1. No category

STAT 2000
Sample Midterm
(actually given in October 2007)
1.
In hypothesis testing with a null hypothesis H 0 versus an alternative hypothesis H a , the
power of the test is:
(A)
(B)
(C)
(D)
(E)
2.
the probability of rejecting H 0 when H a is true.
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the probability of failing to€reject H 0 when H a is true.
the probability of failing to reject H 0 when H 0 is true.
the probability of rejecting H 0 when H 0 is true.
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the probability of failing to reject H 0 .
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€
€
€
€
€
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A box of Teddy Graham’s chocolate
cookies is supposed to weigh 250 grams, on
average. There is some variation in weight from box to box due to the operation of the
packaging machine. The weight of the box is known to be normally distributed with
unknown mean µ and known standard deviation of 3 grams. A consumer agency wishes
to test the hypotheses
H o : µ = 250 versus H a : µ < 250
at 5% significance level. A simple random sample of eight boxes is selected from the
production process. The approximate power of the test under the alternative hypothesis
H a : µ = 245 is:
(A) 0.0011
(B) 0.8493
(C) 0.9374
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(D) 0.95
(E) 0.9989
3.
To compare two methods of teaching reading, groups of elementary school children were
randomly assigned to each teaching method for a 6-month period. The criterion for
measuring achievement was a reading comprehension test. The results on the test scores
were as follows:
# of Children Per Group
x
s2
Method 1
11
64
25
Method 2
14
69
64
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Do the data provide sufficient evidence to indicate a difference
in mean scores on the
comprehension test for the two teaching methods? The value of the test statistic and the
degrees of freedom are, respectively:
(A)
(B)
(C)
(D)
(E)
64 − 69
25 64
+
11 14
,
10
64 − 69
(10)(25) + (13)(64)  1 1 
 + 
11 + 14 − 2
 11 14 
64 − 69
(11)(25) + (14)(64)  1 1 
 + 
11 + 14 − 2
 11 14 
64 − 69
25 64
+
11 14
,
,
23
,
23
13
64 − 69
(10)(25 2 ) + (13)(64 2 )  1 1 
 + 
11 + 14 − 2
 11 14 
2
,
23
4.
5.
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6.
The two types of errors in hypothesis testing are:
(A)
rejecting the null hypothesis when the null hypothesis is true and failing to reject
the null hypothesis when the alternative hypothesis is false.
(B)
failing to reject the null hypothesis when the alternative hypothesis is true and
rejecting the null hypothesis when the null hypothesis is false.
(C)
rejecting the null hypothesis when the null hypothesis is true and failing to reject
the null hypothesis when the alternative hypothesis is true.
(D)
failing to reject the null hypothesis when the alternative hypothesis is false and
rejecting the null hypothesis when the null hypothesis is false.
(E)
none of the above.
Let X 1 and X 2 be the sample means of two independent random samples from
populations that have variance 100 and 25, respectively. If X 1 is based on a sample size
of 10 and X 2 is based on a sample size of 5, then the variance of the difference in the
sample means, X 1 − X 2 , is:
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€
(A) 1125
(B) 100
(C) 16
(D) 84
(E) 15
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One-way ANOVA is performed on independent random samples taken from three
normally distributed populations with equal variances. The following summary statistics
were calculated:
n1 = 7, x1 = 65, s1 = 4.2, n 2 = 8, x 2 = 65, s2 = 4.9, n 3 = 9, x 3 = 65, and s3 = 4.6.
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When testing the equality of the population means at 5% level of significance, the
observed value of the test statistic equals:
(A)
(B)
(C)
(D)
(E)
2.57
3.47
4.42
5.78
0
3
7.
One-way ANOVA is applied to independent random samples taken from three normally
distributed populations with equal variances. The following summary statistics were
calculated:
n1 = 8, x1 = 15, s1 = 2, n 2 = 10, x 2 = 18, s2 = 3, n 3 = 8, x 3 = 20, and s3 = 2.
