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Statistics

STAT 2040

Gary Umphrey

Fall

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G. J. Umphrey STAT*2040 Fall 2012
Some Multiple Choice Questions for Exam Preparation
1. A test statistic value of z = !1.34 was obtained when a test for the equality of binomial proportions, H :
o
p 1 p 2ersus H : 1
1p , 2as conducted. The p-value for this test is closest to:
(A) 0.09
(B) 0.18
(C) 0.41
(D) 0.82
(E) 0.91
2. A nursery advertises that 90% of its stock survives transplanting. An apartment complex purchases 80
trees from this nursery, and 12 do not survive. A 95% confidence interval for the proportion of trees
that survive has the limits:
(A) 0.150 ± 0.066
(B) 0.150 ± 0.078
(C) 0.850 ± 0.066
(D) 0.850 ± 0.078
(E) 0.900 ± 0.066
3. In a particular large population of people, a researcher wished to test her hypothesis that the proportion
of left-handed people in the population was 10%. Based on a random sample, a 95% confidence
interval for the true proportion of (0.092, 0.132) was obtained. Consider the following statements:
I. The confidence interval provides insufficient evidence to reject the researcher's hypothesis.
II. The confidence interval provides sufficient evidence to prove the researcher is correct.
III. The confidence interval suggests that the true proportion of left-handed people may be greater
than 10%.
Which of these statements are true?
(A) I only
(B) III only
(C) I and II
(D) I and III
(E) II and III.
4. Do steroids help build strength? As part of a study to examine this issue, 20 weightlifters volunteered
for the following study. Ten of the weightlifters were chosen at random and treated with steroids, and
the other 10 were administered a placebo. A consistent training program was given to all 20 lifters for
ten weeks. The random variable measured was the increase in the weight (in pounds) that they could
bench press after the training program. The summary statistics were:
placebo group: mean = 14.6, variance = 124
steroid group: mean = 15.1, variance = 146
A 90% confidence interval for the difference in population means has the limits of !0.5 plus and minus
a bound of:
(A) 8.55
(B) 9.01
(C) 9.52
(D) 10.04
(E) 10.92 5. A physician compares the blood pressures of four patients before and after treatment with a drug. The
blood pressures are as follows:
Patient Before After
1 161 140
2 158 148
3 170 150
4 155 140
A 95% confidence interval for the mean decline in blood pressure before and after treatment is obtained
by calculating the limits:
(A) 16.5 ± 4.82
(B) 16.5 ± 7.04
(C) 16.5 ± 8.06
(D) 16.5 ± 9.83
(E) 16.5 ± 10.22
6. Which one of the following statements best describes the general philosophy behind the hypothesis tests
that we have conducted?
(A) The test is conducted under the initial assumption that the null hypothesis is true, and it is then
determined if there is sufficient evidence to reject the null hypothesis and accept the alternative
hypothesis.
(B) The test is conducted under the initial assumption that the alternative hypothesis is true, and it is
then determined if there is sufficient evidence to reject the alternative hypothesis and accept the
null hypothesis.
(C) The test is conducted under the initial assumption that the null hypothesis is true, and it is then
determined if there is sufficient evidence to support the initial assumption and reject the alternative
hypothesis.
(D) The test is conducted under the initial assumption that the alternative hypothesis is true, and it is
then determined if there is sufficient evidence to support the initial assumption and reject the null
hypothesis.
(E) The calculation of the test statistic does not require any initial assumption, it simply supplies
evidence to choose between the null and alternative hypotheses.
7. Suppose I was suspicious that 620 g boxes of Shreddies were being underfilled, on average, by the
manufacturer. I took a sample of 5 boxes of Shreddies and obtained the following (fictitious) data:
635, 640, 615, 625, 605
The decision rule (at the 5% level of significance) for the test to detect underfilling and the value of the
appropriate t test statistic would be:
(A) Reject H of t > 2.132; t = 0.62.
(B) Reject H of t < !2.132; t = 0.62.
(C) Reject H of t > 2.132; t = 0.28.
(D) Reject H of t < !2.132; t = 0.28.
(E) Reject H of | t | > 2.776; t = 0..2 8. The method of least squares obtains an estimated regression equation that minimizes:
(A) the sum of the vertical distances between the points and the estimated regression equation.
(B) the sum of the perpendicular distances between the points and the estimated regression equation.
(C) the sum of the squared vertical distances between the points and the estimated regression equation.
