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Recall the hypothesis test we considered last time in Class Exercise #6(a)-(f) in
Class Handout #3:
6. It is believed that the mean right hand grip strength of men between 20 and 40
years of age in the USA is 86.3 lbs. It is now of interest to perform a hypothesis
test concerning the mean grip strength of men between 20 and 40 years of age in
the country of Techavia.
(a) If we are looking for evidence that the mean grip strength in Techavia is
different from 86.3 lbs., state the null and alternative hypotheses for the
hypothesis test.
H0:  = 86.3 (The mean grip strength is 86.3 lbs.)
H1:   86.3 (The mean grip strength is different from 86.3 lbs.)
(b) Is the hypothesis test one-sided or two-sided?
Since we are looking for evidence that the population mean is different from the
hypothesized value 86.3 in either direction, then the test is two-sided
(c) Describe what it would mean to make a Type I error in this hypothesis test and
what it would mean to make a Type II error in this hypothesis test.
Making a Type I error means the mean grip strength is actually 86.3 lbs., but we
mistakenly conclude that it is different from 86.3 lbs.
Making a Type II error means the mean grip strength is actually different from
86.3 lbs., but we mistakenly conclude that it is equal to 86.3 lbs.
(d) Suppose we plan to measure each right hand grip strength in a random sample of
16 men from Techavia. If we assume that either the grip strengths are normally
distributed or the sample size 16 is sufficiently large so that the sampling
distribution of x is approximately normal, what test statistic would be
appropriate for us to use to decide whether to believe H0 or to believe H1?
x – 86.3
If H0 were true, then
s
—––
16
would be the t-score for x , where df = 15 ,
and we expect this t-score to be within the bounds of random variation.
If H0 were not true, then we would expect the t-score to be outside the bounds
of random variation.
Consequently, we can use this t-score as a test statistic to decide whether to
believe H0 or to believe H1, but we need to choose specific bounds for what
should be considered random variation.
6.-continued
(e) Find the rejection region for the hypothesis test if
(i) a 0.05 significance level were chosen.
 = 0.05
t distribution
with df = 15

— = 0.025
2
1 –  = 0.95

— = 0.025
2
– t0.025 = –2.131
t0.025 2.131
=
The rejection region is defined to be all test statistic values t > 2.131 or t < –2.131 .
(ii) a 0.01 significance level were chosen.
 = 0.01
t distribution
with df = 15

— = 0.005
2
1 –  = 0.99
– t0.005 = –2.947

— = 0.005
2
t0.005 2.947
=
The rejection region is defined to be all test statistic values t > 2.947 or t < –2.947 .
(f) Suppose we actually measure each right hand grip strength in a random sample
of 16 men from Techavia, and we find that x = 91.0 lbs. and s = 7.8 lbs. Find
the test statistic value, and find the p-value for the hypothesis test.
x – 86.3
91.0 – 86.3
The observed test statistic value is t (or t15) =
=
= 2.410
s
7.8
—––
—––
16
16
We now need the definition of a p-value.
Return to the definitions:
p-value (probability value) the probability of obtaining a test statistic value more
supportive of H1 than the test statistic value actually
observed, under the assumption H0 is true
(f) Suppose we actually measure each right hand grip strength in a random sample
of 16 men from Techavia, and we find that x = 91.0 lbs. and s = 7.8 lbs. Find
the test statistic value, and find the p-value for the hypothesis test.
x – 86.3
91.0 – 86.3
The observed test statistic value is t (or t15) =
=
= 2.410
s
7.8
—––
—––
16
16
The p-value is the probability of obtaining a test statistic value more
supportive of H1:   86.3 than the test statistic value actually observed,
under the assumption H0 is true. That is, the p-value is the probability that
x – 86.3
is farther away from zero (0) than the observed test statistic value 2.410.
s
—––
We now see that the p-value must be between 0.02 and 0.05.
16
We denote this by writing 0.02 < p < 0.05.
t distribution
with df = 15
– 2.410
2.410
From Table 3 of the Statistical
Tables, we find that this area
must be between 0.01 and 0.025.
Return to the definition of rejection region for comments on stating the results
of a hypothesis test.
rejection (critical) region a set of test statistic values which lead to rejecting H0 in
favor of H1
(When we find sufficient evidence against H0 in support of H1 , we
say that we “reject H0” or “accept H1”; when we do not find
sufficient evidence against H0 , we say that we “do not reject H0”.)
p-value (probability value) the probability of obtaining a test statistic value more
supportive of H1 than the test statistic value actually
observed, under the assumption H0 is true
6.-continued
(g) What should our conclusion in the hypothesis test be, if
(i) a 0.05 significance level were chosen?
