ACTIVITY SET 13
Activity 13.1
Use Table A.3 to estimate the p-value for each of the following hypothesis testing situations.
Then use the p-value to make a conclusion about the hypotheses. (Note: The value given for t is
the calculated value of the test statistic).
a. H0: = 72, Ha: >72, n=20, t=2.10
p-value  0.015 < p < 0.030
DF = 19
Conclusion : Reject Ho
b. H0: = 0, Ha:   0, n=40, t=2.41
DF = 39 (use 30 in table)
p-value  2*(0.008 < p < 0.013) = 0.016 < p < 0.026
We double the p-value since the alternative is 2-sided
Conclusion : Reject Ho
c. H0: = 98.6, Ha: < 98.6, n=10, t= 1.33
DF = 9
p-value 
0.084 < p < 0.116
Conclusion : Do Not Reject Ho
d. H0: = 100, Ha: > 100, n=16, t= 4.26
DF = 15
p-value  p < 0.004 Conclusion : Reject Ho
Activity 13.2 PSU claims that the average SATM score for the incoming fall 2005 class at
University Park was approximately 610. Use the Class Survey to test whether the current
population of PSU undergrads at UP campus differs from this claim. Again first by hand and
then with Minitab. The descriptive statistics are: sample size is 216; sample mean is 599; and the
sample standard deviation is 85.3
a. Write the null and alternative hypotheses using appropriate statistical notation.
H0: u = 610
Ha:
u ≠ 610
b. Calculate DF and the t-statistic: (You can get the sample mean and n from Activity 12.3)
DF = 215
t=
599 - 610
x - 0
=
= -1.89 (Note this is a negative!)
s
85.34
n
216
c. From Table A3 what is the range of the p-value based on you t-statistic? NOTE: if you
selected a two-sided Ha (i.e. used ≠) then you need to double the p-values found in the table.
Two-sided alternative so p-value range is 2*(0.024 < p < 0.037) = 0.048 < p < 0.074
d. Based on your p-value what is your decision and conclusion? Does this conclusion make sense
based on your confidence interval calculated above? That is, does your confidence interval
contain/not contain 610? If you rejected Ho then your interval should not contain 610 and viceversa.
Based on our p-value we would not reject Ho since the range of values contains our usual alpha
value of 0.05. This agrees with our confidence interval as the interval contains 610 (even though
just barely).
e. Now use Minitab to verify your results. Go to Stat > Basic Statistics > 1-Sample t and
select SATM (column C16). Click the box for “Perform Hypothesis Test” and enter the value
from your hypotheses statements (i.e. uo). Click on Options and select the correct alternative.
Click OK twice and copy and paste your Minitab results. Do your results by hand and those from
Minitab roughly match?
One-Sample T: SATM
Test of mu = 610 vs not = 610
Variable N Mean StDev SE Mean
95% CI
T P
SATM
216 599.005 85.335 5.806 (587.560, 610.449) -1.89 0.060
The t-statistic matches and the p-value from Minitab is in our interval. Notice that if you just
calculated by hand the one sided p-value range that this interval would not have contained the pvalue from Minitab.
Activity 13.3 Now try doing a paired sample. Is there a difference between how students
performed on their SATM and SATV? Since we are considering differences between two
measurements (i.e. SATM and SATV) on the same individual we can consider the data to be
paired. Go to Stat > Basic Statistics > Paired t. Minitab will calculate the difference by taking
the “First Sample” minus the “Second Sample”. So enter SATM as the First Sample and SATV
as the Second Sample. Click Options and be sure that 95 is entered as the confidence interval
and Not Equal is selected as the alternative. Click OK twice and copy and paste your results.
a. Write the null and alternative hypotheses using appropriate statistical notation.
H0: ud = 0
Ha: ud ≠ 0
b. Based on the Confidence interval what is your conclusion?
This interval does not contain 0 so we would reject the
null hypothesis and conclude that a difference does exist in the population between SATM and
SATV scores. Since the bounds of this interval are greater than 0 we could surmise that SATM
scores are higher in the population than SATV.
95% CI for mean difference: (7.4157, 31.5611)
c. Based on your p-value what is your conclusion?
P-Value = 0.002 The p-value
is less than 0.05 so we would reject Ho and draw the same conclusion
as stated above in part b.
d. Use the data from the output to calculate the t-statistic by: t =
which matches the t-value from Minitab.
x d - d 19.4884 - 0
=
 3.18
s
89.8075
n
215