(D) the sum of the squared perpendicular distances between the points and the estimated regression
equation.
(E) the sum of the squared horizontal distances between the points and the estimated regression
equation.
CAUTION: Questions 9 and 10 are based on the same information.
9. One year McDonald's had a promotion offering "30% more fries" for the regular price of a large order of
french fries. Suppose a usual large fries weighs, on average, 150 grams. To test to see if the London
franchises were providing the extra 30%, a sample of 100 of the special promotion large fries were
purchased and weighed. We wish to test the hypotheses (ì is a population mean, p is a binomial
proportion):
(A) H : ì = 195 versus H : ì > 195
o a
(B) H :op = 0.30 versus H : pa> 0.30
(C) H :oì = 195 versus H : ìa< 195
(D) H : p = 0.30 versus H : p < 0.30
o a
(E) The correct answer is not given in (A) to (D).
10. Which one of the following may seriously affect the results of the previous test, due to the violation of
an important assumption?
(A) The weights of orders of french fries may not be normally distributed.
(B) The variability in the lengths of the fries in different packages might not be constant.
(C) It will be very difficult to get a random sample.
(D) The salt on the fries will need to be licked off, but this will leave a thin coating of saliva on each
french fry.
(E) We do not know the population variance.
11. To test the null hypothesis H : ì0= 5000 versus the alternative hypothesis H : ì < 1000, you take a
random sample of size 200 and obtain a sample mean of 4100 and a sample standard deviation of 9000.
What is the approximate p-value (to two decimal places) for this test?
(A) 0.05
(B) 0.08
(C) 0.16
(D) 0.84
(E) 0.92 12. One thousand individuals are classified according to sex and according to whether they are colour blind
(assume these 1000 individuals constitute a random sample from the population of interest). The data
are:
Male Female
Colour blind 30 10
Not colour blind 490 470
To test whether colour blindness depends on sex, a contingency table analysis can be used. The value of
the chi-square test statistic is
(A) 3.84
(B) 4.92
(C) 6.54
(D) 8.83
(E) 9.36
13. In a hypothesis test, a p-value of 0.08 was obtained for the observed value of the test statistic. From this
we can conclude:
(A) there is little evidence against the null hypothesis, since the p-value is so low.
(B) the null hypothesis can only be rejected if the level of significance is exactly 8%.
(C) the null hypothesis should be rejected at the 10% level of significance but not at the 1% level of
significance.
(D) the null hypothesis should be rejected at the 5% level of significance but not at the 10% level of
significance.
(E) we cannot decide, because if a t test was used, we don't know whether it was a one-sided or two-
sided test.
14. A one-way Analysis of Variance (ANOVA) was used to test the equality of four treatment means, where
each treatment was replicated 6 times (thus there are 6 observations per treatment). The sum of squares
for treatments was 72, and the pooled sample variance was 5.1. What was the value of the F test
statistic?
(A) 0.74
(B) 2.42
(C) 3.10
(D) 3.35
(E) 4.71
15. The population of Canada, although only 1/10 as large as the population of the United States, is
nevertheless similar in many ways – including, for example, the proportion who smoke. A random
sample of 1000 Americans gave an estimate for this proportion, with an associated standard error. To
get an equally accurate estimate for Canada, a random sample of Canadians would have to be:
(A) 100 times smaller
(B) 10 times smaller
(C) of equal size
(D) 10 times larger
(E) 100 times larger. 16. To test if tracheal mite infections increase during the summer, 12 bee hives were sampled in May and
again in September. The variable measured was the percentage of bees in the hive with mites. The
following data were obtained:
Hive 1 2 3 4 5 6 7 8 9 10 11 12
May % 15 5 0 0 65 25 0 10 75 25 20 5
Sept % 85 55 5 15 40 70 25 15 80 10 35 45
Which one of the following procedures could be used to test the hypothesis of the experimenter?
(A) chi-square goodness-of-fit test
(B) paired t test
(C) chi-square test of independence (contingency table analysis)
(D) test for equality of binomial proportions
(E) independent sample t test
17. Three independent random samples were made to estimate the population standard deviation, ó. The
summary statistics were:
Sample 1: n1= 10, 1 = 5
Sample 2: n2= 20, 2 = 9
Sample 3: n3= 13, 3 = 8
The pooled estimate of ó based on all three samples is (to three decimal places):
(A) 7.333
(B) 7.800
(C) 7.927
(D) 7.956
(E) 7.988
18. A researcher calculated that a sample size of 600 would assure that a binomial proportion p could be
estimated within a bound of 0.04 with 95% confidence, even if p = 0.5. If this researcher wanted to
lower the bound to 0.01, how many times larger would the sample size need to be?