Since the observed test statistic value t (or t15) = 2.410
is in the rejection region corresponding to  = 0.05,
we say that we reject H0 .
–2.131
We can also tell that H0 should be rejected since
p-value <  .
2.131
2.410
Results can be written formally as follows:
Since t15 = 2.410 and t15;0.025 = 2.131, we have sufficient evidence to reject H0 .
We conclude that the mean grip strength in Techavia is different from 86.3 lbs.
(0.02 < p < 0.05). The results suggest that the mean is larger than 86.3 lbs.
(ii) a 0.01 significance level were chosen?
Since the observed test statistic value t (or t15) = 2.410
is not in the rejection region corresponding to  = 0.01,
we say that we do not reject H0 .
–2.947
We can also tell that H0 should not be rejected since
p-value >  .
Results can be written formally as follows:
2.947
2.410
Since t15 = 2.410 and t15;0.005 = 2.947, we do not have sufficient evidence to reject H0 .
We conclude that the mean grip strength in Techavia is not different from 86.3 lbs.
(0.02 < p < 0.05).
statistically significant difference a difference detected by a hypothesis test
clinically significant difference a difference which is judged to be large enough to
have some practical impact
(h) If we conclude from our hypothesis test that the mean grip strength for Techavia
men is significantly different from the mean of 86.3 lbs. for the USA, what
practical importance does this have?
A hypothesis test is capable only of detecting a statistical significance. In this
hypothesis test, the difference between the hypothesized mean of 86.3 lbs. and
the sample mean of 91.0 lbs. suggests that the mean for Techavia is almost about
5 lbs. higher than for the USA. Whether or not this difference is of practical
significance is a matter of judgment.
Four Steps in a Hypothesis Test
Step 1: State the null and alternative hypotheses, and choose a significance level.
Step 2: Collect data, and calculate the value of an appropriate test statistic.
Step 3: Define the rejection region, decide whether or not to reject the null
hypothesis, and obtain the p-value of the test.
Step 4: State the results (which should include the observed test statistic value,
the tabled value which defines the rejection region, the conclusion, and
the p-value), and perform any further analysis which may be required.
one-sample t test about a mean
The H0 states a hypothesized value 0 for a population mean .
The H1 is a statement that the hypothesized value 0 is not correct.
The test statistic is t (sometimes written tn–1) =
x – 0
s
—–
n
When the H0 about a population mean  is rejected, a confidence interval for 
can be a follow up analysis to the hypothesis test.
???????????Class Handout #4 summarizes hypothesis tests and confidence intervals
concerning one mean, a mean difference, and a difference between two means.
Class Handout #4 (Section 1.9, 1.10, material not in text)
Definitions
Statistical Inference Concerning Means
(assuming that each random sample is selected from population with a normal
distribution or that each sample size is sufficiently large)
one-sample t test about a mean 
The H0 states a hypothesized value 0 for a population mean .
The H1 is a statement that the hypothesized value 0 is not correct.
y – 0
The test statistic is t (sometimes written tn–1)
s
=
—–
n
The data consists of one random sample of n quantitative measurements.
one sample confidence interval for a mean 
We can be (1 – )100% confident that the
population mean  is between
s
s
y – t/2 ——
and y + t/2 —— .
n
n
The data consists of one random sample of n quantitative measurements.
1. Forbes magazine published data on the best small firms in 1993. (Forbes,
November 8, 1993, "America's Best Small Companies,"); these were firms with
annual sales of more than $5 million and less than $350 million. The yearly
salaries ($1000s) of the chief executive officer (CEO) for the first 20 firms listed
are as follows:
145 621 262 208 362 424 339 736 291
58
498 643 390 332 750 368 659 234 396 300
This data is stored in the worksheet CEO_Data of the Excel file M214_Data.
A 0.01 significance level is selected to see if there is any evidence that the mean
salary for the CEOs is larger than 300 thousand dollars.
(a) The firms listed in the FORBES DATA will be treated as a simple random
sample of the best small firms. Complete the four steps of the hypothesis test
by completing the table titled Hypothesis Test About Mean CEO Salaries.
