MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
1
MAT 2379, Introduction to Biostatistics
Sample Calculator Problems for the Final Exam
Note: The exam will also contain some problems based on a printed Minitab output. Sample
problems of this format are not provided. Students are encouraged to review the Minitab instructions
on the web, and the solutions to the Minitab assignment problems.
1. A sample of 25 seven-year-old girls had a sample mean weight of 43.1 pounds and a sample
standard deviation of 6.6 pounds. Assuming normality, find a 90% confidence interval for the
average weight µ of a seven-year-old girl.
(A) [40.4; 45.8]
(E) [38.1; 47.8]
(B) [41.3; 44.8]
(C) [39.5; 46.1]
(D) [40.8; 45.4]
√
Solution: The interval is x̄ ± t(s/ n). In this case, t = 1.711 since P (T24 > 1.711) = 0.05.
We obtain the interval:
!
6.6
43.1 ± 1.711 √
25
that is 43.1 ± 2.258, i.e. [40.8414; 45.3585]. The answer is D.
2. The concentration of active ingredients in a liquid laundry detergent is thought to be affected
by the type of catalyst used in the manufacturing process. Ten observations on concentration
are taken with each catalyst, yielding the following the data:
Catalyst 1 : 57.9, 66.2, 65.4, 65.4, 62.6, 67.6, 63.7, 67.2, 65.2, 71.0
Catalyst 2 : 66.4, 71.7, 70.3, 69.3, 64.8, 69.6, 68.6, 69.4, 65.3, 68.8
Let µ1 and µ2 be the average concentration of active ingredients, when using catalyst 1, respectively catalyst 2. Give a point estimate for µ2 − µ1 .
(A) 2.3
(B) 3.2
(C) 2.0
(D) 1.3
(E) -3.2.
Solution: We have x̄1 = 65.22 and x̄2 = 68.42. An estimate for µ2 − µ1 is x̄2 − x̄1 =
68.42 − 65.22 = 3.2. The answer is B.
3. The height of a male student is a random variable with a normal distribution with mean
µ = 174.5cm and standard deviation σ = 6.75 cm. What is the probability that a randomly
chosen student has a height of more than 188.3 cm?
(A) 0.4772
(B) 0.0207
(C) 0.9772
(D) 0.4778
Solution: Let X be the height of the student. By standardization,
P (Z > 188.3) = P
188.3 − 174.5
X − 174.5
>
= P (Z > 2.04)
6.75
6.75
= 1 − P (Z ≤ 2.04) = 1 − 0.9793 = 0.0207
The answer is B.
(E) 0.3451
MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
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4. In a sample of 300 Canadians, 27 have type B blood. Based on this information, find a 95%
confidence interval for the proportion p of Canadians who have type B blood.
(A) [0.063; 0.117]
(D) [0.071; 0.109]
(B) [0.085; 0.095]
(E) [0.011; 0.139]
(C) [0.058; 0.122]
Solution: We have p̂ = 27/300 = 0.09. The interval is:
s
0.09 ± 1.96
(0.09)(0.91)
300
that is 0.09 ± 0.032, or [0.058; 0.122]. The answer is C.
5. Two companies manufacture a rubber material intended for use in automobiles. 25 samples
of material produced by each company are tested in an abrasion test, and the amount of wear
is observed. The sample means and sample standard deviations for company 1 and company
2 are summarized below:
Company 1 : x̄1 = 20 mg,
s1 = 2 mg,
Company 2 : x̄2 = 15 mg,
s2 = 8 mg.
Construct a 99% confidence interval for µ1 − µ2 , where µ1 and µ2 are the average amounts of
wear for the two companies. Assume that the two populations are normally distributed with
unequal variances. Hint: Use the T distribution with γ = 26 degrees of freedom.
(A) [0.42; 9.58]
(D) [2.29; 7.71]
(B) [1.77; 8.23]
(E) [3.14; 6.86]
(C) [2.03; 7.03]
Solution: The populations are independent. The number of degrees of freedom of T is:
ν=
(4/25 + 64/25)2
= 26.98
[(4/25)2 + (64/25)2 ]/24
We use γ = 26. From Table VI, we find t = 2.779, since P (T26 < 2.779) = 0.995. Note that
x̄1 − x̄2 = 5 and the interval is:
s
4
64
5 ± (2.779)
+
25 25
that is 5 ± 4.58, or [0.42; 9.58]. The answer is A.
