Download ht_one

Hypothesis Testing – Proportion and One Mean
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. In Minitab use 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.”
What value is given for the test statistic Z in the output? _______
What is the p-value? ______
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?
1
g. With the data given in part f, use Minitab to do a hypothesis test of whether the “true” success rate for
the new treatment is not or is greater than .65. That is, repeat part c but change the number of trials and
events to 50 and 36 respectively.
What value is given for the test statistic Z 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?
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. In Minitab use 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
2
“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.”
What value is given for Z 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
p-value 
DF =
Conclusion :
b. H0: = 0, Ha:   0, n=40, t=2.41
p-value 
DF =
Conclusion :
c. H0: = 98.6, Ha: < 98.6, n=10, t= 1.33
p-value 
DF =
Conclusion :
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 2005 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
3
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. 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?
4