Hypothesis Test About Mean CEO Salaries
Step 1
H0:  = 300
H1:  > 300
 = 0.01 (one sided)
n = 20
y = 400.8
y – 0 400.8 – 300
t=
=
= 2.318
s
194.503
These statistics can all be obtained by using the
—–
———–
n
20
Excel spreadsheet named Summary_Statistics,
Step 2
s = 194.503
Step 3
t distribution
with df = 19
do not reject H0
p-value
0.01 < p < 0.025
t0.01 = 2.539
Step 4 Since t19 = 2.318 and t19;0.01 = 2.539, we do not have sufficient evidence to
reject H0. We conclude that the mean CEO salary is not larger than 300
thousand dollars (0.01 < p < 0.025).
Step 1
H0:  = 300
H1:  > 300
 = 0.01 (one sided)
n = 20
y = 400.8
y – 0 400.8 – 300
t=
=
= 2.318
s
194.503
These statistics can all be obtained by using the
—–
———–
n
20
Excel spreadsheet named Summary_Statistics,
Step 2
s = 194.503
Step 3
t distribution
with df = 19
do not reject H0
p-value
0.01 < p < 0.025
t0.01 = 2.539
Step 4 Since t19 = 2.318 and t19;0.05 = 2.539, we do not have sufficient evidence to
reject H0. We conclude that the mean CEO salary is not larger than 300
thousand dollars (0.01 < p < 0.025).
(b) Considering the results of the hypothesis test, decide which of the Type I or
Type II errors is possible, and describe this error.
Since H0 is not rejected, the Type II error is possible, which is concluding
that  = 300 when actually  > 300.
1.-continued
(c) Decide whether H0 would have been rejected or would not have been rejected
with each of the following significance levels: (i)  = 0.05, (ii)  = 0.10.
H0 would be rejected with  = 0.05 and with  = 0.10.
(d) Use SPSS to do the calculations necessary for the hypothesis test and to create
an appropriate graphical display. This data is stored in the SPSS data file ceo.
Section E.6 in the appendix of the textbook illustrates how to use SPSS to do
the calculations necessary for a one-sample t test about a mean. After selecting
the Analyze > Compare Means > One Sample T Test options to display One
Sample T Test dialog box, the variable and the hypothesized mean must be
entered.
Clicking on the Options button allows one to set the confidence level of the
confidence interval for the mean displayed by SPSS.
When we reject H0 in a hypothesis test about , a confidence interval can be
used to estimate . (In the current hypothesis test, we enter a 99% confidence
level, since we used  = 0.01.)
A box plot or histogram would be an appropriate graphical display for one
sample of quantitative measurements.
One-Sample Statistics
N
s alary
20
Mean
400.80
Std. Deviation
194.503
Std. Error
Mean
43.492
the sample size, sample
mean, and sample standard
deviation
s
the estimated standard error of the mean —–
n
One-Sample Test
hypothesized mean
Tes t Value = 300
s alary
t
2.318
df
19
Sig. (2-tailed)
.032
Mean
Difference
100.800
99% Confidence
Interval of the
Difference
Lower
Upper
-23.63
225.23
the t statistic and
degrees of freedom
The p-value displayed on the SPSS output is for a two sided test;
this must be divided by 2 when doing a one sided test.
Consequently, the exact p-value for the hypothesis test in part (a)
is 0.032/2 = 0.016.
