Question

Rosenthal and Jacobson (1968) informed classroom teachers that some of their students showed unusual potential for intellectual gains. Eight months later the students identified to teachers as having potentional for unusual intellectual gains showed significantly greater gains performance on a test said to measure IQ than did children who were not so identified. 

Answer 1

Scores for first Graders

First group

$$35, 40, 12,15,21, 14, 46,10,28, 48,$$

$$16,30,32,48,31,22, 12,39,19,25$$

$$mean_1=\bar{x}_1=\dfrac{1}{20}(35+40+12+15+21+14$$

$$+46+10+28+48+16+30+32+48+31$$

$$+22+12+39+19+25)=27.15$$

$$s_1^2=\dfrac{1}{20-1}((35-27.15)^2+(40-27.15)^2$$

$$+(12-27.15)^2+(15-27.15)^2+(21-27.15)^2$$

$$+(14-27.15)^2+(46-27.15)^2+(10-27.15)^2$$

$$+(28-27.15)^2+(48-27.15)^2+(16-27.15)^2$$

$$+(30-27.15)^2+(32-27.15)^2+(48-27.15)^2$$

$$+(31-27.15)^2+(22-27.15)^2+(12-27.15)^2$$

$$+(39-27.15)^2+(19-27.15)^2+(25-27.15)^2)$$

$$=156.45$$

$$s_1=\sqrt{s_1^2}=\sqrt{156.45}\approx12.50800$$

Second group

$$2,27,38,31,1,19,1,34,3, 1,$$

$$2,3,2,1,2,1,3,29,37,2$$

$$mean_2=\bar{x}_2=\dfrac{1}{20}(2+27+38+31+1+19$$

$$+1+34+3+1+2+3+2+1+2$$

$$+1+3+29+37+2)=11.95$$

$$s_2^2=\dfrac{1}{20-1}((2-11.95)^2+(27-11.95)^2$$

$$+(38-11.95)^2+(31-11.95)^2+(1-11.95)^2$$

$$+(19-11.95)^2+(1-11.95)^2+(34-11.95)^2$$

$$+(3-11.95)^2+(1-11.95)^2+(2-11.95)^2$$

$$+(3-11.95)^2+(2-11.95)^2+(1-11.95)^2$$

$$+(2-11.95)^2+(1-11.95)^2+(3-11.95)^2$$

$$+(29-11.95)^2+(37-11.95)^2+(2-11.95)^2$$

$$=\dfrac{4056.95}{19}\approx213.52368$$

$$s_2=\sqrt{s_2^2}=\sqrt{\dfrac{4056.95}{19}}\approx14.61245$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\leq\mu_2$

$H_1:\mu_1>\mu_2$

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{156.45}{\dfrac{4056.95}{19}}\approx0.7327$$

The critical values are $F_L=0.396$ and $F_U=2.526,$ and since $F=0.7327,$ then the null hypothesis of equal variances is not rejected.

Based on the information provided, the significance level is $\alpha=0.05,$ and the degrees of freedom are $df=n_1+n_2-2=38.$

Hence, it is found that the critical value for this right-tailed test is $t_c=1.685954,$ for $\alpha=0.05,$

$df=38$ degrees of freedom.

The rejection region for this right-tailed test is $R=\{t:|t|>1.685954\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{27.15-11.95}{\sqrt{\dfrac{(20-1)156.45+(20-1)213.524}{20+20-2}(\dfrac{1}{20}+\dfrac{1}{20})}}$$

$$\approx33.601$$

Since it is observed that $t=33.601>1.685954=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed, $\alpha=0.05, df=38,t=1.685954,$ is 

$p=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is greater than $\mu_2,$ at the $\alpha=0.05$ significance level.