The error sum of squares equals:
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(A)
(B)
(C)
(D)
(E)
137
55
154
60
5.96
Questions 8 and 9 refer to the following:
The marketing manager of a pizza chain is in the process of examining some of the demographic
characteristics of her customers. In particular, she would like to investigate the belief that the
ages the customers of pizza parlors, hamburger emporiums, and fast-food chicken restaurants are
different. As an experiment, the ages of eight customers of each of the restaurants are recorded
and listed below. From previous analyses, we know that the ages are normally distributed with
equal variances.
Customer's Ages
Pizza Hamburger Chicken
23
26
25
19
20
28
25
18
36
17
35
23
36
33
39
25
25
27
28
19
38
31
17
31
Source of Variation
Treatments
Error
Total
df
SS
MS
F
41.131
1067.33
Do these data provide enough evidence at the 5% significance level to conclude that there
are differences in ages among the customers of the three restaurants?
4
8.
At the 5% level of significance (α=0.05), the critical value of the test and the observed
value of the test statistic are, respectively:
(A)
(B)
(C)
(D)
(E)
9.
2.57, 2.475
3.47, 2.475
4.42, 2.475
41.131, 2.475
3.47, 41.131
Refer to question #8. The margin of error of the 95% confidence interval for the mean
age difference for Pizza and Chicken customers is:
(A) 6.413
1 1
+
8 8
(B) (1.721)(6.413)
1 1
+
8 8
(C) (2.120)(6.413)
1 1
+
8 8
(D) (2.080)(6.413)
1 1
+
8 8
(E) None of the above
10.
Consider a binomial parameter p and a test of hypotheses of the form:
H o : p = 0.4 versus H a : p ≠ 0.4
Let X represent the number of successes in 15 trials of a binomial experiment, with p
being the probability of success. Consider the use of the following critical region (that is,
rejection region) for testing the above hypotheses:
CR = RR = {Reject H o if X = 0, 1, 2, 13, 14, 15}.
The probability of a Type I error for this test is:
(A) 0.0274
(B) 0.0093
(C) 0.0364
(D) 0.0548
5
(E) 0.0271
Questions 11 and 12 refer to the following:
Records from the past years of freshmen admitted to the university showed that their mean score
on an aptitude test was 125 with a standard deviation of 20. An administrator wants to know
whether the new freshman population is similar to previous years with respect to the aptitude
test. To test the hypotheses H o : µ = 125 versus H a : µ ≠ 125 , she decides to choose 25
students at random and reject H o if x ≤ 118 or x ≥ 132 .
11.
The level of significance for her test is:
€
(A) 0.0802 €(B) 0.0401
12.
(C) 0.1000
(D) 0.0500
(E) 0.3085
The probability of a Type II error against the alternative hypothesis H a : µ = 116 is:
(A) 0.082
(B) 0.0401
(C) 0.9198
(D) 0.6915
(E) 0.3085
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13.
Before planting a crop for the next year, a producer does a risk assessment. According to
her assessment, she concludes that there are three possible net outcomes: a $7,000 gain, a
$4,000 gain, or a $10,000 loss with probabilities 0.55, 0.20, and 0.25 respectively. The
expected profit is:
(A)
(B)
(C)
(D)
(E)
14.
$3,850
$7,150
$2,150
$2,500
$2,350
A simple random sample of size 25 is selected and a test of the hypotheses
H o : µ = 85 versus H a : µ ≠ 85 for a population mean is to be conducted at the 5% level
of significance. Under the alternative hypothesis Ha : µ = 90 , the power of the test is
determined to be 0.9203. Which of the following would lead to a larger power?
(i) Increase the sample size to 50.
€.
(ii) Use the alternative hypothesis Ha : µ = 72
(iii) Use a 10% level of significance.
(A) (i) only
(B) (i) and (ii) only
(C) (i) and (iii) only
(D) (ii) and (iii) only
(E) (i), (ii), and (iii)
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6
15.
