Download One Sample Tests of Hypothesis Problems and Answers for Chapter 6

Problems and Answers for Chapter 6
One Sample Tests of Hypothesis
1. A plant breeder has been selecting Crispin apple trees for many years in an attempt to decrease
variability in fruit size. When he began selecting, the variance was 400 g2 for mature fruit.
A random sample of 30 apples from his most recent crop had a variance of 300 g2 . Has the
breeder been successful in making Crispin apples less variable? What is the expected value
of the test statistic here, if H0 is true? Find the P value of this test.
1. No. The hypotheses are H0 : σ 2 ≥ 400 g2 versus Ha : σ 2 < 400 g2 . E(χ2 ) = n − 1 = 29 and
the actual χ2 = 21.75, so 0.10 < P < 0.25. The chi-square statistic is not significantly less
than expected under H0 .
2. As part of a benthic community survey of Lady Elliot Island, sixteen sea stars, Linckia laevigata, were collected and their longest arm was measured to the nearest tenth of a centimeter.
Assume normality for this population.
10.3
12.1
11.0
9.4
10.5
11.3
10.0
12.0
11.3
11.5
14.5
9.3
13.0
10.1
12.1
7.6
(a) For these data calculate the sample mean and variance.
(b) Calculate the 95% confidence interval for the population mean. In your own words explain
what this confidence interval represents.
(c) Use a t test with α = 0.05 to determine whether the mean is significantly different from
12 cm. Discuss the relationship between the confidence interval in (b) above and this
two-tailed t test.
(d ) Calculate the 95% confidence interval for the population variance. Again, in your own
words, explain what this confidence interval means.
(e) Use a χ2 test with α = 0.05 to determine whether the variance is significantly different
from 5.0 cm2 . Discuss the relationship between the confidence interval in (d ) above and
this two-tailed χ2 test.
Problems and Answers for Chapter 6. One Sample Tests of Hypothesis
40
2. (a) X = 11.00 cm, s2 = 2.644 cm2 , and n = 16.
(b) C(10.13 cm ≤ µ ≤ 11.87 cm) = 0.95. I am 95% confident that the parametric mean µ for
arm length lies between 10.13 cm and 11.87 cm
(c) The hypotheses are H0 : µ = 12 cm versus Ha : µ = 12 cm. t = −2.46 and the c.v. =
±2.131. Since −2.46 < −2.131, reject H0 The confidence interval in (b) uses ±2.131 √sn to
generate limits. Algebraically, the two-tailed test of hypothesis at α and the confidence
interval at 1 − α are equivalent.
(d ) C(1.442 cm2 ≤ σ 2 ≤ 6.335 cm2 ) = 0.95. I am 95% confident that the parametric variance,
σ 2 , for arm length lies between 1.442 cm2 and 6.335 cm2 .
(e) The hypotheses are H0 : σ 2 = 5.0 cm2 versus Ha : σ 2 = 5.0 cm2 . Calculate χ2 = 7.932.
The c.v.’s = 6.26, 27.5. Since 6.26 < 7.932 < 27.5, accept H0 . The variance cannot be
shown to be significantly different from 5.0 cm2 . Again algebraically, the two-tailed test of
hypothesis at α and the confidence interval at 1 − α are equivalent. The sample variance
of 5.0 cm2 lies within the confidence limits at 95%.
3. In healthy adult males the sciatic nerve conduction velocity values are normally distributed
with a mean of 65 cm/msec and a standard deviation of 5 cm/msec. The conduction velocities
of 16 subjects admitted to the poison control center of a metropolitan hospital with a diagnosis
of methylmercury poisoning had a mean conduction velocity of 55 cm/msec and a variance of
49 (cm/msec)2 .
(a) Do these data provide sufficient evidence to indicate that the conduction velocities are
significantly slower in the poisoned individuals?
(b) Do these data provide sufficient evidence to indicate that poisoned individuals are more
variable in their sciatic nerve conduction velocities than normal individuals?
3. (a) The hypotheses are H0 : µ ≥ 65 cm/msec versus Ha : µ < 65 cm/msec. Using α = 0.05,
the c.v. = −1.753. But t = −5.71. Since −5.71 < −1.753, reject H0 . Or note that the
P value < 0.0005. The average conduction velocity in poisoned individuals is significantly
slower than in normal individuals.