(A) 2
(B) 4
(C) 8
(D) 16
(E) 64 CAUTION: Questions 19 and 20 refer to the following information.
Krefting and Roe (1949) conducted an experiment to investigate the rates of seed survival when the seed
passes through the digestive system of two species of birds, ring-necked pheasant and bob-white quail.
Recovery rates of whole grey dogwood seeds from the guano (bird shit) were 35% for pheasants and
46% for quail. Suppose that 500 seeds were fed to each species, and we can assume that the usual
conditions for binomial proportions apply.
19. A 95% confidence interval for the difference in seed survival proportions between the two species of
birds would have a bound of:
(A) 0.02
(B) 0.03
(C) 0.04
(D) 0.05
(E) 0.06
20. Suppose it was hypothesized, prior to gathering the data, that seed survival would be higher in pheasants
than in quail. Let p and p be the true proportion of grey dogwood seeds that will survive passing
1 2
through pheasants and quail, respectively. Against the null hypothesis of no difference, the alternative
hypothesis would be:
(A) p ! p > 0
1 2
(B) p 1 p <20
(C) p 1 p
20
(D) p ! p $ 0
1 2
(E) p 1 p #20
21. Let á be the significance level of a hypothesis test. Why would one choose a small value of á?
(A) To increase the power of the test.
(B) To reduce the chance of falsely accepting the null hypothesis.
(C) To seldom reject a true null hypothesis.
(D) To minimize the chance of any kind of error.
(E) To ensure the p-value will be small.
22. The type of graph used in linear regression and correlation analysis to illustrate the bivariate data is
called a:
(A) box plot.
(B) relative frequency histogram.
(C) pie chart.
(D) cumulative frequency polygon.
(E) scatter diagram.
23. In simple linear regression, the units of the intercept must be:
(A) the same as the units of the dependent variable.
(B) the same as the units of the independent variable.
(C) the same as the units of the slope.
(D) the units of dependent variable divided by the units of the independent variable.
(E) dimensionless (has no units). NOTE: The next two questions refer to the following information.
To test if a coin was biased, a prisoner of war actually flipped a coin 10,000 times and observed 5067
heads.
24. What is the p-value for testing whether or not the coin was biased?
(A) 0.09
(B) 0.18
(C) 0.36
(D) 0.41
(E) 0.82
25. At the 5% level of significance, do you reject the null hypothesis that the coin is fair?
(A) yes
(B) no
(C) –––
(D) –––
(E) –––
26. Does the amount of sleep obtained before a test affect a student's performance on the test? To study this
question, students in a certain statistics exam were asked to record how many hours they had slept in the
24 hour period prior to the exam. A random sample of 6 students gave the following results:
Hours of Sleep Mark (Y)
(X)
8 66
1 36
7 90
10 73
3 54
4 87
[Some possibly useful calculations are: 3i = 33, Gyi= 406, Sum of Squares for x is Sxx= 57.5, Sum of
Squares for y is Syy= 2093.33, Sum of Cross-Products for xy is SPxy 201.]
When a simple linear regression model is fit using the method of least squares, the estimates of the Y-
intercept and slope are, respectively:
(A) 3.50, 48.4
(B) 48.4, 3.50
(C) 4.12, 38.6
(D) 38.6, 4.12
(E) 14.6, 18.2 27. To investigate the impact of free trade, an economist took a random sample of Canadian and U.S. tool-
and-die firms. Seven out of 20 Canadian firms, and 18 out of 30 U.S. firms, used computer-controlled
machine tools (robots). What is the approximate p-value for testing the null hypothesis that there is no
difference between the countries in the proportion of firms using robots? (The alternative hypothesis is
that there is a difference.)
(A) 0.04
(B) 0.06
(C) 0.08
(D) 0.10
(E) 0.12
28. For a particular species of tree, a simple linear regression analysis was conducted to investigate the
association between leaf biomass (in kg) and tree age (in years). Leaf biomass is treated as the
dependent variable and tree age as the independent variable. The slope will have the following units:
(A) kg.
(B) years.
(C) kg/year.
(D) years/kg.
(E) it will be unitless.
