Hypothesis Testing and Confidence Intervals

Hypothesis Testing and Confidence Intervals
Question 1
Part (a)
We are going to use a two-tailed paired sample t-test to compare the difference between the scores obtained on
the first and second attempt. A t-test is used because the sample size is small i.e. n<30. While a paired sample t-
test is employed because the group is the same for both attempts; the only difference is that the test attempts are
conducted at different time periods.
Null hypothesis (H0): The means for the two test attempts are the same i.e. μ1=μ2
Alternative hypothesis (H1): There is significant difference between the means for the two test attempts i.e.
μ1 ≠ μ2
T-score calculation
D.F..= n-1=15-1=14
t= d
√s2 /n
t= 6.5333
√ 144.2667 /15
t=2.106675
P-value=0.05366
We will therefore not reject the null hypothesis and conclude that the average scores of students in the first and
second attempt are the same.
Part (b)
Calculating the effect size for the given test scores using the t-score
Effect ¿ t
√n
Effect ¿ 2.106675
√15
Effect ¿ 0.14
The effect size i.e. Cohen’s d is less than 0.2 which test us the difference between the two means for first
attempt and second attempt is quite trivial in spite of the fact that it is significant.

Part (c)
Confidence interval for this is given by
d ± t0.0 2 5,14 ( Sd
n )
hence
6.5333 ± 2.145∗(12.0111
15 )
95 % Confidence Interval=(8.25 , 4.82)
The 95% confidence interval implies that the mean for test scores on the second attempts is between 8.25 and
4.82 percent large than that of test scores on the first attempt.
Question 2
Part (a)
We are going to use a two-tailed independent samples t-test to compare the difference in results between
Professor A’s class and Professor B’s class (assuming equal variance). The hypothesis here is:
Null hypothesis (H0): The means for the two classes are the same i.e. μ1=μ2
Alternative hypothesis (H1): There is significant difference between the means for the two classes i.e. μ1 ≠ μ2
T-score calculation
t= μA −μB
√ [ ( ∑ A2− ( ∑ A )
2
nA ) + ( ∑ B2− ( ∑ B )
2
nB )
nA +nB−2 ]∗
[ 1
nA
+ 1
nB ]
t= −6.5333
√ [ 3633.733+2515.333
28 ]∗0.13333
Using the formula above
We get that
t=-1.207368562
P-value=0.237