In: Statistics and Probability
You want to estimate the difference between the average grades on a certain math exam before the students take the associated math class and after they take the associated math class. You take the random samples of 5 students who have taken the class and 5 students who have not taken the class. You get the following results:
Not Taken | Taken |
54 | 82 |
25 | 76 |
73 | 98 |
23 | 43 |
42 | 38 |
A) Determine the population(s) and parameter(s) being discussed.
B) Determine which tool will help us find what we need (one sample z test, one sample t test, two sample t test, one sample z interval, one sample t interval, two sample t interval).
C) Check if the conditions for this tool hold.
D) Whether or not the conditions hold, use the tool you choose in part B. Use C=95% for all confidence intervals and alpha=5% for all significance tests.
* Be sure that all methods end with a sentence describing the results *
Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.776
since our test is two-tailed
reject Ho, if to < -2.776 OR if to > 2.776
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 24
We have d = 24
pooled variance = calculate value of Sd= √S^2 = sqrt [
4426-(120^2/5 ] / 4 = 19.66
to = d/ (S/√n) = 2.73
critical Value
the value of |t α| with n-1 = 4 d.f is 2.776
we got |t o| = 2.73 & |t α| =2.776
make Decision
hence Value of |to | < | t α | and here we do not reject
Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.7297 )
= 0.0525
hence value of p0.05 < 0.0525,here we do not reject Ho
ANSWERS
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A.
the difference between the average grades on a certain math exam
before the students take the associated math class
and after they take the associated math class.
B.
Paired sample t test
C.
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: 2.73
critical value: reject Ho, if to < -2.776 OR if to >
2.776
D.
decision: Do not Reject Ho
p-value: 0.0525
we do not have enough evidence to support the claim that take the
random samples of 5 students who have taken the class and 5
students who have not taken the class