6. A patient’s condition in a hospital is classified as “surgical” or ”medical”. In the United
States, some patients have health insurance coverage under a national plan called Medicare,
while others do not have. The data obtained for 177 patients is given in the following table:
Surgical Medical
Medicare: yes
46
52
Medicare: no
36
43
MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
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Is there an association between the patient’s condition and the patient’s heath insurance coverage? Give the range of the p-value of the test, and report your conclusion at level α = 0.05.
(A) 0.1 <p-value< 0.2; there is no association
(B) 0.025 <p-value< 0.05; there is an association
(C) 0.05 <p-value< 0.10; there is no association
(D) 0.10 <p-value< 0.20; there is no association
(E) p-value> 0.20; there is no association
Solution: We have to test H0 which says that “there is no association between the two
variables”. We fill-in the following table:
Surgical Medical Total
Medicare: yes
46
52
98
(45.40) (52.60)
Medicare: no
36
43
79
(36.30) (42.40)
Total
82
95
177
The test statistic is:
(46 − 45.4)2 (52 − 52.6)2 (36 − 36.6)2 (43 − 42.4)2
+
+
+
= 0.033
45.4
52.6
36.6
42.4
P-value= P (Y > 0.033) is greater than 0.20. Since the p-value is larger than 0.05, we fail to
reject H0 . We conclude that there is no association. The answer is E.
7. 85% of the adults living in a region are non-smokers, and 24% are non-smokers with emphysema (a chronic lung disease). What is the probability that a randomly selected person has
emphysema, given that this person is a non-smoker?
(A) 0.15
(B) 0.182
(C) 0.625
(D)0.282
(E) 0.931
Solution: Let A be the event that the person is a non-smoker and B be the event that the
person has emphysema. We know that P (A) = 0.85 and P (A and B) = 0.24. The desired
probability is:
P (A and B)
0.24
=
= 0.282
P (B|A) =
P (A)
0.85
The answer is D.
8. Ten adult males between the age of 35 and 50 participated in a study to evaluate the effect of
diet and exercise on lowering the blood cholesterol level. The cholesterol level was measured
in each subject before and after participating in an aerobic exercise program. The data is
given in the following table:
MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
Subject (i)
1
2
3
4
5
6
7
8
9
10
4
Before (xi ) After (yi )
265
229
240
231
258
227
295
240
251
238
287
234
314
256
260
247
283
246
240
218
The differences d1 = x1 − y1 , d2 = x2 − y2 , . . . , d10 = x10 − y10 have a sample mean d¯ = 32.7
and sample standard deviation sd = 18.36. Construct a 95% confidence interval for the average difference µd = µX − µY , where µX and µY are the average cholesterol levels before the
program, respectively after the program. Does the data support the claim that the program
has been efficient in reducing the cholesterol level?
(A) [-13.56;14.57]; not enough evidence
(C) [-7.37; -1.28]; no
(E) [19.57; 45.83]; yes
(B) [13.13; 26.26]; yes
(D) [7.38; 14.54]; yes
Solution: The two populations are dependent. From Table 4, we find t = 2.262, since P (T9 >
2.262) = 0.975. The interval is:
18.36
32.7 ± 2.262 √
10
!
that is 32.7 ± 13.13, or [19.57; 45.83]. Since the interval contains only positive values, we
conclude that µd > 0, i.e. µX > µY . The program has been efficient. The answer is E.
9. A medical test is conducted to determine whether someone has a disease. Suppose that 8% of
population has this disease. The test has a 95% chance of detecting the disease, if the person
has the disease, and a 90% chance of indicating that the disease is absent, if the person does
not have the disease. What is the probability that a person has the disease, given that this
person has tested positive?
(A) 0.52
(B) 0.32
(C) 0.45
(D) 0.90
(E) 0.12.
Solution: Let D be the event that the person has the disease, T + be the event that the
test is positive, and T − the event that the test is negative. We know that P (D) = 0.08,
P (T + |D) = 0.95 and P (T − |Dc ) = 0.90. From here, we infer that P (T + |Dc ) = 0.10. Using
the addition rule, and the multiplicative rule, we get:
P (T +) =
=
=
=
P (T + and D) + P (T + and Dc )
P (D)P (T + |D) + P (Dc )P (T + |Dc )
(0.08)(0.95) + (0.92)(0.10)
0.076 + 0.092 = 0.168
MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
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The desired probability is:
P (D|T +) =
P (T + and D)
0.076
=
= 0.45
P (T +)
0.168
The answer is C.