2. Measurements of body temperature (BT) in degrees Fahrenheit and heart rate (HR)
in beats per minute were derived from a data set presented in Mackowiak, P. A.,
Wasserman, S. S., and Levine, M. M. (1992), "A Critical Appraisal of 98.6 Degrees
F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl
Reinhold August Wunderlich," Journal of the American Medical Association, 268,
1578-1580. The resulting data is as follows:
Males
BT 96.3
HR 70
96.7
71
96.9
74
97.0
80
97.1
73
97.1
75
97.1
82
97.2
64
97.3
69
97.4
70
97.4
68
BT
HR
97.4
72
97.4
78
97.5
70
97.5
75
97.6
74
97.6
69
97.6
73
97.7
77
97.8
58
97.8
73
97.8
65
BT
HR
97.8
74
97.9
76
97.9
72
98.0
78
98.0
71
98.0
74
98.0
67
98.0
64
98.0
78
98.1
73
98.1
67
BT
HR
98.2
66
98.2
64
98.2
71
98.2
72
98.3
86
98.3
72
98.4
68
98.4
70
98.4
82
98.4
84
98.5
68
BT
HR
98.5
71
98.6
77
98.6
78
98.6
83
98.6
66
98.6
70
98.6
82
98.7
73
98.7
78
98.8
78
98.8
81
BT
HR
98.8
78
98.9
80
99.0
75
99.0
79
99.0
81
99.1
71
99.2
83
99.3
63
99.4
70
99.5
75
Females
BT 96.4
HR 69
96.7
62
96.8
75
97.2
66
97.2
68
97.4
57
97.6
61
97.7
84
97.7
61
97.8
77
97.8
62
BT
HR
97.8
71
97.9
68
97.9
69
97.9
79
98.0
76
98.0
87
98.0
78
98.0
73
98.0
89
98.1
81
98.2
73
BT
HR
98.2
64
98.2
65
98.2
73
98.2
69
98.2
57
98.3
79
98.3
78
98.3
80
98.4
79
98.4
81
98.4
73
BT
HR
98.4
74
98.4
84
98.5
83
98.6
82
98.6
85
98.6
86
98.6
77
98.7
72
98.7
79
98.7
59
98.7
64
BT
HR
98.7
65
98.7
82
98.8
64
98.8
70
98.8
83
98.8
89
98.8
69
98.8
73
98.8
84
98.9
76
99.0
79
BT
HR
99.0
81
99.1
80
99.1
74
99.2
77
99.2
66
99.3
68
99.4
77
99.9 100.0 100.8
79
78
77
A 0.10 significance level is selected to see if there is any evidence that the mean
heart rate for males is different from 72 beats per minute.
2.-continued
(a) The 65 males in the data set will be treated as a random sample. Use SPSS to
do the calculations necessary for the hypothesis test and to create an
appropriate graphical display. Then, complete the four steps of the hypothesis
test by completing the table titled Hypothesis Test About Mean Heart Rate of
Males. The data is stored in the SPSS data file metabolism. Before using the
Analyze > Compare Means > One Sample T Test options in SPSS, we must
first select only the males in that data set as follows:
Select the Data> Select Cases options to display the Select Cases dialogue
box, and select the If condition is satisfied option.
Click on the If button to display the Select Cases If dialogue box.
From the list of variables on the left, select the variable sex, and click on the
arrow button pointing to the right.
Either by use of the buttons in the dialog box or by direct typing, finish the
formula so that it reads sex = 1 .
Click on the Continue button, and click on the OK button, after which you
will now notice that a new variable has been added to indicate which cases are
to be included and which are to be excluded.
In case we reject H0 and want to estimate the mean with a confidence interval,
set the confidence level in SPSS to be 90%, since we have  = 0.10.
A box plot or histogram
would be an appropriate
graphical display for one
sample of quantitative
measurements.
One-Sample Statistics
N
heart_rt
65
Mean
73.37
Std. Deviation
5.875
One-Sample Test
Tes t Value = 72
heart_rt
t
1.879
df
64
Sig. (2-tailed)
.065
Mean
Difference
1.369
90% Confidence
Interval of the
Difference
Lower
Upper
.15
2.59
Std. Error
Mean
.729
Hypothesis Test About Mean Heart Rate of Males
Step 1
H0:  = 72
H1:   72
Step 2
 = 0.10 (two sided)
n = 65
y = 73.37
s = 5.875
t = 1.879
These statistics can all be obtained from the SPSS output.
Step 3
reject H0
t distribution
with df = 64
–1.671
t0.05 = 1.671
p-value
0.05 < p < 0.10
p = 0.065
from the Student’s t
distribution table
from the SPSS output
Step 4 Since t64 = 1.879 and t64;0.05 = 1.671, we have sufficient evidence to reject H0.
We conclude that the mean heart rate for males is different from 72 beats per
minute (0.05 < p < 0.10). The data suggest that the mean heart rate for males is
larger than 72 beats per minute.
or (P = 0.065)
2.-continued
(b) Considering the results of the hypothesis test, decide which of the Type I or
Type II errors is possible, and describe this error.
For the next class, see if you can finish this exercise by answering parts (b) to (e).
(c) Decide whether H0 would have been rejected or would not have been rejected
with each of the following significance levels: (i)  = 0.05 , (ii)  = 0.01 .
(d) Does the difference between the sample mean heart rate and the hypothesized
mean heart rate represent a clinically significant difference? Why or why not?
(e) Considering the results of the hypothesis test, explain why a 90% confidence
interval for the mean heart rate for males would be of interest. Then find and
interpret the confidence interval.
Before submitting Homework #4, check some of the answers (if you haven’t
done so already) from the link on the course schedule:
http://srv2.lycoming.edu/~sprgene/M214/Schedule214.htm