29. A 95% confidence interval for the mean of a certain population was calculated to be (56.2, 69.4).
Consider the following statements:
I. In repeated sampling, we know that 95% of possible sample means will fall in this interval.
II. This confidence interval provides evidence to reject the hypothesis that the population mean is 55.
III. This confidence interval either contains the population mean or it does not.
Which of these statements are TRUE?
(A) I and II
(B) I and III
(C) II and III
(D) All three of the statements are true
(E) Only one of the statements is true.
30. For a particular hypothesis test, the null hypothesis is rejected at the 5% level of significance. At the 1%
level of significance, which one of the following is true?
(A) The null hypothesis will definitely be rejected.
(B) The null hypothesis will definitely be accepted.
(C) The null hypothesis may or may not be rejected.
(D) Whether or not the null hypothesis is rejected depends on whether the test is one-sided or two-
sided.
(E) Whether or not the null hypothesis is rejected depends on whether the population standard
deviation is known or unknown. CAUTION: Questions 31 and 32 are based on the following information.
R.A. Fisher’s famous data set on iris flowers consists of 50 measurements on four variables for each of
three species (Iris setosa, Iris versicolor, and Iris virginica). A one-way Analysis of Variance was
conducted to compare the means of the three species for the variable Sepal Width. Key parts of the S-
Plus output is given below.
*** Analysis of Variance Model ***
Df Sum of Sq Mean Sq F Value Pr(F)
Species 2 1134.493 567.2467 49.16004 0
Residuals 147 1696.200 11.5388
Tables of means
Grand mean
30.573
Species
Setosa Versicolor Virginica
34.28 27.7 29.74
Standard errors for differences of means
Species
0.67938
replic. 50.00000
95 % non-simultaneous confidence intervals for specified
linear combinations, by the Fisher LSD method
critical point: 1.9762
response variable: Sepal.W.
intervals excluding 0 are flagged by '****'
Estimate Std.Error Lower Bound Upper Bound
Setosa-Versicolor 6.58 0.679 5.24 7.920 ****
Setosa-Virginica 4.54 0.679 3.20 5.880 ****
Versicolor-Virginica -2.04 0.679 -3.38 -0.697 ****
31. If you squared the difference between each observation and the grand mean of 30.573 and then summed
these squared differences you should get a value of:
(A) 1134.493
(B) 1696.200
(C) 2830.693
(D) 567.2467
(E) 11.5388
32. The results of Fisher’s least significant difference procedure at the 5% level of significance are:
(A) All pairs of means differ significantly.
(B) None of the pairs of means differ significantly.
(C) Setosa differs from the other two species, but versicolor and virginica do not differ significantly.
(D) Only versicolor and virginica differ significantly.
(E) Only setosa and virginica differ significantly. 33. Industrial wastes and sewage dumped into our rivers and streams absorb oxygen and thereby reduce the
amount of dissolved oxygen available for fish and other aquatic life. Suppose that provincial
regulations state that water must have, on average, a minimum of 5 parts per million (ppm) of dissolved
oxygen to support aquatic life. Six water specimens taken from a river near a large city gave readings
(in ppm) for the amount of dissolved oxygen of: 4.7, 4.3, 5.0, 4.1, 3.6, 5.2
Assuming oxygen readings are approximately normal and represent a random sample, we can test the
hypothesis that minimum dissolved oxygen standards are not being maintained in this river at the 5%
level of significance. The appropriate set of null and alternative hypotheses are:
(A) H : 0 $ 5; H : A < 5
(B) H : 0 # 5; H : A > 5
(C) H : ì > 5; H : ì # 5
0 A
(D) H : 0 < 5; H :Aì $ 5
(E) none of the above.
34. In a test for no linear association in a simple linear regression analysis, a p-value of 0.03 was obtained
for the observed value of the test statistic. From this we can conclude:
(A) the null hypothesis should be rejected at the 1% level of significance but not at the 5% level of
significance.
(B) the null hypothesis should be rejected at the 5% level of significance but not at the 1% level of
significance.
(C) there is little evidence against the null hypothesis, since the p-value is so low.
(D) the null hypothesis can only be rejected if the level of significance is exactly 3%.
(E) we cannot decide, because if a t test was used, we don't know whether it was a one-sided or two-
sided test.
