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B. HYPOTHESIS TESTS FOR ONE SAMPLE
1. The Meaning of Hypothesis Testing
8.1, 8.6
2. Steps for Testing a Hypothesis Applied to testing for a Population Mean
8.17† (Assume that 60 is a population standard deviation ), 8.18†, 8.44†, 8.52†
3. The Use of p-value instead of Significance Levels.
p-value for 8.17†, 8.29†*, 8.30†, 8.40†( 5.10   ) , 8.39†, 8.48†, B6
Graded Assignment 2 (Will be posted)
4. Type One and Type Two Errors
8.12†, 8.13†, 8.14*
5. Hypotheses about a Proportion
8.58†, 8.61†, 8.66†
6. The Sign Test
B1, B2, 15.1†, 15.2*, 15.4†, 15.7†, 15.8†*
7. Hypothesis Test for Means - Rare Events
B3, B4
8. Hypothesis Tests for a Variance.
B5
Problem B6: Test the following using (i) a test ratio with a p-value, (ii) a confidence interval for the
sample mean and (iii) a critical value for the sample mean.   .05 .
a. H 0 :   5, H 1 :   5 when n  49 ,  x  8.40 and x  5.92 .
b. H 0 :   5, H 1 :   5 when n  49 ,  x  8.40 and x  5.92 .
c. H 0 :   5, H 1 :   5 when n  49 ,
d. H 0 :   5, H 1 :   5 when n  49 ,
e. H 0 :   5, H 1 :   5 when n  49 ,
f. H 0 :   5, H 1 :   5 when n  49 ,
Solution: From the formula table.
Interval for
Confidence
Interval
Mean (
  x  z 2  x
Known)
Mean (
Unknown)
  x  t 2 s x
 x  8.40 and x  5.92 .
s x  8.40 and x  5.92 .
s x  8.40 and x  5.92 .
s x  8.40 and x  5.92 .
Hypotheses
Test Ratio
H0 :   0
z
H1 :    0
H0 :   0
t
x  0
x
x  0
sx
Critical Value
xcv   0  z 2  x
xcv   0  t 2 s x
H1 :    0
DF  n 1
A p-value is a measure of the credibility of the null hypothesis and is defined as the probability that a test
lower
 low 






statistic or ratio as extreme  as or more extreme  than the observed statistic or ratio could occur,
 high 
 higher 




assuming that the null hypothesis is true.
The rule on p-value:
If the p-value is less than the significance level (alpha) reject the null hypothesis.
If the p-value is greater than or equal to the significance level, do not reject the null hypothesis.
For the following 3 sections  x 
x
n