A study was conducted to compare five different training programs for improving
endurance. Thirty-five subjects were randomly divided into five groups of seven subjects
in each group. A different training program was assigned to each group. After two
months, the improvement in endurance was recorded for each subject. An ANOVA
procedure is used to compare the five training programs, and the resulting F-statistic is
3.69. What can we say about the P-value for this F-test?
(A)
(B)
(C)
(D)
(E)
16.
P-value < 0.001
0.001 < P-value < 0.01
0.01 < P-value < 0.025
0.025 < P-value < 0.05
P-value > 0.05
A discrete random variable X has the following probability distribution:
P( X = x) =
2x + 1
, x = 1, 2, 3, 4, 5
c
What is the value of the constant c?
(A) 18
(B) 35
(C) 34
(D) 15
(E) 36
Questions 17 and 18 refer to the following:
The water diet requires one to drink two cups of water every half an hour from when one gets up
until one goes to bed, but otherwise allows one to eat whatever one likes. Four adult volunteers
agree to test the diet. They are weighed prior to beginning the diet and after six weeks on the
diet. The weights (in pounds) are:
Person
Weight before the diet
Weight after six weeks
Difference (before - after)
1
180
170
10
2
125
130
-5
3
240
215
25
4
150
152
-2
size
4
4
4
mean standard deviation
173.75
49.56
166.75
36.09
7
13.64
For the population of all adults, let µ be the mean weight loss after six weeks on the diet (weight
before beginning the diet minus weight after six weeks on the diet). To determine if the diet leads
to weight loss, we test the hypotheses
H 0 : µ = 0 versus Ha: µ > 0.
17.
The value of the test statistic for the appropriate test is closest to:
(A) 0.06 € (B) 0.23
(C) 1.03
(D) 2.05
7
(E) 0.51
18.
The 5% rejection region for the test statistic used in Question # 17 is:
(A) t > 2.353 (B) t > 3.182 (C) t > 1.943 (D) Z > 1.645 (E) Z > 1.96
19.
Suppose that we test H 0 : µ = 10 versus H a : µ ≠ 10 at level of significance α based on a
random sample of size n from a normally distributed population N(µ, σ), where σ is
known. If we increase the level of significance from α = 5% to α = 10%, then the power
of the test against Ha : µ = 20 will:
€
€
(A) increase
(B) decrease
(C) double
(D) decrease by half
(E) not change
Questions 20 and 21 refer to the following information:
A USA political analyst was curious if younger adults were becoming more conservative. He
decided to see if the mean age of registered Republicans was lower than that of registered
Democrats. He selected a simple random sample (SRS) of 15 registered Republicans from a list
of registered Republicans and determined the mean age to be x1 = 39 years with a standard
deviation s1 = 4 years. He also selected an independent SRS of 27 registered Democrats from a
list of registered Democrats and determined the mean age to be x2 = 44.5 years with a standard
deviation s2 = 10 years. Let µ1 and µ2 represent the mean ages of the populations of all registered
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Republicans and Democrats, respectively.
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20.
A 90% confidence interval for µ1 - µ2 is:
(A)
(39 − 44.5) ± (1.684)
14 × 16 + 26 × 100
40
(B)
(39 − 44.5) ± (1.684)
16 100
+
15 27
(C)
(39 − 44.5) ± (1.761)
4 10
+
15 27
(D)
(E)
14 × 16 + 26 × 100
40
14 ×16 + 26 ×100
(39 − 44.5) ± (1.697)
40
(39 − 44.5) ± (1.761)
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8
1
1
+
15 27
1
1
+
15 27
1
1
+
15 27
21.
Refer to Question # 20. Suppose the researcher had wished to test the hypotheses
H0: µ1 = µ2
versus
Ha: µ1 < µ2
The P-value for the test is:
(A)
(B)
(C)
(D)
(E)
between 0.02 and 0.025
between 0.025 and 0.05
between 0.01 and 0.02
below 0.01
above 0.05
Use the following to answer questions 22 and 23:
The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures the
motivation and attitude toward school. A university professor selected independent random
samples of male and female first-year students. The professor wished to see if male first-year
students differ from female first-year students with regard to motivation and attitude toward
school.