35. A one-way ANOVA (Analysis of Variance) was used to test the equality of treatment means for an
experiment that had six treatments and five replications (experimental units) per treatment. The
following quantities were calculated: SS for Treatments = SSTr = 99.0, Total SS = 215.0. When the
ANOVA table is completed, the value of the F test statistic and the decision on the null hypothesis
(using a 5% level of significance) is:
(A) 0.85, accept null hypothesis
(B) 2.2, accept null hypothesis
(C) 2.2, reject null hypothesis
(D) 4.1, accept null hypothesis
(E) 4.1, reject null hypothesis.
36. Consider the following distributions: binomial, normal, t, F, chi-square. For how many of these
distributions is it impossible to obtain a negative value for the random variable?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 37. To test if two bee species differ in their preferences for two species of flowers, 250 visits by different
individual bees were recorded (assume these 250 visits constitute a random sample from the population
of interest). The data are summarized in the following table.
Bee Species 1 Bee Species 2
Flower Species 1 52 83
Flower Species 2 68 47
A contingency table analysis can be used to test the null hypothesis that bee species do not differ in their
choice of flower species. The value of the chi-square test statistic is
(A) 6.82
(B) 8.46
(C) 10.57
(D) 12.46
(E) 14.97
38. In a simple linear regression analysis, the proportion of variation in the dependent variable Y which
cannot be explained by its linear association with the independent variable X is measured by:
(A) the sum of squares for regression, divided by the total sum of squares.
(B) the sum of squares for regression, divided by the sum of squares for error.
(C) the mean square for regression, divided by the mean square for error.
(D) the coefficient of determination.
(E) one minus the coefficient of determination.
39. A chi-square test of independence was conducted at the 5% level of significance on a contingency table
that had an equal number of rows and columns. The decision rule was to reject the null hypothesis if the
value of the test statistic exceeds 16.92. The dimensions of the contingency table must be:
(A) 2×2
(B) 3×3
(C) 4×4
(D) 5×5
(E) 6×6
40. The t-table provided is not ideal for calculating a p-value for a t test statistic value. Nonetheless, we can use the
t-table to obtain a range of possible values for the p-value. If a two-sided t test was conducted, the t test statistic
had 16 degrees of freedom, and a test statistic value of !2.183 was obtained, the t-table would allow us to say:
(A) 0.005 < p-value < 0.01.
(B) 0.01 < p-value < 0.025.
(C) 0.02 < p-value < 0.05.
(D) 0.05 < p-value < 0.10.
(E) 0.01 < p-value < 0.02. CAUTION: Questions 41 and 42 refer to the following information.
The resin in the bark of some tropical trees seems to confer protection against termites. Researchers
conducted an experiment that investigated the effect of one type of resin on termites. The resin was
dissolved in a solvent and placed on filter paper in different doses (5 mg and 10 mg). For each dosage
level, eight dishes were set up with 25 termites in each dish. After three days, the number of surviving
termites was counted.
41. To test (at the 5% level of significance) if a higher dosage of the resin will reduce termite survival, the
alternative hypothesis and decision rule for the appropriate t test would be (we will subtract the 10 mg
mean from the 5 mg mean):
(A) H : a !5ì > 010reject H if to> 1.895
(B) H : a !5ì > 010reject H if to> 1.761
(C) H : ì ! ì < 0; reject H if t < !1.895
a 5 10 o
(D) H : a !5ì < 010reject H if to< !1.761
(E) H : a !5ì < 010reject H if to< !2.145
42. For the 5 mg dosage, the mean number of termites surviving per dish was 22.4 and the sample variance
was 0.36. For the 10 mg dosage, the mean number of termites surviving per dish was 16.4 and the
sample variance was 0.44. The value of the t test statistic was:
(A) 2.6
(B) 3.9
(C) 4.6
(D) 5.4
(E) 19.0
43. One thousand individuals are classified according to sex and according to whether they are colour blind
(assume these 1000 individuals constitute a random sample from the population of interest). The data
are:
Male Female
Colour blind 30 10
Not colour blind 490 470
A 95% confidence interval for the proportion of males who are colour blind is:
(A) (0.0194, 0.0406)
(B) (0.0377, 0.0777)
(C) (0.0400, 0.0824)
(D) (0.0422, 0.0802)
(E) (0.0424, 0.0730)
44. Consider the following distributions: binomial, normal, t, F, chi-square. For how many of these
distributions is the random variable discrete?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 Note: The next two questions are based on the following information.
