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B. HYPOTHESIS TESTS FOR ONE SAMPLE
1. The Meaning of Hypothesis Testing
Text 9.1-9.12 [9.1 – 9.12] (9.1 – 9.12), 9.16** [9.17], (9.17)
2. Steps for Testing a Hypothesis Applied to testing for a Population Mean
Text 9.20, 9.28a,c,d, 9.29 a,c,e [9.18, 9.26a,b,d,e, 9.27a,b,d,e], (9.18, 9.26a,b,d,e, 9.27a,b,d,e)
Text 9.48-9.51, 9.59 [9.46 - 9.49, 9.52] (9.44 – 9.47, 9.50)
3. The Use of p-value instead of Significance Levels.
Text 9.28b, 9.29b [9.26c, 9.27c], B7 (B6), Text 9.34-9.44 [9.32 – 9.39, 9.40*, 9.41*, 9.44] (9.26c, 9.27c, B6, 9.32 – 9.39, 9.40)
[9.40* reads ‘Suppose that in a one-tail hypothesis test where you reject H 0 only in the lower tail, you compute the value of the test
statistic z as 1.38, what is the p-value?’ 9.41 reads ‘In Problem 9.40, what would be your statistical decision if you tested the null
hypothesis at the 0.01 level of significance?’] B8.
Graded Assignment 2 will be posted.
4. Type One and Type Two Errors
Text 9.82-9.86 [9.86 – 9.90] (9.81 – 9.85)
5. Hypotheses about a Proportion
Text 9.66-9.69, 9.70**, 9.72**[9.62 – 9.64, 9.66*, 9.67*], (9.57 – 9.59)
6. The Sign Test
B1*, B2, B3 (B1, B2)
7. Hypothesis Test for Means - Rare Events
B4, B5 (B3, B4)
8. Hypothesis Tests for a Variance.
Text 12.45[9.80] (9.75), B6 (B5)
------------------------------------------------------------------------------------------------------------------
Sections 1-2 are in this document.
Hypothesis Testing
Answers to 9.1-9.12 come from the Instructor’s Solution Manual.
Exercise 9.1: H0 is used to denote the null hypothesis.
Exercise 9.2: H1 is used to denote the alternative hypothesis.
Exercise 9.3:  is used to denote the significance level, or the chance of committing a Type I error.
Exercise 9.4:  is used to denote the consumer’s risk, or the chance of committing a Type II error.
Exercise 9.5: 1    represents the power of a statistical test – that is, the probability of correctly rejecting
the null hypothesis when in reality the null hypothesis is false and should be rejected.
Exercise 9.6:  is the probability of making a Type I error – that is, the probability of incorrectly rejecting
the null hypothesis when in reality the null hypothesis is true and should not be rejected.
Exercise 9.7:  is the probability of making a Type II error – that is, the probability of incorrectly failing to
reject the null hypothesis when in reality the null hypothesis is false and should be rejected.
Exercise 9.8: The power of a test is the complement 1    of the probability  of making a Type II error.
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Exercise 9.9: It is possible to incorrectly reject a true null hypothesis because it is possible for the mean of a
single sample to fall in the rejection region even though the hypothesized population mean is true.
This is a type I error!
Exercise 9.10: It is possible to incorrectly fail to reject a false null hypothesis because it is possible for the
mean of a single sample to fall in the nonrejection region even though the hypothesized population
mean is false. This is a Type II error.
Exercise 9.11: If  goes from 0.05 to 0.01, what happens to  ?  will increase.
Exercise 9.12: Other things being equal, the closer the hypothesized mean is to the actual mean, the larger
is the risk of committing a Type II error.
Exercise 9.16 in 10th edition: The registrar wants to evaluate the assumption that 20 minutes between
scheduled classes is sufficient. Presumably students are polled as to how much time they need so
that x represents the amount of time needed. What are the null and alternative hypotheses?
H0:   20 Average amount of time required is 20 minutes.
H1:   20 Average amount of time required is not 20 minutes.
But please note that the text may be wrong on this one. The unspoken agenda here seems to be that
the time is inadequate. This implies that the alternate hypothesis is H1:   20 This cannot be a
null hypothesis since it does not contain an equality, so the null hypothesis would be H0:   20
Average amount of time required is at least 20 minutes.
Exercise 9.17 in 8th and 9th editions:
H0:   140 Average amounts withdrawn from ATMs are equal to $140.
H1:   140 Average amounts withdrawn from ATMs are not equal to $140.
Tests for a Population Mean
Exercise 9.20 (9.18 in 8th and 9th edition): If you do a 2-tail hypothesis test with a 5% significance level,
what would you decide if the computed value of your test ratio was z  2.21 ?
Since z .025  1.960 and 95% of the values of z are thus between -1.960 and 1.960, our decision rule is to
reject H0 if z  – 1.960 or z  + 1.960. Make a diagram. Draw a “Normal curve centered at zero. Show a
‘do not reject’ zone between -1.960 and +1.960. Now put 2.21 in the diagram.
Decision: Since your value of z = + 2.21 is greater than z cv  z  1.960 , reject H0.
Exercise 9.28 a, c, d (9.26a, b, d, e in 8th and 9th editions): Manufacturer’s specifications say breaking
strength is 70 lbs. with a std. deviation of 3.5 lbs. A sample of 49 pieces gives sample mean
breaking strength of 69.1 lbs.
a) Is there evidence that machine is not meeting specifications?   .05  c) Do a) again with
  1.75. d) Do a) again with x  69 and   3.5 .
Solution: Part a)
(Was a) H 0 :   70 pounds. The cloth has an average breaking strength of 70 pounds.
H 1 :   70 pounds. The cloth has an average breaking strength that differs from 70
pounds.
(Was b) Given:  0  70,   3.5, n  49 , x  69.1 and   .05 .
H 0 :    0
According to Table 3, if we wish to test 
, and  is known, we can use
H 1 :    0
x  0
one of three methods: (i) A test ratio z 
; (ii) a critical value xcv   0  z  x ;
x
2
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or (iii) a confidence interval   x  z  x . Each of these is done below. Each one
2
should give the same conclusion.  x 
(i) Test Ratio: z 
x  0
x