10. Let X be the number of visits to the dentist’s office of a randomly chosen student. Suppose
that X has a probability distribution function given by the following table:
x
0
P (X = x) 0.15
1
0.5
2
0.35
Find P (X ≤ µ), where µ = E(X) is the average number of visits.
(A) 0
(B) 0.15
(C)1
(D) 0.65
(E) 0.15
Solution: We have µ = E(X) = 0(0.15) + 1(0.5) + 2(0.35) = 1.2. The desired probability is:
P (X ≤ 1.2) = P (X ≤ 1) = P (X = 0) + P (X = 1) = 0.15 + 0.5 = 0.65
The answer is D.
11. A meteorologist measures the temperature (in degrees Celsius) in downtown Ottawa at noon
on 5 consecutive days, and obtains the following data:
−15,
−10,
−9,
+1,
+3
Based on this data, give an estimate for: (i) the average temperature µ at noon; and (ii) the
variance σ 2 of the temperature at noon.
(A) (i) −6.0;
(C) (i) −7.8;
(E) (i) −5.0;
(B) (i) −6.0;
(D) (i) −2.1;
(ii) 59.0
(ii) 76.2
(ii) 36.4
(ii) 54.4
(ii) 54.4
Solution: (i) An estimate for the average temperature µ is
x̄ =
(−15) + (−10) + (−9) + (+1) + (+3)
= −6.
5
(ii) An estimate for σ 2 is:
1
s2 = {[−15 − (−6)]2 + [−10 − (−6)]2 + [−9 − (−6)]2 + [+1 − (−6)]2 + [+3 − (−6)]2 } = 59.0
4
The answer is A.
12. In a recent poll of 1,215 Canadians, 643 are in favor of Canada withdrawing its troops from
Afghanistan. Give a point estimate of the percentage p of Canadian who are not in favor of
Canada withdrawing its troops from Afghanistan.
(A) 0.53
(B) 0.42
(C) 0.47
(D) 0.55
(E) 0.50
Solution: The estimate for p is p̂ = (1215 − 643)/1215 = 572/1215 = 0.47. The answer is C.
MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
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13. 10% of the residents of a certain city test positive for a certain antibody. What is the probability that in a randomly selected group of 5 residents, at least one will test positive?
(A) 0.4095
(B) 0.5905
(C) 0.5
(D) 0.9040
(E) 0.7651
Solution: Let X be the number of people who will test positive. X has a binomial distribution
with n = 5 and p = 0.1. The desired probability is:
P (X ≥ 1) = 1 − P (X = 0) = 1 − (0.9)5 = 1 − 0.5905 = 0.4095
The answer is A.
14. A population includes 40% females and 60% males. For the women, the probability of having
a hand size smaller than 100 cm2 is 0.31. For the men, this probability is 0.08. What is the
probability that a randomly chosen person will have a hand size smaller than 100 cm2 ?
(A) 0.352
(B) 0.452
(C) 0.172
(D) 0.202
(E) 0.912.
2
Solution: Let H be the event that the person has a hand smaller than 100 cm . We know
that P (H|F ) = 0.31 and P (H|M ) = 0.08. Using the rule of total probability,
P (H) = P (H|F )P (F ) + P (H|M )P (M ) = (0.31)(0.4) + (0.08)(0.6) = 0.172
The answer is C.
15. Suppose A and B are events such that P [A] = 0.2, P [B] = 0.6 and
P [A or B] = 0.68.
Are A and B independent events?
(A) Yes
(B) No
(C) Insufficient information given
Solution: Note that
P [A and B] = P [A] + P [B] − P [A or B] = 0.2 + 0.6 − 0.68 = 0.12
P [A]P [B] = (0.2)(0.6) = 0.12
Since P [A and B] = P [A]P [B], we conclude that the events A and B are independent.
16. Suppose that A and B are events such that P [A] = 0.6, P [B] = 0.4 and
P [A and B] = 0.3.
Find P [B|Ac ],
occur.
