Business, 23.08.2020 14:01 mrashrafkotkaat

Data on fifth-grade test scores (reading and mathematics) for 420 school districts in California yield \bar{y} = 646.2 and standard deviation s_{y} = 19.5 a. Construct a 95% confidence interval for the mean test score in the population.
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}) n
Small 657.4 19.4 238
Large 650.0 17.9 182
Is there statistically significant evidence that the districts with smaller classes have higher average test scores? Explain (set up your hypothesis, and clearly state your decision rule).

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Answer from: izzyisawesome5232

a. At 95% confidence interval, the mean test score in the population =

[ 648.06 , 644.34 ]

Null hypothesis;

$H_o : \mu_1 - \mu_2 = 0$

Alternative hypothesis

$H_1 : \mu_1 - \mu_2 0$

t $\simeq$ 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 $\overline{Y} = 646.2$

standard deviation $s_{y}$ = 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:

$[ \overline Y \pm Z \times SE( \overline Y)]$

Z value at 95% C.I = 1.96

The level of significance = 1 - 0.95

The level of significance = 0.05

$SE(\overline Y ) = \dfrac{s_y}{\sqrt{n}}$

∴

⇒ $[ 646.2 \pm 1.96 \times \dfrac{19.5}{\sqrt{420}}]$

= $[ 646.2 \pm 1.96 \times \dfrac{19.5}{20.4939}]$

= $[ 646.2 \pm 1.96 \times 0.9515]$

= $[ 646.2 \pm 1.86494]$

= [ 646.2 + 1.86494 , 646.2 - 1.86494 ]

= [ 648.06494 , 644.33506 ]

$\simeq$ [ 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}) n

Small $\overline{Y_1}$ = 657.4 19.4 238

Large $\overline{Y_2}$ = 650.0 17.9 182

The null hypothesis and the alternative hypothesis for this given data can be computed as follows:

Null hypothesis;

$H_o : \mu_1 - \mu_2 = 0$

Alternative hypothesis

$H_1 : \mu_1 - \mu_2 0$

The t- test statistics score can be expressed by using the formula:

$t = \dfrac{\overline Y_1 -\overline Y_2 }{SE( \overline Y_1 - \overline Y_2)}$

where;

$SE ( \overline Y_1 - \overline Y_2) = {\sqrt{\dfrac{s^2_y_1}{n_1} + \dfrac{s^2_y_2}{n_2}}}$

∴

$t = \dfrac{\overline Y_1 -\overline Y_2 }{\sqrt{\dfrac{s^2_y_1}{n_1} + \dfrac{s^2_y_2}{n_2}}}$

$t = \dfrac{657.4-650.0 }{\sqrt{\dfrac{19.4^2}{238} + \dfrac{17.9^2}{182}}}$

$t = \dfrac{7.4 }{\sqrt{\dfrac{376.36}{238} + \dfrac{320.41}{182}}}$

$t = \dfrac{7.4 }{\sqrt{1.581344538+ 1.760494505}}$

$t = \dfrac{7.4 }{\sqrt{3.341839043}}$

$t = \dfrac{7.4 }{1.828069759}$

t = 4.04799

t $\simeq$ 4.048

To determine the P - Value; we have:

P - value = 1 - $\mathtt{\phi}$ ( t)

P - value = 1 - $\mathtt{\phi}$ ( 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.

Answer from: Quest

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Answer from: Quest

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