n

3.5
 0.5. .
49
x  70
. Our decision rule is to reject H 0 if
0.5
z  z.025  1.960 or z   z.025  1.960 . Make a diagram showing a Normal curve
with a mean at 0 and shaded 'reject' zones above 1.960 and below -1.960. Since
69 .1  70
z
 1.80 is between the two critical values of z , we do not reject the Null
0.5
hypothesis.
(ii) Critical Value: xcv   0  z  x is the formula for a two sided critical value.
2
So we use xcv   0  z  x  70  1.9600.5  70  0.98 Make a diagram showing a
2
Normal curve with a mean at 70 and shaded 'reject' zones below the critical values 69.02
and above 70.98. Since x  69.1 is between these critical values, we do not reject H 0 .
(iii) Confidence interval:   x  z  x is the formula for a two sided confidence
2
interval. This becomes   69.1  1.960 0.5  69.1  0.98. Make a diagram showing a
Normal curve with a mean at x  69.1 and, to represent the confidence interval, shade the
area between 68.12 and 70.08. Since  0  70 lies within the confidence interval, do not
reject H 0 .
c) What is the answer to a) if the std. deviation is 1.75 lbs?

1.75

 0.25 . If we use the Test Ratio method,
Solution: (was d) Now   1.75. So  x 
n
49
x   0 x  70
z

, our decision rule is still to reject H 0 if z  z.025  1.960 or
x
0.25
z   z.025  1.960 . Use your a diagram showing a Normal curve with a mean at 0 and
69 .1  70
 3.60 is not
0.25
between the two critical values of z , we reject the Null hypothesis.
shaded 'reject' zones above 1.960 and below -1.960. Since z 
Solution:
d) What is the answer to a) if the sample mean is 69 lbs and the std. deviation is 3.5 lbs?
(Was e.)
The easiest method here is the critical value method. We already know
that
xcv   0  z  x is the formula for a two sided confidence interval. So we repeat
2
xcv   0  z  x  70  1.960 0.5  70  0.98 We use our diagram showing a
Normal curve with a mean at 70 and shaded 'reject' zones below the critical values 69.02
and above 70.98. Since x  69.0 is not between these critical values, we reject H 0 .
Note: if you computed z , its value is -2.00.
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Exercise 9.29 a, c, d (9.27a, b, d, e in 8th and 9th editions): Manufacturer’s specs say mean amount of
paint in can is 1 gal. with std deviation of 0.2 gal. For a sample of 50 cans the mean is 0.995 gal.
a) Is there evidence that mean amount is different from 1 at 1% significance level? c) Construct a
95% confidence interval for the mean. d) Compare results of a) and c).
a) (Was a) H 0 :   1.00 . The average amount of paint per one-gallon can is one gallon.
H 1 :   1.00 The average amount of paint per one-gallon can differs from one gallon.
(Was b) Given:  0  1.00,   0.02 , n  50, x  0.995 and   .01 . So
x 

