedom, and the critical value for a two-tailed test is tc=2.032244.t_c =2.032244.

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

The t-statistic is computed as follows:


t=xˉμs/n=43.53.5/350.845154t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{4-3.5}{3.5/\sqrt{35}}\approx0.845154

Since it is observed that t=0.8451542.032244=tc,|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=34df=34 degrees of freedom, two-tailed, t=0.845154t=0.845154 is p=0.403933,p = 0.403933, and since p=0.403933>0.05=α,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 μ\mu

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


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