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Hypothesis Testing – Proportion and One Mean
Data Set Needed: Class Survey
NOTE: for those using SPSS you will need to do hand calculations in parts 1 and 2.
P-value Guidelines when using Standard Normal Table (i.e. the Z-table):
Keep this in mind: The method for finding the p-value is based on the alternative hypothesis.
Minitab will provide the p-value. If doing by hand, then find p-value from Standard Normal
Table by observe the following methods:
For Ha: p ≠ po then the p-value = 2P(Z ≥ |z|) That is, find 1 – P(Z < |z|) and then
multiply this p-value by 2.
For Ha: p > po then the p-value = P( Z ≥ z)
For Ha: p < po then the p-value = P( Z ≤ z)
1 The present success rate in the treatment of a particular psychiatric disorder is 0.65 (65%). A research
group creates a new treatment for this disorder. Their null hypothesis is that the success rate for the new
treatment is 0.65 (no different from the standard). The alternative hypothesis is that the success rate is
better than 0.65 for the new treatment.
a. Let p = true success rate of the new treatment. Using statistical notation, write null and alternative
hypotheses about p.
b. A clinical trial is done in which 144 of 200 patients who use the new treatment are successfully treated.
What is the value of p̂ = success rate for the sample? How does it compare to 0.65 (the old standard)?
c. Minitab Users: Go to Stat>Basic Stats>1 proportion, click Summarized Data, enter 200 for
number of trials and 144 for Number of events. Click on Perform Hypothesis Test and enter 0.65 where
it says “Hypothesized proportion” AND click Options to select the alternative hypothesis as “greater
than” AND also click on “Use test and interval based on normal distribution.”
SPSS Users: Open the Excel Summarized Procedures and select the tab “Z test of a Proportion”.
Enter 0.72 as the Sample Proportion; 200 for the Sample Size; and 0.65 for the Hypothesized Proportion.
NOTE: The resulting p-value is for a two-sided test (i.e. “not equal” alternative hypothesis – if you are
conducting a one-sided test, where your alternative is specified as either “less than” or “greater than” you
will need to cut this p-value in half to arrive at the proper p-value for the one-sided test).
What value is given for the test statistic in the output? _______
What is the p-value? ______
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d. Decide between the null hypothesis and the alternative hypothesis. Explain your decision.
e. Write a conclusion about how the new treatment compares to previous treatment(s).
f. Suppose the data had been that 50 patients used the new treatment, with 36 successes. What is the value
of p̂ = success rate for this sample? How does it compare to the success rate for the sample used in parts
b-e?
g. With the data given in part f, conduct a hypothesis test of whether the “true” success rate for the new
treatment is greater than .65. That is, repeat part c but change the number of trials and events to 50 and
36 respectively. For Excel Summarized Procedures just change the Sample Size from 200 to 50.
What value is given for the test statistic in the output? _______
What is the p-value? ______
h. Refer to the previous two parts. Decide between the null hypothesis and the alternative hypothesis.
Explain your decision.
i. For the trial with only 50 patients (and 36 success), write a conclusion about how the new treatment
compares to previous treatment(s).
j. Briefly explain what this activity illustrates about how sample size affects the statistical significance of
an observed result. As a starting points, note that the observed success rate was .72 for both samples, and
we wish to determine if this is “significant” evidence that the true proportion is greater than .65.
2 In a marketing survey for an automobile manufacturer, 90 randomly selected adults are asked which car
color they would choose, if a particular car were available in either blue or red body colors. Of the 90
respondents, 53 said “blue.”
a. Let p = population proportion that would say “blue.” The manufacturer wants to learn if a majority of
buyers would pick blue. Keeping in mind that a majority is p>0.5, write a null and alternative hypothesis
about p in this situation. (Hint: What somebody wants to “prove” is usually the alternative.)
H0:
Ha:
b. What is the value of p̂ =sample proportion that picked blue?
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c. Test the hypotheses stated in part a above. By hand, calculate the test statistic by using (Notice that this
statistic is sensitive to the difference between the sample result and the null hypothesis value):
z
sample p - null hyp. p

null std. error
p̂  p 0
p 0 (1  p 0 )
n
(round your final value to two decimal places)
d. Use the Standard Normal Table to find the p-value associated with this test statistic. Use the p-value
guidelines found at the beginning of this activity.
e. Minitab users: Go to Stat>Basic Stats>1 proportion, click Summarized Data, enter 90 for number
of trials and 53 for Number of events. Click on Perform Hypothesis Test and enter 0.50 where it says
“Hypothesized proportion” AND click Options to select the alternative hypothesis as “greater than” AND
also click on “Use test and interval based on normal distribution.”
SPSS Users: Open the Excel Summarized Procedures and select the tab “Z test of a Proportion”.
Enter 0.59 as the Sample Proportion; 90 for the Sample Size; and 0.50 for the Hypothesized Proportion.
NOTE: The resulting p-value is for a two-sided test (i.e. “not equal” alternative hypothesis – if you are
conducting a one-sided test, where your alternative is specified as either “less than” or “greater than” you
will need to cut this p-value in half to arrive at the proper p-value for the one-sided test).
What value is given for test statistic in the output? _______
What is the p-value? ______
i. Do the Z test statistic you found by hand in part c and the p-value from part d approximately
equal to the Z statistic found in part e when using Minitab?
ii. Decide whether the result is significant based on the p-value from Minitab and report a
conclusion in the context of this situation.
iii What would the p-value have been if manufacturer wanted to test if a equal proportion of red
and blue cars would be purchased? That is, test Ho: p = 0.5 versus Ha: p ≠ 0.5
3 Use the T-Table 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
DF =
p-value 
Conclusion :
b. H0: = 0, Ha:   0, n=40, t=2.41
DF =
p-value 
Conclusion :
c. H0: = 98.6, Ha: < 98.6, n=10, t= 1.33
DF =
p-value 
Conclusion :
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d. H0: = 100, Ha: > 100, n=16, t= 4.26
DF =
p-value 
Conclusion :
4 PSU claims that the average SATM score for the incoming fall class at University Park was
approximately 610. Use the Class Survey data in the Datasets folder 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:
Ha:
b. Calculate DF and the t-statistic:
DF =
t=
x - 0
=
s
n
c. From the T-table what is the range of the p-value based on you t-statistic? That is, first find between
what two t-values does this test statistic fall, and then find the tail probability area for each of these
values. NOTE: if you selected a two-sided Ha (i.e. used ≠) then you need to double the p-values found in
the table. That is, first find between what two t-values does this test statistic fall, and then find the tail
probability area for each of these values.
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 vice-versa.
e. Use software to verify your results. Do your results by hand and those from Minitab roughly match?
Minitab Users: 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?
SPSS Users: Go to Analyze > Compare Means > Ones Sample T Test. Select SATM and move to the
Test Variables box. Enter 610 in the box for Test Mean. Click Options and verify that 95 is the
Confidence Level Percentage. Click Continue then OK.
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