0.02

n
50
 0.002828 . Note that z  z.005  2.576.
2
(i) Test Ratio: z 
x  0
x

x  1.00
. Our decision rule is to reject H 0 if
0.002828
z  z .005  2.576 or z   z.005  2.576 . Make a diagram showing a Normal curve
with a mean at 0 and shaded 'reject' zones above 2.576 and below -2.576. Since
0.995  1.00
z
 1.768 is between the two critical values of z , we do not reject the
0.002828
Null hypothesis. (There is not enough evidence to conclude that the average amount of
paint per one-gallon can differs from one gallon.)
(ii) Critical Value: xcv   0  z  x is the formula for a two sided critical value.
2
So we use xcv   0  z  x  1.000  2.576 0.002828  1.000  0.00728 . Make a
diagram showing a Normal curve with a mean at 1.00 and shaded 'reject' zones below
the critical values 0.993 and above 1.007. Since x  0.995 is between these critical
values, we do not reject H 0 .
c) Repeat a) using a 95% confidence interval. d) Compare the result in c) with the result in a).
(Was d and e) Same decision.   x  z  x is the formula for a two sided confidence interval. This
2
becomes   0.995  2.576 0.002828   0.995  0.00728 . Make a diagram showing a
Normal curve with a mean at x  0.995 and, to represent the confidence interval, shade the
area between 0.998 and 1.002. Since  0  1.00 lies within the confidence interval, do not
reject H 0 .
Exercise 9.48 [9.46 in 9th edition] (9.44 in 8th edition): What is value of t if we are testing H 0 :   50
and s  12 , n  16, x  56 . We are now assuming that  is unknown.
Since the null hypothesis is H 0 :   50 , we have  0  50, s  12 , n  16, x  56 . Thus
sx  s
n
 12
16
 3 and t 
x   0 x   0 56  50 6


  2.00 .
s
12
sx
3
16
n
Exercise 9.49 [9.47 in 9th edition] (9.45 in 8th edition: What is the number of degrees of freedom in 9.48?
df = n – 1 = 16 – 1 = 15
Exercise 9.50 [9.48 in 9th edition] (9.46 in 8th edition): What are the critical values of t in 9.48 if   .05
and the alternate hypothesis is (a) H 1 :   50 H 1 :   50 ?
We use t n 1 or  t n1 for a one-tailed test and t n 1 for a two-tailed test. n  1  15.



2
  .05 . (In these problems the text used ‘crit’ for critical values of t from the t table and ‘calc’ for
the value of t calculated in 9.48)
15
 2.131 .
(a) H 1 :   50 For a two-tailed test with a 0.05 level of confidence, tcrit = t .025
15
 1.753 .
(b) H 1 :   50 For an upper-tailed test with a 0.05 level of confidence, tcrit = t .05
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Exercise 9.51 [9.49 in 9th] (9.47 in 8th edition): What is the conclusion if the alternate hypothesis is (a)
H 1 :   50 (b) H 1 :   50 ?
From the Instructor’s Solution Manual and using the test ratio method,
(a) Since tcalc = 2.00 is between the critical bounds of tcrit =  2.131, do not reject H0.
(b) Since tcalc = 2.00 is above the critical bound of tcrit = + 1.753, reject H0.
Exercise 9.54 [9.52 in 9th] (9.50 in 8th edition): Problem Summary: From a sample of 100 students we find
that the mean expenditure for books is $315.40 with a standard deviation of $43.20. In a) you
are asked to test whether the (population) mean expenditure is above $300 using 10%
significance level. b) What if   .05 and s  75 ? c) What if   .10 , x  305 .11 and
s  43 .20 .
Solution: Since the problem asks if the mean is above $300   300  , and this does not contain an equality, it
must be an alternate hypothesis. Our hypotheses are H 0 :   300 , (The average cost of
textbooks per semester at a large university is no more than $300.) and H 1 :   300 (The
average cost of textbooks per semester at a large university is more than $300.) This is a
right-sided test.
(a)
Given:  0  300 .00, s  43 .20, n  100 , df  n  1  99, x  315 .40 and
  .10 . (   .01 in the 8th edition.) So s x 
s

n
43 .20
 4.320 . Note that
100

n 1  t 99  2.365 in the 8th edition.)
tn 1  t ..99
.01
10  1.290 . ( t 
x   0 315 .40  300 .00

