Question

3) Suppose you are an educational researcher who wishes to examine the effect of a school district's class size on its student achievement. Specifically, you are interested in whether in the U.S., on average, school districts with smaller class sizes perform differently on test scores than school districts with larger class sizes. To test this, you have conducted two surveys. The first survey randomly sampled 238 school districts with small class sizes. This survey found a sample average test score of 657.4 points and a sample standard deviation of 19.4 points. The second survey uses a random sample of 182 school districts with large class sizes. That survey found a sample average test score of 650 points and a sample standard deviation of 17.9 points. Let the random variable y denote the test score of a school district with a large class size in the U.S., and let the random variable X denote the test score of a school district with a small class size in the U.S. You wish to test the null hypothesis that the average test score of school districts with a large class size (My) is equal to the average test score of school districts with a small class size (px) against the alternative that the average test scores differ. For this problem, suppose that the maximum probability of a Type 1 error that you are willing to tolerate is 5%. (a) State the null and alternative hypotheses (1 points) (b) Calculate that t-statistic (2 points) (c) Calculate the p-value (2 points) (d) Compute the 95% confidence interval for My – Hx (2 points) (e) Based on either the p-value or the t-statistic, do you reject or fail to reject the null hypothesis? For full credit explain how you arrived to this conclusion. You cannot simply state "Reject" or "Fail to Reject." You need to explain why you are taking the specific action. (2 points) (f) Suppose you find evidence in your sample that school districts with small class sizes have higher test scores than school districts with large class sizes. Does this mean every school district should then strive to shrink its class size? Why or why not? (2 points)

Answer 1

3.

Given that,
mean(x)=657.4
standard deviation , s.d1=19.4
number(n1)=238
y(mean)=650
standard deviation, s.d2 =17.9
number(n2)=182
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =1.973
since our test is two-tailed
reject Ho, if to < -1.973 OR if to > 1.973
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =657.4-650/sqrt((376.36/238)+(320.41/182))
to =4.048
| to | =4.048
critical value
the value of |t α| with min (n1-1, n2-1) i.e 181 d.f is 1.973
we got |to| = 4.04799 & | t α | = 1.973
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 4.048 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 != u2
b.
test statistic: 4.048
critical value: -1.973 , 1.973
decision: reject Ho
c.
p-value: 0
we have enough evidence to support the claim that difference of means between Y and X.
d.
TRADITIONAL METHOD
given that,
mean(x)=657.4
standard deviation , s.d1=19.4
number(n1)=238
y(mean)=650
standard deviation, s.d2 =17.9
number(n2)=182
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((376.36/238)+(320.41/182))
= 1.828
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 181 d.f is 1.973
margin of error = 1.973 * 1.828
= 3.607
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (657.4-650) ± 3.607 ]
= [3.793 , 11.007]
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DIRECT METHOD
given that,
mean(x)=657.4
standard deviation , s.d1=19.4
sample size, n1=238
y(mean)=650
standard deviation, s.d2 =17.9
sample size,n2 =182
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 657.4-650) ± t a/2 * sqrt((376.36/238)+(320.41/182)]
= [ (7.4) ± t a/2 * 1.828]
= [3.793 , 11.007]
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interpretations:
1. we are 95% sure that the interval [3.793 , 11.007] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
e.
we have enough evidence to support the claim that difference of means between Y and X.
f.
Given that,
mean(x)=657.4
standard deviation , s.d1=19.4
number(n1)=238
y(mean)=650
standard deviation, s.d2 =17.9
number(n2)=182
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.653
since our test is right-tailed
reject Ho, if to > 1.653
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =657.4-650/sqrt((376.36/238)+(320.41/182))
to =4.048
| to | =4.048
critical value
the value of |t α| with min (n1-1, n2-1) i.e 181 d.f is 1.653
we got |to| = 4.04799 & | t α | = 1.653
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:right tail - Ha : ( p > 4.048 ) = 0.00004
hence value of p0.05 > 0.00004,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 4.048
critical value: 1.653
decision: reject Ho
p-value: 0.00004
we have enough evidence to support the claim that average test score of school district with large size is higher than small class sizes .