(A) 0.5
the
(B) 0.1
probability
(C) 0.4
that
B
(D) 0.25
occurs
given
that
A
does
not
(E) insufficient information given
MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
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Solution: We have P [Ac ] = 1 − P [A] = 1 − 0.6 = 0.4 and P [B and Ac ] = P [B] − P [A and B] =
0.4 − 0.3 = 0.1. Hence
0.1
P [B and Ac ]
c
=
= 0.25
P [B|A ] =
c
P [A ]
0.4
17. There are 500 students writing this examination. Experience tells us that the marks on the
exam are normally distributed with mean 70 and standard deviation 10. What is the expected
number of students that will have a mark of 80 or more on this exam?
(A) 70
(B) 79.35
(C) 15.87
(D) 84.13
(E) 420.65
Solution: Let X be the grade of a randomly chosen student. The probability that this student’s
grade is greater than 80 is
80 − 70
X − 70
≥
= P [Z ≥ 1] = 1 − 0.8413 = 0.1587
P [X ≥ 80] = P
10
10
Let Y be the number of students (among the 500) which will have a grade greater than 80.
Then Y : Binomial(500, 0.1587) and
E[Y ] = (500)(0.1587) = 79.35.
18. Let X1 , ..., X7 be a random sample from a normal distribution with mean 5. Let X and S be
the sample mean and sample standard deviation, respectively. Find a constant c such that
!
P
(A) 2.576
X −5
√ < c = 0.01
S/ 7
(B) -2.576
(C) -2.998
(D) 2.998
(E) -3.143
√
Solution: The variable Y = (X̄ − 5)/[S/ 7] has a t distribution with 6 d.f. From Table 4, we
find P [Y ≥ 3.143] = 0.01 and hence c = −3.143.
19. The blood pressure of a male between 50 and 60 is known to be normally distributed and to
have a standard deviation of 10 mm. Hg. The blood pressure of 36 men (all between 50 and
60) is measured. The average pressure in the sample was observed to be 118 mm. Hg., and
the sample standard deviation was 12 mm. Hg. Find a 95% confidence interval for the mean
blood pressure µ of men aged 50-60.
(A) 118 ± 2.74
(E) 118 ± 0.54
(B) 118 ± 3.27
(C) 118 ± 4.06
(D) 118 ± 3.38
Solution: Because σ = 10 is know, we use the confidence interval based on z. From Table 3
we read z = 1.96. The confidence interval is
10
118 ± (1.96)
6
or equivalently 118 ± 3.27
MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
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20. It is known that monozygotic twins occur with probability 2/9 and dizygotic twins occur
with probability 7/9. In the monozygotic case the probability that both twins are male is
1/2, whereas in the dizygotic case this probability is 1/4. What is the probability that in a
randomly selected pair of twins, both are males?
(A) 1/4
(B) 7/36
(C) 5/18
(D) 11/36
(E) 1/2
Solution: Let A be the event that the twins are monozygotic, and B the event that the both
twins are male. We know that P [A] = 2/9, P [Ac ] = 7/9, P [B|A] = 1/2 and P [B|Ac ] = 1/4.
Hence
11
2 1 7 1
P [B] = P [A] · P [B|A] + P [Ac ] · P [B|Ac ] = · + · =
9 2 9 4
36
21. Continuing with the situation described in Problem 20, if a couple has a pair of male twins,
what is the probability that these twins are monozygotic?
(A) 4/11
(B) 4/9
(C) 15/36
(D) 7/18
(E) 2/9
Solution: Using the answer calculated in Problem 20, we get
P [A|B] =
(2/9) · (1/2)
4
P [A] · P [B|A]
=
=
P [B]
11/36
11
22. The following table gives the cumulative distribution function of a discrete random variable
X. Find the expected value of X.
x
F (x)
(A) 7.5
(B) 6.3
5
0.4
6
7
10
0.6 0.9 1.00
(C) 21.9
(D) 5.8
(E) 8.1
Solution: We have the following table of the density function of X:
x
f (x)
5
6
7
10
0.4 0.2 0.3 0.1
The expected value of X is
E[X] = (5)(0.4) + (6)(0.2) + (7)(0.3) + (10)(0.1) = 2 + 1.2 + 2.1 + 1 = 6.3
23. The following graph represents the density function of a continuous random variable X. Calculate the height h on the graph.