 3.5648 . The larger the sample mean is,
sx
4.320
the more positive will be this ratio. We will reject the null hypothesis if the ratio is larger

than tn 1  t ..99
10  1.290 . Make a diagram showing a Normal curve with a mean at 0
(i) Test Ratio: t 
and a shaded 'reject' zone above 1.290. Since the test ratio is above 1.290, we reject H 0 .
(For the 8th edition, the critical bound is 2.365 and we reject H 0 .)
(If you are reviewing and wish to use a p-value approach, since this is a right-side test,
the p-value is the probability of getting a value above 315.40 when the null hypothesis is
true. If we use the test ratio above we get. pval  Px  315 .40   Pt  3.5648  . To find
this approximately, go to the df  99 line of the t table. 3.5648 is not between any

values, but we can see that t ..99
001  3.175 . This means that Pt  3.175   .001 . Since
3.5648 is even further away from zero than 3.175, the probability above it must be even
lower than .001. We conclude that pval  .001 . Since the p-value is below   .10 , we
reject the null hypothesis.
(ii) Critical value: We need a critical value for x above $300. Common sense says that if
the sample mean is too far above $300, we will not believe H 0 :   300 . The formula
for a critical value for the sample mean is x    t n1 s , but we want a single value
cv
0

2
x
above 300, so use xcv   0  tn1 s x  300 .00  1.290 4.320   305 .57 . Make a
diagram showing an almost Normal curve with a mean at 300 and a shaded 'reject' zone
above 305.57. Since x  315 .40 is above 305.57, we reject H 0 . (In the 8th edition, the
critical value is x    t n1 s  300 .00  2.365 4.320   310 .22 , so we reject H )
cv
0

x
0
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(iii) Confidence interval:   x  t  s x is the formula for a two sided interval. The rule
2
for a one-sided confidence interval is that it should always go in the same direction as the
alternate hypothesis. Since the alternative hypothesis is H 1 :   300 , the confidence
interval is   x  t n1 s  315 .40  1.290 4.320   309 .83 . Make a diagram

x
showing an almost Normal curve with a mean at x  315 .40 and, to represent the
confidence interval, shade the area above 309.83 in one direction. Then, on the same
diagram, to represent the null hypothesis, H 0 :   300 , shade the area below 300 in the
opposite direction. Notice that these do not overlap. What the diagram is telling you is
that it is impossible for   309.83 and H 0 :   300 to both be true. So we reject H 0 .
(If we use the numbers in the 8th edition, the confidence interval is   x  t n1 s

 315 .40  2.365 4.320   305 .18 with the same results.)
(b)
x
Summary: Repeat a) with a 5% significance level and a sample standard deviation of $75.
Our hypotheses are still H 0 :   300 , (The average cost of textbooks per semester at a
large university is no more than $300.) and H 1 :   300 (The average cost of textbooks
per semester at a large university is more than $300.) Given:  0  300 .00, s  75 .00,
n  100 , df  n  1  99, x  315 .40 and
  .05 . So s x 

Note that tn 1  t ..99
05  1.660 .
If we use the Test Ratio method t 
s

n
75 .00
 7.500 .
100
x   0 315 .40  300 .00

 2.0533 . We will reject
sx
7.500

the null hypothesis if the ratio is larger than tn 1  t ..99
05  1.660 . Make a diagram
showing a Normal curve with a mean at 0 and a shaded 'reject' zone above 1.660. Since
the test ratio is above 1.290, we reject H 0 . ( There is enough evidence to conclude that
the average cost of textbooks per semester at a large university is more than $300.)
(For the p-value approach, note that 2.0533 is between 1.984 and 2.365 on the df  99
line of the t table, since these are the .025 and .01 values, we can say .01  pval  .025 .
Since this p-value is below the significance level, we reject the null hypothesis.
(c)
Problem summary: Repeat a) with a 10% significance level and a sample standard deviation
of $43.20.
H 0 :   300 . Given:  0  300 .00, s  43 .20, n  100 , df  n  1  99, x  305 .11 and
  .10 . (   .01 in the 8th edition.) So s x 
s
n

43 .20
 4.320 . Note that
100

n 1  t 99  2.365 in the 8th edition.) It is probable easiest to use a
tn 1  t ..99
.01
10  1.290 . ( t 
critical value for the sample mean. We already know x    t n1 s  305 .57 . Since
cv
0

x
x  305 .11 is not above 305.57, we do not reject H 0 . (In the 8th edition, the critical value is
x    t n1 s  310 .22 , so we do not reject H ) There is not enough evidence to
cv
0

x
0
conclude that the average cost of textbooks per semester at a large university is more than
$300.