8.40
49
 1.20 , z 
x  0
x

5.92  5
 0.7667
1.20
a) H 0 :   5, H 1 :   5 when n  49 ,  x  8.40 and x  5.92 .
Solution: This is a one-sided problem where the 'reject' zone is to the right.
(i) Test ratio: We already know that z  0.7667 , In this right-sided problem
pval  Px  5.92   Pz  0.77   Pz  0  P0  z  0.77   .5  .2794  .2206 .
Make diagrams. A diagram for the standardized Normal distribution has zero in the middle. For the pvalue shade the area above 0.77. We already know that is .2206. Since .2206 is above   .05, do not reject
the null hypothesis.
A diagram for the conventional test would show a shaded 'reject' zone above z  z .05  1.645 . Since
0.7667 does not fall in this region, do not reject the null hypothesis.
(ii) The confidence interval has the same direction as the alternate hypothesis, so it has the form
  x  z  x  5.92  1.645 1.20   3.946 . Make a diagram. Put x  5.92 in the middle and shade the
area above 3.946. Since  0  5 is in the confidence interval, do not reject the null hypothesis
(iii) Since we are afraid that the mean may be above 5, the critical value for the sample mean must be
above 5. The formula we use is xcv   0  z  x  5  1.645 1.20   6.974 . Make a diagram. Shade the
area above 6.974. Since x  5.92 does not fall in this 'reject region, do not reject the null hypothesis.
b) H 0 :   5, H 1 :   5 when n  49 ,  x  8.40 and x  5.92 .
Solution: This is a one-sided problem where the 'reject' zone is to the left.
(i) Test ratio: We already know that z  0.7667 , In this left-sided problem
pval  Px  5.92   Pz  0.77   Pz  0  P0  z  0.77   .5  .2794  .7206 .
Make diagrams. A diagram for the standardized Normal distribution has zero in the middle. For the pvalue shade the area below 0.77. We already know that is .7206. Since .7206 is above   .05, do not
reject the null hypothesis.
A diagram for the conventional test would show a shaded 'reject' zone below  z   z.05  1.645 . Since
0.7667 does not fall in this region, do not reject the null hypothesis.
(ii) The confidence interval has the same direction as the alternate hypothesis, so it has the form
  x  z x  5.92  1.645 1.20   7.894 . Make a diagram. Put x  5.92 in the middle and shade the
area below 7.894. Since  0  5 is in the confidence interval, do not reject the null hypothesis
(iii) Since we are afraid that the mean may be below 5, the critical value for the sample mean must be
below 5. The formula we use is xcv  0  z x  5  1.645 1.20   3.026 . Make a diagram. Shade the
area below 3.026. Since x  5.92 does not fall in this 'reject region, do not reject the null hypothesis.
c) H 0 :   5, H 1 :   5 when n  49 ,  x  8.40 and x  5.92 .
Solution: This is a two-sided problem where the 'reject' zone is in both tails of the distribution.
(i) Test ratio: We already know that z  0.7667 , In this two-sided problem, take the probability to the
nearest corner and double it.
pval  2Px  5.92   2Pz  0.77   2Pz  0  P0  z  0.77   2.5  .2794   2.2206   .4412 .
Make diagrams. A diagram for the standardized Normal distribution has zero in the middle. For the pvalue shade the area above 0.77 and the area below -.77. We already know that is .4412. Since .4412 is
above   .05, do not reject the null hypothesis.
A diagram for the conventional test (also with zero in the middle) would show a shaded 'reject' zone above
z  z.025  1.960 and a second 'reject region below  z   z.025  1.960 Since 0.7667 does not fall in
2
2
these regions, do not reject the null hypothesis.
(ii) The confidence interval has the form   x  z  x  5.92  1.96 1.20   5.92  2.352 , or 3.568 to
2
8.272.. Make a diagram. Put x  5.92 in the middle and shade the area between 3.568 and 8.272. Since
 0  5 is in the confidence interval, do not reject the null hypothesis
(iii) The critical value for the sample mean has the formula xcv   0  z  x . So the two critical values
2
are xcv  5  1.960 1.20   5  2.352 or 2.648 to 7.352. Make a diagram. Put  0  5 in the middle.
Shade the areas below 2.648 and above 7.352 to show two 'reject regions. Since x  5.92 does not fall in
either 'reject region, do not reject the null hypothesis.
For the following 3 sections s x 
sx

