Question

An educational psychologist studies the effect of frequent testing on retention of class material. In one section of an introductory course, students are given quizzes each week. A second section of the same course receives only two tests during the semester. At the end of the semester, both sections receive the same final exam, and the scores are summarized below:

                                    Frequent Quizzes       Two Exams

                                            n = 20                    n = 20

                                           M = 73                   M = 68

1. Determine the correct t-test needed. If the first sample variance is s² = 38 and the second sample has s² = 42, do the data indicate that testing frequency has a significant effect on performance? Use a two-tailed test at the .05 level of significance. (Note: because the two samples are the same size, the pooled variance is simply the average of the two sample variances.) Draw your distribution. (2 pts)
2. Determine the correct t-test needed. If the first sample variance has s² = 84 and the second sample has s² = 96, do the data indicate that testing frequency has a significant effect? Again, use a two-tailed test with α = .05. Draw your distribution. (2 pts)
3. Using symbols, write up your results. Describe how the size of the variance affects the outcome of the hypothesis test. (1 pt)

Answer 1

a.
Given that,
mean(x)=73
standard deviation , s.d1=6.164
number(n1)=20
y(mean)=68
standard deviation, s.d2 =6.48
number(n2)=20
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.024
since our test is two-tailed
reject Ho, if to < -2.024 OR if to > 2.024
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (19*37.995 + 19*41.99) / (40- 2 )
s^2 = 39.993
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=73-68/sqrt((39.993( 1 /20+ 1/20 ))
to=5/2
to=2.5
| to | =2.5
critical value
the value of |t α| with (n1+n2-2) i.e 38 d.f is 2.024
we got |to| = 2.5 & | t α | = 2.024
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 2.5002 ) = 0.0167
hence value of p0.05 > 0.0167,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.5
critical value: -2.024 , 2.024
decision: reject Ho
p-value: 0.0167
we have enough evidence to support the claim that testing frequency has a significant effect on performance
b.
Given that,
mean(x)=73
standard deviation , s.d1=9.165
number(n1)=20
y(mean)=68
standard deviation, s.d2 =9.797
number(n2)=20
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.024
since our test is two-tailed
reject Ho, if to < -2.024 OR if to > 2.024
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (19*83.997 + 19*95.981) / (40- 2 )
s^2 = 89.989
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=73-68/sqrt((89.989( 1 /20+ 1/20 ))
to=5/3
to=1.667
| to | =1.667
critical value
the value of |t α| with (n1+n2-2) i.e 38 d.f is 2.024
we got |to| = 1.667 & | t α | = 2.024
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 != 1.6668 ) = 0.1036
hence value of p0.05 < 0.1036,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.667
critical value: -2.024 , 2.024
decision: do not reject Ho
p-value: 0.1036
we do not have enough evidence to support the claim that testing frequency has a significant effect on performance
c.
yes,
the size of the variance affects the outcome of the hypothesis test, because if the sample variance is increases then the out come of result is changes from reject
the null hypothesis to fails to reject the null hypothesis.