Download Activity 10: Confidence Interval for 1 – Proportion

Activity 3: Hypothesis Tests
Obtain the data for this activity on the website. When you click the link, the data should
open up automatically in Minitab.
1. The first column of the data gives observations on the breakdown voltage (kV) for
a particular insulating liquid. Suppose we want to test whether the mean
breakdown voltage for this insulating liquid is more than 53 kV.
a) What are the null and alternative hypotheses?
Ho: μ = 53
Ha: μ > 53
b) Define the parameter (the one in your hypotheses above) in words. Hint:
The answer already appears somewhere above on this page.
μ = mean breakdown voltage for this insulating liquid
c) Perform the rest of the steps of the hypothesis test by hand using α=.05.
Do you reject the null hypothesis? To obtain the summary statistics that
you need, go to Stat > Basic Statistics > Display Descriptive
Statistics…. Enter Breakdown Voltage as the variable. Use the summary
statistics to perform the hypothesis test (Case II in Section 8.2). SHOW
YOUR WORK!
z = (54.708 – 53)/.755 = 2.26
Rejection Region (RR):
z > 1.645 (since z.05=1.645)
The test statistic is in the RR and the p-value < .05 so we can reject the
null. We have enough evidence at the .05 level to conclude that the
mean breakdown voltage for this insulating liquid is more than 53 kV.
d) Now read the following definition, the gist of which you should memorize
before the next exam:
The p-value is the probability, calculated assuming H0 is true, of
obtaining a test statistic at least as supportive of Ha as the statistic
actually observed.
In this case, your Z test statistic should be standard normal assuming H0 is
true. Therefore (because Ha uses >), you just need to find the probability that
a standard normal variable would be greater than your observed z statistic.
Find this p-value by hand, explaining your work:
p-value = P(Z > 2.26) = .0119
d) Now check your work in Minitab. Go to Stat > Basic Statistics > 1
Sample t… Enter Breakdown Voltage as the variable and 53 as the ‘test
mean’. Under ‘Options’, adjust the alternative to match your answer to
part (a). Copy and paste the p-value from Minitab. (Note: This value will
not match exactly what you calculated by hand (for various reasons) but it
should be close.)
Variable
Breakdown Vo
95.0% Lower Bound
53.442
T
2.26
P
0.014
e) Read the alternative definition of p-value at the bottom of p. 346. How
can you tell at a glance, by looking at your p-value, whether to reject H0
using α=.05? Would you have rejected using α=.01? Why or why not?
Since the p-value is smaller than .05, we would reject H0 using α=.05.
Since the p-value is larger than .01, we would not reject H0 using
α=.01.
2. The second column of the data gives responses to the question, “Are you a
member of a fraternity or sorority?” The question was answered by a sample of
PSU students. Assuming this is a random sample of students, we want to test
whether the percentage of PSU students who are part of the Greek system is
different than 20%.
a) What are the appropriate null and alternative hypotheses?
Ho: p = .2
Ha: p ≠ .2
b) Define the parameter (the one in your hypotheses above) in words. Again,
it’s written somewhere above.
p = proportion of all PSU students who are part of the Greek system
c) Perform the rest of the steps of the hypothesis test by hand and obtain a pvalue. SHOW YOUR WORK! (You can obtain the info you need by
selecting Stat > Tables > Tally….) You should also know that all tests of
proportions use Z statistics as long as np0≥10 and n(1-p0) ≥10 and that the
formula for Z appears on p. 340 of the textbook.
Note in this case that your alternative is 2-tailed, which means that your pvalue should be the probability that a standard normal variable is less than
the observed z OR greater than –z. (in other words, by symmetry, you’ll
need to multiply by two after you use Table A.3.)
.2(1  .2)
=.0139
830
z = (.1892 - .2)/.0139 = -0.78
Standard error =
p-value = 2*P(Z < -0.78) = 2(.2177) = .4354
d) Now check your work in Minitab. Go to Stat > Basic Statistics > 1 –
Proportion… and enter “Greek?” as the variable. Under ‘Options’, make
sure to choose the appropriate alternative and null value. Check the box
beside ‘Use test and interval based on normal distribution’. Report the test
statistic and p-value.
Variable
Greek?
X
157
N
830
Sample p
0.189157
95.0% CI
(0.162513, 0.215800)
Z-Value
-0.78
P-Value
0.435
(Note in this case that the p-value is so large that we would not reject H0 for ANY
reasonable value of α.)