(b) The hypotheses are H0 : σ 2 ≤ 25 (cm/msec)2 versus Ha : σ 2 > 25 (cm/msec)2 . Using
α = 0.05, the c.v. = 25.0. But χ2 = 29.4 (and 0.01 < P value < 0.025). Since 29.4 > 25.0,
reject H0 . Conduction velocities are significantly more variable in individuals who have
methylmercury poisoning.
4. A television documentary on overeating claimed that adult American males are on average
15 lb overweight. To test this claim, 25 randomly selected adult males were examined, and
the average excess weight was found to be 18 lb with a standard deviation of 10 lb. Do these
data give us reason to believe the claim of 15 lb is incorrect? What type of error could you
have made here? Explain.
4. The hypotheses are H0 : µ = 15 lb versus Ha : µ = 15 lb. Use α = 0.05 with c.v.’s = ±2.064.
Calculate t = 1.5 (and 0.10 < P value < 0.20). Since −2.064 < 1.5 < 2.064, accept H0 . The
sample mean is not sufficiently different from the claim of 15 lb to reject H0 . By accepting
H0 we may be making a Type II Error—accepting a false null hypothesis. Probability of this
type of error is unknowable.
Problems and Answers for Chapter 6. One Sample Tests of Hypothesis
41
5. Before purchasing a large corn farm you wish to determine its profitability by analyzing its
productivity. One claim made by the present owner is that the fields are very uniform in their
yield per acre. In fact, it is claimed that the standard deviation in bushels per acre of corn
yield is 5. A random sample of 25 one-acre plots yields an average of 140 bushels/acre and
a standard deviation of 6.5 bushels/acre. Are the fields significantly more variable than is
claimed? What type of error could you have made here? Explain.
5. The hypotheses are H0 : σ 2 ≤ 25 (bushels/acre)2 versus Ha : σ 2 > 25 (bushels/acre)2 . Here
n = 25, and s = 6.5 bushels/acre. Using α = 0.05, the c.v. = 36.4. But χ2 = 40.56. Since
40.56 > 36.4, reject H0 . Evidence suggests the yields are more variable than claimed. Type I
error is possible here. 0.01 < P value < 0.025.
6. The data below represent the systolic blood pressures (in mmHg) of 14 patients undergoing
drug therapy for hypertension. Assuming normality of systolic blood pressures, on the basis
of these data can you conclude that the mean is significantly less than 165 mmHg?
183
194
152
163
178
114
157
178
194
152
163
118
144
158
6. The hypotheses are H0 : µ ≥ 165 mmHg versus Ha : µ < 165 mmHg. t = −0.671 and the
P value = 0.2550. So accept H0 ; we cannot conclude the mean systolic blood pressure is
significantly less than 165 mmHg.
7. Analysis of the amniotic fluid from a random sample of 16 pregnant women early in gestation
yielded the following measurements of uric acid present (in mg/100 ml). Do these data provide
evidence to indicate that the population variance is greater than 0.92 (mg/100 ml)2 ?
4.2
4.7
3.3
4.7
3.5
3.0
3.1
4.8
3.8
5.5
3.3
4.0
3.5
3.6
4.0
7.6
7. The hypotheses are H0 : σ 2 ≤ 0.92 (mg/100 ml)2 versus Ha : σ 2 > 0.92 (mg/100 ml)2 . s2 =
1.15 (mg/100 ml)2 . χ2 = 21.67 and the P value = 0.117 > α = 0.05. Accept H0 , there is not
sufficient evidence to claim the variance is greater than 0.92 (mg/100 ml)2
8. In one population of patients, the mean hemoglobin level is 13.5 g/100 ml of whole blood.
The standard deviation of this variable is 1.4 g/100 ml of whole blood. A series of nine tests
on a patient revealed a sample standard deviation of 2.1 g/100 ml of whole blood. Is the
hemoglobin level in this patient significantly more variable than normal?
8. The hypotheses are H0 : σ 2 ≤ 1.4 (g/100 ml])2 versus Ha : σ 2 > 1.4 (g/100 ml)2 . Here
s = 2.1 g/100 ml and n = 9. Testing at α = 0.05, the c.v. = 15.5. But χ2 = 18.0. Since
18.0 > 15.5, reject H0 . The patient’s hemoglobin level is significantly more variable than
normal.