Group
Male
Female
Measurements
154, 109, 137, 115, 152, 140, 154, 101, 178
108, 140, 114, 91, 180, 115, 126
size n
9
7
x
137.78
124.86
s
25.13
28.63
Let µ1 and µ2 represent respectively the mean score on SSHA for male and female first-year
students at this university. Assume the measurements for male and female students are normally
distributed with unknown variances.
22.
The margin of error for the 99% confidence interval of the difference in the population
means, µ1 - µ2 , is:
(A) (3.707)
25.13 2 28.63 2
+
9
7
(D)
(B) (3.707)
25.13 28.63
+
9
7
(E) (2.977)
(C) (2.977)
8 × (25.13 2 ) + 6 × (28.63 2 )
14
9
1 1
+
9 7
8 × (25.13 2 ) + 6 × (28.63 2 )
14
8 × 25.13 + 6 × 28.63
14
1 1
+
9 7
1 1
+
9 7
23.
At 5% level of significance, we:
(A) would reject H 0 because P-value < 0.05
(B) would reject H 0 because P-value > 0.05
(C) would not reject H 0 because P-value < 0.05
(D) would not reject H 0 because P-value > 0.05
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(E) would reject H 0 because α = 0.05 is very small
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€
Questions 24 and€
25 refer to the following information:
€
Suppose the number
of accidents at a given intersection has a Poisson distribution with a mean
equal to 0.7 accidents per day.
24.
What is the probability that in a 5-day work week there are exactly three accidents?
e−0.7 (0.7) 5
5!
−3.5
e (3.5) 3
(D)
3!
e−0.7 (0.7) 3
3!
−2.1
e (2.1) 5
(E)
5!
(A)
€
25.
€
(B)
(C)
e−5 5 3
3!
€
€
The standard deviation of the number of accidents in two weeks (14 days) is:
(A) 3.74
(B) 2.94
(C)
€ 3.13
(D) 9.80
(E) 1.18
Questions 26-28 refer to the following information:
In order to see whether different age groups have different mean body temperatures, independent
random samples were collected from each of the three age groups. Assume that conditions for
ANOVA are satisfied. Data are presented in the following table of body temperatures (in oF)
categorized by age. A partially completed ANOVA table is also included.
Means and Standard Deviations
Level
18-20 Years
21-29 Years
30 & older
Group
1
2
3
Number
4
8
5
Mean
98.20
98.60
97.56
Std Dev
0.42
0.67
0.59
Analysis of Variance
Source of Variation
Group
Error
Total
df
SS
3.328
5.0639
MS
Let µ i be the population mean body temperature for Group i, i = 1, 2, 3.
10
F
26.
27.
28.
The correct statement of both the null and alternative hypotheses is:
(A) H 0 : X1 = X 2 = X 3 versus H a : X1, X 2 and X 3 are all different
(B) H 0 : X1 = X 2 = X 3 versus H a : not all of X1, X 2 and X 3 are the same
(C) H 0 : µ1 = µ2 = µ3 versus H a : µ1, µ2 and µ3 are all different
(D) H 0 : µ1 = µ2 = µ3 versus H a : not all of µ1, µ2 and µ3 are the same
€ €
€ €
(E) H 0 : µ1 = µ2 = µ3 versus H a : µ1 ≠ µ2 ≠ µ3
€ €
€
€
€
€
€Refer to Question # 26.€At 5% level of significance, the statistical decision is to:
€
€
(A) not reject H 0 because the value of the test statistic is less than 3.74
(B) reject H 0 because the value of the test statistic is larger than 3.74
(C) not reject H 0 because the value of the test statistic is smaller than 4.86
(D) reject H 0 because the value of the test statistic is larger than 4.86
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(E) reject H 0 because the value of the test statistic is larger than 3.34
€
€
€
Refer to Question # 26. The margin of error for a 95% confidence interval for the
€
population
mean body temperature of the 21-29 Years group is closest to:
(A) 1.03
(B) 0.14
(C) 0.75
(D) 0.46
11
(E) 0.96
Answer Key
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