8.40
 1.20 , df  n  1  48 , t 
n
49



n 1
48
n 1
48
note that t
 t.025
 2.011 .
 t.05  1.677 and t
x   0 5.92  5

 0.7667 , and
sx
1.20
2
d) H 0 :   5, H 1 :   5 when n  49 , s x  8.40 and x  5.92 .
Solution: This is a one-sided problem where the 'reject' zone is to the right.
(i) Test ratio: We already know that t  0.7667 , In this right-sided problem
48
48
pval  Px  5.92   Pt  0.77  . If we look on the t table, we find that t.25
 0.680 and t.20
 0.849 .
This means that Pt  0.680   0.25 and Pt  0.849   0.20 . So the probability that t is above 0.77 is
between .20 and .25. We can say that .20  p  val  .25.
Make diagrams. A diagram for the t distribution has zero in the middle. For the p-value shade the area
above 0.77. We already know that is between .20 and .25. Since these numbers are both   .05, do not
reject the null hypothesis.
A diagram for the conventional test would show a shaded 'reject' zone above t  t.05  1.677 . Since
0.7667 does not fall in this region, do not reject the null hypothesis.
(ii) The confidence interval has the same direction as the alternate hypothesis, so it has the form
  x  t x  5.92  1.677 1.20   3.908 . Make a diagram. Put x  5.92 in the middle and shade the
area above 3.908. Since  0  5 is in the confidence interval, do not reject the null hypothesis
(iii) Since we are afraid that the mean may be above 5, the critical value for the sample mean must be
above 5. The formula we use is xcv  0  t x  5  1.677 1.20   7.012 . Make a diagram. Shade the
area above 7.012. Since x  5.92 does not fall in this 'reject region, do not reject the null hypothesis.
e) H 0 :   5, H 1 :   5 when n  49 , s x  8.40 and x  5.92 .
Solution: This is a one-sided problem where the 'reject' zone is to the left.
(i) Test ratio: We already know that t  0.7667 , In this left-sided problem
pval  Px  5.92   Pt  0.77 . . We already know that the probability that t is above 0.77 is between
.20 and .25. So the probability below t must be between .75 and .80.
Make diagrams. A diagram for the t distribution has zero in the middle. For the p-value shade the area
below 0.77. We already know that it is between .75 and .80. Since both probabilities are above   .05, do
not reject the null hypothesis.
A diagram for the conventional test would show a shaded 'reject' zone below t  t.05  1.677 . Since
0.7667 does not fall in this region, do not reject the null hypothesis.
(ii) The confidence interval has the same direction as the alternate hypothesis, so it has the form
  x  t x  5.92  1.677 1.20   7.932 . Make a diagram. Put x  5.92 in the middle and shade the
area below 7.932. Since  0  5 is in the confidence interval, do not reject the null hypothesis
(iii) Since we are afraid that the mean may be below 5, the critical value for the sample mean must be
below 5. The formula we use is xcv  0  t x  5  1.677 1.20   2.988 . Make a diagram. Shade the
area below 2.988. Since x  5.92 does not fall in this 'reject region, do not reject the null hypothesis.
f.) H 0 :   5, H 1 :   5 when n  49 , s x  8.40 and x  5.92 .
Solution: This is a two-sided problem where the 'reject' zone is in both tails of the distribution.
(i) Test ratio: We already know that t  0.7667 , In this two-sided problem, take the probability to the
nearest corner and double it. pval  2Px  5.92   2Pz  0.77  . We already know that the probability
that t is above 0.77 is between .20 and .25. If we double the probability, we can say that
.40  p  val  .50.
Make diagrams. A diagram for the t distribution has zero in the middle. For the p-value shade the area
above 0.77 and the area below -.77. We already know that the total area is between .40 and .50.. Since these
probabilities are above   .05, do not reject the null hypothesis.
A diagram for the conventional test would show a shaded 'reject' zone above t n 1  t 48  2.011 and a

2
.025
48
second 'reject region below  tn1  t.025
 2.011 Since 0.7667 does not fall in these regions, do not
2
reject the null hypothesis.
(ii) The confidence interval has the form   x  t  x  5.92  2.0111.20   5.92  2.413 , or 3.507 to
2
8.333. Make a diagram. Put x  5.92 in the middle and shade the area between 3.507 and 8.333 Since
 0  5 is in the confidence interval, do not reject the null hypothesis
(iii) The critical value for the sample mean has the formula xcv   0  z  x . So the two critical values
2
are xcv  5  2.0111.20   5  2.413 or 2.587 to 7.413. Make a diagram. Shade the areas below 2.587
and above 7.413 to show two 'reject regions. Since x  5.92 does not fall in either 'reject region, do not
reject the null hypothesis.