9. A geneticist, studying flower color in Mirabilis jalapa, the four-o’clock flower, crossed pink
and white individuals together and predicted 50% white-flowered offspring and 50% pinkflowered offspring. Of the 100 seeds collected and germinated, 60 resulted in pink-flowered
plants and 40 resulted in white-flowered plants. Was this result significantly different from
the expectation? (Use a normal approximation here.)
Problems and Answers for Chapter 6. One Sample Tests of Hypothesis
42
9. The hypotheses are H0 : p = 0.50 versus Ha : p =
0.50. From the Z distribution, the c.v. =
60
±1.960. pˆ = 100
= 0.60.
0.10
0.60 − 0.50
=
Z=
= 2.0.
0.05
0.50(1−0.50)
100
Since 2.0 > 1.960, reject H0 . The result is significantly different from the expectation.
10. Recently there have been concerns about the effects of phthalates on the development of the
male reproductive system. Phthalates are common ingredients in many plastics. In a pilot
study a researcher gave pregnant rats daily doses of 750 mg/kg of body weight of DEHP (di-2ethylhexyl phthalate) throughout the period when their pups’ sexual organs were developing.
The newly born male rat pups were sacrificed and their seminal vesicles were dissected and
weighed. Below are the weights for the eight males (in mg).
1710
1630
1580
1670
1350
1650
1600
1650
If untreated newborn males have a mean of 1700 mg, can you say that rats exposed to
DHEP in utero have a significantly lower weight?
10. The hypotheses are H0 : µ ≥ 1700 mg versus Ha : µ < 1700 mg. For a t test, the c.v. = −1.895.
Calculate that t = −2.44. Since −2.44 < −1.895, reject H0 . Exposure to DEHP significantly
decreases seminal vesicle weight.
11. Suppose that for years an investigator has used a stock of inbred rats whose weights have
σ = 30 g. He is considering switching to a cheaper source of supply of rats, except he suspects
that the new rats will show greater variability in weight. Measurement of a random sample of
20 rats from the cheaper source yielded a corrected sum of squares of 30,400 and a standard
deviation of 40 g. Statistically confirm or refute the investigator’s suspicions.
11. The hypotheses are H0 : σ 2 ≤ 900g2 versus Ha : σ 2 > 900g2 . At α = 0.05, the c.v. = 30.1.
χ2 = 33.78, so reject H0 . There is statistical evidence that the new rats are more variable
than the old stock.
12. (a) The ancient Greeks used the term
“golden rectangle” to describe those rectangles whose
√
1+ 5
height-to-width ratio is 1 : 2 ≈ 0.618034. They found such rectangles especially
pleasing to the eye and often used them in architectural design (e.g., the Parthenon).
Other artists and artisans, including musicians, have made explicit use of this ratio in
their work. The ratio, unbelievably, arises in biological growth processes. For example,
there are claims that the ratio of navel height-to-total height is this same golden ratio.
The data below are breadth-to-length ratios of beaded rectangles used by the Shoshoni
Indians to decorate leather goods. Is it reasonable to assume that they are implicitly using
golden rectangles? Use the sign test here. (HSDS, #150)
0.693
0.749
0.654
0.670
0.662
0.672
0.615
0.606
0.690
0.628
0.668
0.611
0.606
0.609
0.601
0.553
0.570
0.844
0.576
0.933
(b) Find a 95% confidence interval for the median height-to-width ratio.
Problems and Answers for Chapter 6. One Sample Tests of Hypothesis
43
12. (a) The hypotheses are H0 : M = 0.618034 versus Ha : M = 0.618034. Use the sign test.
S− = 9, S+ = 11, so the test statistic is S− = 9 for the two-tailed test. F (9) = 0.4119
from Table C.1. Since 0.4119 > 0.025 accept H0 . Shoshoni Indians appear to be using the
golden rectangle in their bead work.
(b) A 95% confidence interval is [0.606, 0.672].
