Answer to Question #224125 in Statistics and Probability for josh

Question

A test can be conducted to determine the length of time required for a student to read a specified amount of material. In this test, students were instructed to read at the maximum speed at which they could still comprehend the material. A random sample of sixteen students look the test, with the following results (in minutes): 

18 27 29 20 19 25 24 21 24 19 23 28 31 22 27 21

Assume that the results of the test are normally distributed. Is there sufficient evidence to claim that the mean length of time (in minutes) required to read is not equal to 25? Test at 5% level of significance

Answer 1

$$n=16$$

$$mean=\bar{x}=\dfrac{1}{16}(18+27+29+20+19+25$$

$$+24+21+24+19+23+28+ 31 +22+ 27+ 21)=23.625$$

$$s^2=\dfrac{1}{16-1}((18-23.625)^2+(27-23.625)^2$$

$$+(29-23.625)^2+(20-23.625)^2+(19-23.625)^2$$

$$+(25-23.625)^2+(24-23.625)^2+(21-23.625)^2$$

$$+(24-23.625)^2+(19-23.625)^2+(23-23.625)^2$$

$$+(28-23.625)^2+(31-23.625)^2+(22-23.625)^2$$

$$+(27-23.625)^2+(21-23.625)^2=15.45$$

$$s=\sqrt{s^2}=\sqrt{15.45}\approx3.930649$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=25$

$H_1:\mu\not=25$

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 $\alpha=0.05,$

$df=n-1=16-1=15$ degrees of freedom, and the critical value for a two-tailed test is $t_c=2.131449.$  

The rejection region for this two-tailed test is $R=\{t:|t|>2.131449\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}\approx\dfrac{23.625-25}{3.930649/\sqrt{16}}\approx-1.399260$$

Since it is observed that $|t|=1.399260<2.131449=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for two-tailed, $\alpha=0.05, df=15, t=-1.3999260$ is  $p=0.181878,$ and since $p=0.181878>0.05=\alpha,$ 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 25, at the $\alpha=0.05$ significance level.