Answer:
a. At 95% confidence interval, the mean test score in the population =
[ 648.06 , 644.34 ]
b.
Null hypothesis;
Alternative hypothesis
t 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
standard deviation = 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:
Z value at 95% C.I = 1.96
The level of significance = 1 - 0.95
The level of significance = 0.05
∴
⇒
=
=
=
= [ 646.2 + 1.86494 , 646.2 - 1.86494 ]
= [ 648.06494 , 644.33506 ]
[ 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 = 657.4 19.4 238
Large = 650.0 17.9 182
The null hypothesis and the alternative hypothesis for this given data can be computed as follows:
Null hypothesis;
Alternative hypothesis
The t- test statistics score can be expressed by using the formula:
where;
∴
t = 4.04799
t 4.048
To determine the P - Value; we have:
P - value = 1 - ( t)
P - value = 1 - ( 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.