13. The wall skink, Cryptoblepharus virgatus, is a common insectivore in the Brisbane area. They
can be seen sunning themselves on vertical surfaces such as the brick walls of buildings. They
can also be heard scurrying into the leaf litter as one walks along a path. Suppose we suspect
that their median length (including tail) is 7 cm. Test the hypothesis H0 : M = 7 cm versus
Ha : M = 7 cm at the 0.05 level. The data consist of a random sample of 66 skinks that were
captured and measured: 41 have length greater than 7 cm, 23 are shorter than 7 cm, and 2
have length equal to 7 cm.
13. The hypotheses are H0 : M = 7 cm versus Ha : M = 7 cm. There are n = 66 observations
with 41 larger than 7 cm, 23 shorter than 7 cm, 2 equal to 7 cm. Use the sign test with
n = 64 usable observations. S− = 23, S+ = 41, so the test statistic for a two-tailed test is
S− = 23. Test with α = 0.05. Use a normal approximation with µ =
F (23) = P (S ≤ 23) = P
23.5 − 32
Z≤
4
n
2
= 32 and σ =
64
4 .
= P (Z ≤ −2.13) = 0.0166.
Since 0.0166 < 0.025, reject H0 . Evidence suggests the median is different (larger) than 7 cm.
14. (a) The angle that a spider’s web makes with a line vertical to the earth’s surface varies by
species. For example, for Isoxya cicatricosa this angle has been shown to have a median
of 28.12◦ . The angles for ten webs were observed. Based on these data can you reject the
hypothesis that I. cicatricosa constructed these webs? (HSDS, #194)
25◦
12◦
31◦
26◦
17◦
15◦
24◦
10◦
16◦
12◦
(b) Find a 95% confidence interval for the median angle.
14. (a) The hypotheses are H0 : M = 28.12◦ versus Ha : M = 28.12◦ . Test with α = 0.05. S− = 9,
S+ = 1, so the test statistic for a two-tailed test is S+ = 1. F (1) = 0.0107 from Table C.1.
Since 0.0107 < 0.025, reject H0 . The webs don’t appear to be made by I. cicatricosa.
(b) A 95% confidence interval is [12, 26].
15. These data give pulse rates (in beats/minute) of 39 Peruvian Indians. Construct a 95%
confidence interval for the median pulse rate. (HSDS, #345)
88
68
72
64
√
76
68
88
60
84
72
80
60
64
76
76
72
60
72
64
92
64
52
72
80
60
72
60
72
64
64
76
64
68
60
88
68
74
56
72
n
= 12.88 ≈ 13. A 95% confidence interval is [64, 72]. More conservatively, with
15. d = n−1−1.96
2
d = 12, the interval is [64, 74].
Problems and Answers for Chapter 6. One Sample Tests of Hypothesis
44
16. A Secchi disc (see figure below) is a simple but effective device to measure water transparency.
The disc is attached to a rope and slowly lowered through the water until it just disappears
from view. This depth is noted. It is lowered some more, then slowly raised until the disc
is just observed. This depth is also noted. The Secchi depth is the average of these two
depths. The depth is proportional to water transparency and inversely proportional to water
turbidity, but it is also influenced by cloudiness, position of the sun, roughness of the water,
and observer bias. Plankton biomass and suspended mud particles are the main contributors
to water turbidity. In Seneca Lake (New York), the depth is primarily a function of plankton
biomass. In 1997, the reported median Secchi depth at a station at the north end of the lake
was 8.0 m. The following depths were recorded on 16 similar dates in 1999. The question
is whether there has been a change in Secchi depth since 1997. Use a sign test. (Based
on data reported in John Halfman, http://people.hws.edu/Halfman/Zebras/Zebras.htm,
July, 2000.)
.............
....... ..................................................................
...........................................
....
......................................................
...
..................................................................................................................
........................................................................
..
..
..................................................................
..................................................... .......
............................... .........
....
9.0
9.0
10.5
8.0
9.2
7.0
8.3
6.0
7.1
4.7
7.8
6.0
7.9
6.5
8.3
7.4
16. The hypotheses are H0 : M = 8 m versus Ha : M = 8 m. Notice that n = 15 since one of
the observations equals 8.0 m. The test statistic is the minimum of S+ = 6 and S− = 9.
F (6) = 0.3036 and since this is a two-tailed test, the P value is 0.6072. So we accept H0 .
There does not appear to be a change in the median Secchi depth from 1997.
