Respuesta :
Answer:
a. At 95% confidence interval, the mean test score in the population =
[ 648.06 , 644.34 ]
b.
Null hypothesis;
[tex]H_o : \mu_1 - \mu_2 = 0[/tex]
Alternative hypothesis
[tex]H_1 : \mu_1 - \mu_2 > 0[/tex]
t  [tex]\simeq[/tex] 4.048
P - value = 0.000003
Decision Rule: To reject  the null hypothesis if P - value is less than the level of significance
Conclusion: We reject the null hypothesis and conclude that there is sufficient statistical evidence that smaller class sizes do give rise to higher test averages.
Explanation:
Given that:
sample size n = 420
mean [tex]\overline{Y} = 646.2[/tex]
standard deviation [tex]s_{y}[/tex] = 19.5
a. Construct a 95% confidence interval for the mean test score in the population.
The 95% confidence interval for the mean test score can be expressed as:
[tex][ \overline Y \pm Z \times SE( \overline Y)][/tex]
Z value at 95% C.I = 1.96
The level of significance = 1 - 0.95
The level of significance = 0.05
[tex]SE(\overline Y ) = \dfrac{s_y}{\sqrt{n}}[/tex]
∴
⇒ [tex][ 646.2 \pm 1.96 \times \dfrac{19.5}{\sqrt{420}}][/tex]
= [tex][ 646.2 \pm 1.96 \times \dfrac{19.5}{20.4939}][/tex]
= [tex][ 646.2 \pm 1.96 \times 0.9515][/tex]
= [tex][ 646.2 \pm 1.86494][/tex]
= [ 646.2 + 1.86494 , 646.2 - 1.86494 ]
= [ 648.06494 , 644.33506 ]
[tex]\simeq[/tex] [ 648.06 , 644.34 ]
Thus at 95% confidence interval, the mean test score in the population =
[ 648.06 , 644.34 ]
b. When the districts were divided into districts with small classes (< 20 students per teacher) and large classes (? 20 students per teacher), the following results were found:
Class Size Average (\bar{Y}) Standard Deviation (s_{y}) Â Â Â n
Small        [tex]\overline{Y_1}[/tex]= 657.4         19.4                 238
Large        [tex]\overline{Y_2}[/tex]= 650.0         17.9                 182
The null hypothesis and the alternative hypothesis for this given data can be computed as follows:
Null hypothesis;
[tex]H_o : \mu_1 - \mu_2 = 0[/tex]
Alternative hypothesis
[tex]H_1 : \mu_1 - \mu_2 > 0[/tex]
The t- test statistics score can be expressed by using the formula:
[tex]t = \dfrac{\overline Y_1 -\overline Y_2 }{SE( \overline Y_1 - \overline Y_2)}[/tex]
where;
[tex]SE ( \overline Y_1 - \overline Y_2) = {\sqrt{\dfrac{s^2_y_1}{n_1} + \dfrac{s^2_y_2}{n_2}}}[/tex]
∴
[tex]t = \dfrac{\overline Y_1 -\overline Y_2 }{\sqrt{\dfrac{s^2_y_1}{n_1} + \dfrac{s^2_y_2}{n_2}}}[/tex]
[tex]t = \dfrac{657.4-650.0 }{\sqrt{\dfrac{19.4^2}{238} + \dfrac{17.9^2}{182}}}[/tex]
[tex]t = \dfrac{7.4 }{\sqrt{\dfrac{376.36}{238} + \dfrac{320.41}{182}}}[/tex]
[tex]t = \dfrac{7.4 }{\sqrt{1.581344538+ 1.760494505}}[/tex]
[tex]t = \dfrac{7.4 }{\sqrt{3.341839043}}[/tex]
[tex]t = \dfrac{7.4 }{1.828069759}[/tex]
t = 4.04799
t  [tex]\simeq[/tex] 4.048
To determine the P - Value; we have:
P - value = 1 - [tex]\mathtt{\phi}[/tex] ( t)
P - value = 1 - [tex]\mathtt{\phi}[/tex]( 4.048)
P - value = 1 - 0.99997
P - value = 0.000003
Decision Rule: To reject  the null hypothesis if P - value is less than the level of significance
Conclusion: We reject the null hypothesis and conclude that there is sufficient statistical evidence that smaller class sizes do give rise to higher test averages.