(a) h = 1/3
(b) h = 1/4
hP
0
PP
PP
(c) h = 2/3
PP
P
PP
PP
3
(d) h = 7/8
(e) h = 1/2
MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
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Solution: The total area under the graph is equal to 1. We have
1 = Area(triangle) =
h·3
2
and hence h = 2/3.
24. A study is conducted to compare the effect of green light, in comparison to red light, on the
growth of bean plants. The following table shows data on the heights of plants (in inches)
after two weeks. Test the hypothesis H0 : µ1 = µ2 versus HA : µ1 > µ2 . What are the
observed value of the test statistic and the p-value?
Green Light
n1 = 9
x̄1 = 9.8
s21 = 5.4
Red Light
n2 = 5
x̄2 = 7.3
s22 = 2.5
(A) The test statistic is 4.52; the p-value is smaller than 0.0005
(B) The test statistic is 1.43; the p-value is in the interval [0.05; 0.10]
(C) The test statistic is 2.38; the p-value is in the interval [0.01; 0.02]
(D) The test statistic is 1.27; the p-value is larger than 0.10
(E) none of the preceding
Solution: The number of degrees of freedom is
ν=
1.21
(0.6 + 0.5)2
=
= 11.25 ;
(0.36)/8 + (0.25)/4
0.1075
we take ν = 11.
The observed value of the test statistic is
9.8 − 7.3
2.5
√
= 2.38.
=√
0.6 + 0.5
1.1
The p-value lies in the interval [0.01 ; 0.02].
25. The maximum temperature (in degrees Fahrenheit) on December 1st in Boston has a normal
distribution with mean µ = 40 and variance σ 2 = 100. We know that x degrees Fahrenheit
equals 5(x−32)/9 degrees Celsius. What is the probability of having a maximum temperature
of at most 0 degrees Celsius on December 1st in Boston?
(A) 0.2119
(B) 0.5000
Solution:
P(
(C) 0.5320
(D) 0.7881
(E) less than 0.0002
5(X − 32)
X − 40
32 − 40
≤ 0) = P (X ≤ 32) = P (
≤
)
9
10
10
= P (Z ≤ −0.8) = 0.2119.
26. In the USA, 30% of adults have been tested for HIV. In the emergency room of a small American hospital, it is standard procedure to test all patients for HIV if they have not already
been tested. If 15 patients are seen in the emergency room today, what is the probability that
at least 10 will need to be tested?
MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
(A) 0.0007
(B) 0.2784
(C) 0.5155
10
(D) 0.7216
(E) 0.0039
Solution: The probability that a person have not already been tested in 0.7. Let X be
the number of persons (among the 15) which have not already been tested for HIV. Then
X : Binomial(15, 0.7). From Table I, the desired probability is
P [X ≥ 10] = 1 − P [X ≤ 9] = 1 − 0.2784 = 0.7216.
27. In the “double cohort” year, the students in first year calculus had different mathematics
courses in their final year of high school. It was of interest to see if there was an association
between their high school math courses (A and B) and the likelihood of passing first year
calculus. A sample of 93 first year students yielded the following data:
Course A
Passed Calculus
34
19
Failed Calculus
Total
53
Course B Total
16
50
24
43
40
93
Test the hypothesis H0 that there was no association between high school math and first
year calculus against the alternative HA that there was an association. Give the range of the
p-value of the test, and state your conclusion at the level α = 0.05
(A) 0.005 <p-value< 0.01, reject H0 , there is some association
(B) 0.01 <p-value< 0.02, reject H0 , there is some association
(C) 0.02 <p-value< 0.05, reject H0 , there is some association
(D) 0.05 <p-value< 0.1, fail to reject H0 , no association
(E) 0.2 <p-value, fail to reject H0 , no association
Solution: We fill up the contingency table as follows:
Course A
Passed Calculus
34
(28.5)
Failed Calculus
19
(24.5)
Total
53
Course B
16
(21.5)
24
(18.5)
40
Total
50
43
93
The observed value of the test statistic is:
χ2 =
(34 − 28.5)2 (16 − 21.5)2 (19 − 24.5)2 (24 − 18.5)2
+
+
+
=
28.5
21.5
24.5
18.5
30.25 30.25 30.25 30.25
+
+
+
= 5.34
28.5
21.5
24.5
18.5
The p-value=P (Y > 5.34) lies in the interval [0.02; 0.05]. We reject H0 at level 0.05. We
decide that there is some association between the two variables.
MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
11
28. The effectiveness of a new method of teaching semi-literate adults to read is being assessed.
A group of 15 such adults is randomly selected and a reading test is administered to evaluate
their reading levels. The resulting sample mean is x̄1 = 6.7 with sample standard deviation
s1 = 1.8. The 15 people then undergo the new teaching method and are re-tested, with
x̄2 = 7.2 and s2 = 2.1. We want to test H0 : µ1 − µ2 ≥ 0 against H1 : µ1 − µ2 < 0, at level
α = 0.05. Which of the following statements is correct?
(A) t = −0.700, reject H0
(B) t = −0.700, do not reject H0
(C) There is not enough information provided to carry out the appropriate test procedure.
Solution: This is a paired comparison test. In order to perform the test we need the value of
sd (the standard deviation of the differences) and this value is not given. The correct answer
is C.
29. Air Canada is trying to determine the proportion p of “no-shows” (i.e. the proportion of
passengers who book flights but who do not show up). In a three day period, 850 out of
10,000 booked passengers did not show up for their flights. Find a 95% confidence for p.
(A) [0.0795, 0.0905]
(D) [0.9095, 0.9205]
(B) [0.0804, 0.0896]
(E) none of the preceding.
(C) [0.9104, 0.9196]
Solution: p̂ = 850/1000 = 0.085. The interval is
s
0.085 ± 1.96
(0.085)(0.915)
1000
that is 0.085 ± 0.0055, or [0.0795; 0.0905].
30. The coach of the University of Ottawa weight-lifting team wants the 10 men to increase
their mean body weight, and so he puts them on a protein diet for two months and then
records their weights. If it can be assumed the the weights are still normally distributed with
standard deviation 10, find a 95% confidence interval for the mean weight of a weight-lifter
on the protein diet if the weights yield the following data:
x = 105, s = 11.
(A) 105 ± 1.96
(E) 105 ± 6.20
(B) 105 ± 0.72
(C) 105 ± 2.26
(D) 105 ± 0.52
√
Solution: Since σ = 10 is known, we use the formula x̄ ± z(σ/ n). The interval is:
10
105 ± 1.96 √
10
that is 105 ± 6.20.
MAT 2379, Introduction to Biostatistics, Sample Calculator Questions
12
31. Triglycerides are blood constituents that are thought to play a role in coronary artery disease. To see whether regular exercise could reduce triglyceride levels, researchers measured
the concentration of triglycerides in the blood serum of 5 male volunteers, before and after
participation in a ten-week exercise program. The results are shown in the table below:
Subject
1
2
3
4
5
Before (x1 )
2.1
1.8
2.4
1.6
2.0
After (x2 )
1.9
1.7
2.5
1.3
1.5
We want to test H0 : µ1 = µ2 against H1 : µ1 > µ2 , at level α = 5%. Which of the following
statements is correct?
(A) t = 2.00, reject H0
(B) t = 2.00, do not reject H0
(C) t = 0.49, reject H0
(D) t = 0.49, do not reject H0
(E) none of the preceding
Solution: In this case, the populations are dependent. We calculate the column of differences
d1 , d2 , d3 , d4 , d5 . For these differences, we obtain: d¯ = 0.2 and sd = 0.2236. The test statistic
is:
0.2 − 0
d¯ − 0
√ = 2.00
√ =
sd / n
0.2236/ 5
p-value= P (T4 > 2.00) lies in the interval 0.05 and 0.10. We do not reject H0 .
32. If the serum cholesterol levels for a certain population are approximately normally distributed
with a mean of 200 mg/100 ml and a standard deviation of 20 mg/100 ml, find the probability
that a randomly chosen individual from this population will have a cholesterol level greater
than 225 mg/100 ml.
(A) 0.1056
(B) 0.0885
(C) 0.2573
(D) 0.1374
(E) less than 0.0029
Solution: Let X be the serum cholesterol level. By standardization,
P (X > 225) = P
225 − 200
X − 200
= P (Z > 1.25)
>
20
20
= 1 − P (Z ≤ 1.25) = 1 − 0.8944 = 0.1056