17. The fasting blood glucose levels of nine women who had been using oral contraceptives for at
least six months prior to sampling are listed below. It is thought that the mean fasting blood
glucose level in women not using contraceptives is 3.8 mmol/l. Test whether there is evidence
that the use of contraceptives raises glucose levels. Assume symmetry.
4.71
3.61
4.92
4.85
4.66
3.92
5.10
3.55
4.78
17. The hypotheses are H0 : M ≤ 3.8 mmol/L versus Ha : M > 3.8 mmol/L. Use the Wilcoxon
signed-rank test with n = 9. The test statistic is W− = 5 (see table below). F (5) = 0.0195
so reject H0 . Data support the hypothesis that median fasting blood glucose level exceeds
3.8 mmol/L.
Xi − 3.8
Signed rank
0.91
5
−0.19
−2
1.12
8
1.05
7
0.86
4
0.12
1
1.30
9
−0.25
−3
0.98
6
18. The carpet python, Morelia spilota variegata, is widely distributed and is common in rainforests, wet or dry eucalypt forests, and pastoral and urban areas. They locate warm-blooded
prey by means of heat-sensory pits located along the lower jaw that can detect temperature changes of less than one-thirtieth of a degree (◦ C). Their mean length is approximately
250 cm. Because of an extended drought from 1994 through early 1996 at Lamington National
Park, it was thought that the mean lengths of these snakes would be smaller than 250 cm.
A survey was conducted and the lengths (in cm) below were recorded. Clearly state the null
and alternative hypotheses. Assuming that lengths of snakes are symmetrically distributed
about the mean, carry out an appropriate test of the hypothesis at the 0.05 level. Interpret
the result.
Problems and Answers for Chapter 6. One Sample Tests of Hypothesis
269
221
204
258
211
198
276
223
45
207
242
18. The hypotheses are H0 : M ≥ 250 cm versus Ha : M < 250 cm. Use the Wilcoxon signed rank
test with n = 10 and α = 0.05. The test statistic is W+ = 8.5 (see below). The corresponding
P value from Table C.6 is between 0.0244 and 0.0322. So P < 0.05 = α, reject H0 . The
median is significantly less than 250 cm.
Xi
Xi − 250
|Xi − 250|
Signed rank
269
19
19
3
221
−29
29
−6
204
−46
46
−9
258
8
8
1.5
211
−39
39
−7
198
−52
52
−10
276
26
26
4
223
−27
27
−5
207
−43
43
−8
242
−8
8
−1.5
19. The following measurements of wind velocities (in km/hr) during 20 consecutive 15 minute
intervals (starting at 1430) were monitored during a thunderstorm on 9 July 2000 in Geneva,
NY. Test whether the median wind velocity significantly exceeded 30 km/hr. (Based on data
from Robert Seem, New York State Agricultural Experiment Station, August, 2000, Personal
communication.)
24.6
38.5
27.5
36.7
39.6
34.4
42.5
33.3
41.8
33.3
47.2
36.7
52.3
33.3
44.7
29.3
37.3
28.6
42.5
24.1
19. The hypotheses are H0 : M ≤ 30 km/hr versus Ha : M > 30 km/hr. Use the sign test with
n = 20. The test statistic is S− = 5. F (5) = 0.0059. Since 0.0059 < 0.05, reject H0 . Data
support the hypothesis that wind velocity exceeded 30 km/hr.
20. Create a table giving the density and cumulative distribution function for W+ for a sample
size of n = 4. Use the same techniques that were used for a sample size of n = 5 in Section 6.6.
20. The density and cumulative distribution function for W+ for n = 4 are
W
P (W = w)
P (W ≤ w)
0
1
2
3
4
5
6
7
8
9
10
1
16
1
16
1
16
2
16
1
16
3
16
2
16
5
16
2
16
7
16
2
16
9
16
2
16
11
16
2
16
13
16
1
16
14
16
1
16
15
16
1
16
16
16
which are derived from 42 = 16 possible signed rankings below.
Problems and Answers for Chapter 6. One Sample Tests of Hypothesis
Ranking
−1
+1
−1
−1
−1
+1
+1
+1
−1
−1
−1
+1
+1
+1
−1
+1
−2
−2
+2
−2
−2
+2
−2
−2
+2
+2
−2
+2
+2
−2
+2
+2
46
W+
−3
−3
−3
+3
−3
−3
+3
−3
+3
−3
+3
+3
−3
+3
+3
+3
−4
−4
−4
−4
+4
−4
−4
+4
−4
+4
+4
−4
+4
+4
+4
+4
0
1
2
3
4
3
4
5
5
6
7
6
7
8
9
10
21. Bornstein and Bornstein (see “The pace of life,” Nature, 1976, 259: 556–558) proposed a
correlation between the population of a city or town and the pace at which people walk in
that location. For a particular sized town their data predict a time of 13.9 s to walk 50 ft. To
investigate this prediction the paces of 16 randomly chosen people were measured. Making
no special assumptions about the distribution of individual paces, set up the appropriate
hypotheses and determine whether or not the null hypothesis should be rejected at α = 0.05.
8.4
11.6
9.8
11.9
10.3
12.4
10.6
12.5
10.8
13.9
11.0
14.0
11.1
15.4
11.5
15.7
21. The hypotheses are H0 : M = 13.9 s versus Ha : M = 13.9 s. Use the sign test with n = 15.
S− = 12 and S+ = 3. F (3) = 0.0176. Since 0.0176 < 0.025, reject H0 . Data indicate a
significant difference from 13.9 s.
22. Redo Problem 4 in Chapter 1 as a test of hypothesis question.
22. The hypotheses are H0 : µ ≤ 12.5 yr versus Ha : µ > 12.5 yr. Use a t test with α = 0.05.
Calculate that X = 14.75 yr, n = 10, and s = 0.84 yr. The c.v. = 1.833.
t=
X −µ
√s
n
=
14.75 − 12.5
0.84
√
10
=
2.25
= 8.33.
0.27
Since 8.33 > 1.833, reject H0 . Menarche is significantly later in world-class swimmers.
23. Redo Problem 6 in Chapter 1 as a test of hypothesis question.
Problems and Answers for Chapter 6. One Sample Tests of Hypothesis
47
23. The hypotheses are H0 : µ ≤ 24 hours versus Ha : µ > 24 hours. For a t test with α = 0.05.
Calculate that X = 14.75 yr, n = 15, and s2 = 0.849 hr2 . The c.v. = 1.761.
t=
X −µ
√s
n
=
25.87 − 24
0.92
√
15
=
1.87
= 7.79.
0.24
Since 7.79 > 1.761, reject H0 . The average day for bunkered people is significantly longer
than 24 hours.
24. (a) The fasting blood glucose (FBG) levels of a population of women is thought to have a
standard deviation of 0.75 mmol/l. A study is being designed to determine whether FBG
levels are increased in women who have taken oral contraceptives for at least six months.
How large a sample is required to detect a 0.50 mmol/l increase if α = 0.05 and the power
is 0.95?
(b) The researchers also wish to determine whether oral contraceptives have any effect on
cholesterol levels. If the standard deviation for cholesterol is thought to be 0.80 mmol/l,
how large a sample is required to detect a 0.30 mmol/l difference if α = 0.05 and the
power is 0.95?
24. (a) Assuming σ = 0.75 mmol/l,
σ 0.75 =
µ0 − µ 0.50 = 1.5
Utilize Table 6.1 to find the minimum required sample size of 25 for this one-sided test.
(b) Assuming σ = 0.80 mmol/l,
σ 0.80 =
µ0 − µ 0.30 = 2.67.
To be conservative, round this value up to 2.70. Utilize Table 6.1 to find the minimum
required sample size of 95 for this two-sided test.
25. A machine used to produce individual serving packages of ketchup is normally run at 75
units/minute and at that speed the σ = 0.12 g. The production manager is considering
raising the production speed to the maximum of 100 units/minute, but believes the servings
will be more variable at that production rate. Measurement of a random sample of 25 serving
packages produced at the higher rate yielded a standard deviation of 0.15 g. Statistically
confirm or refute the production manager’s suspicions.
25. H0 : σ 2 = 0.0144 g2 versus Ha : σ 2 > 0.0144 g2 . At α = 0.05, the c.v. = 36.4. χ2 = 37.5, so
reject H0 . There is statistical evidence that running the machine at its maximum capacity
will cause the serving packages to be more variable.