A Canadian study compared % butterfat content for breeds of cattle. The following edited S-Plus
analysis compares mature cattle of four breeds, with the milk of 10 randomly selected cattle of each
breed being measured for butterfat content, using a one-way ANOVA. The breed codes are: A =
Ayrshire, G = Guernsey, H = Holstein, J = Jersey.
*** Analysis of Variance Model ***
Df Sum of Sq Mean Sq F Value Pr(F)
Breed 3 15.21607 5.072022 29.10722 9.848996e-010
Residuals 36 6.27311 0.174253
Tables of means
Grand mean
4.4543
Breed
A G H J
4.003 4.848 3.721 5.245
95 % non-simultaneous confidence intervals for specified
linear combinations, by the Fisher LSD method
critical point: 2.0281
response variable: Butterfat
intervals excluding 0 are flagged by '****'
Estimate Std.Error Lower Bound Upper Bound
A-G -0.845 0.187 -1.2200 -0.4660 ****
A-H 0.282 0.187 -0.0966 0.6610
A-J -1.240 0.187 -1.6200 -0.8630 ****
G-J -0.397 0.187 -0.7760 -0.0184 ****
H-J -1.520 0.187 -1.9000 -1.1500 ****
45. The ANOVA table F-test allows us to test the null hypothesis that the four breeds have equal population
means. Which one of the following is the most appropriate conclusion from the F test, at the 5% level of
significance?
(A) All means must differ from one another, because the p-value is so small.
(B) Most means must differ from one another, because the p-value is so small.
(C) At least one mean must differ from the others, because the p-value is so small.
(D) The null hypothesis must be accepted, because the p-value is so small.
(E) The null hypothesis must be accepted, because the p-value is so large.
46. The results of the LSD comparisons allow us to conclude, at the 5% level of significance, all of the
following except:
(A) Mean butterfat is higher in Jerseys than in the other three breeds.
(B) Mean butterfat is higher in Guernseys than it is in Holsteins or Ayrshires.
(C) Mean butterfat does not differ significantly between Ayrshires and Holsteins.
(D) Mean butterfat is lower in Holsteins than in any of the other three breeds.
(E) Mean butterfat is lower in Ayrshires than in Guernseys or Jerseys. 47. An important parameter in botanical studies is the germination rate, defined as the proportion of seeds in
a population that will germinate under standardized test conditions. A random sample of 800 seeds from
a certain population were tested for germination; 520 germinated and 280 did not. The 95% confidence
interval for estimating the proportion of seeds in the population that will germinate has the limits:
(A) 0.650 ± 0.017
(B) 0.650 ± 0.028
(C) 0.650 ± 0.033
(D) 0.650 ± 0.043
(E) 0.650 ± 0.048
CAUTION: Questions 48 and 49 are based on the following information.
A company’s fitness director established an employee health program. Part of the program evaluation
required measuring the body mass index (BMI) both at the beginning of the program (baseline BMI) and
then two years into the program (two-year BMI). A regression analysis of two-year BMI on baseline
BMI was performed on the data from a random sample of 12 employees using S-Plus. The output is
given below.
*** Linear Model ***
Call: lm(formula = Two.Year.BMI ~ Baseline.BMI, data = BMI, na.action = na.exclude)
Residuals:
Min 1Q Median 3Q Max
-0.6946 -0.3253 -0.0126 0.304 0.9367
Coefficients: Value Std. Error t value Pr(>|t|)
(Intercept) 4.1093 6.8114 0.6033 0.5597
Baseline.BMI 0.8382 0.2629 3.1886 0.0097
Residual standard error: 0.4927 on 10 degrees of freedom
Multiple R-Squared: 0.5041
F-statistic: 10.17 on 1 and 10 degrees of freedom, the p-value is 0.009677
48. The linear regression equation predicts what value of two-year BMI for an employee with a baseline
BMI of 25.80?
(A) 25.41
(B) 25.73
(C) 25.96
(D) 26.20
(E) 26.57
49. Is there a statistically significant linear relationship between the two-year BMI and baseline BMI at the
5% level of significance, and why or why not?
(A) No, because the intercept’s t statistic value has a p-value greater than 0.05.
(B) Yes, because the intercept’s t statistic value has a p-value greater than 0.05.
(C) No, because the coefficient of baseline BMI’s t statistic value has a p-value less than 0.05.
(D) Yes, because the coefficient of baseline BMI’s t statistic value has a p-value less than 0.05.
(E) No, because the F statistic value has a p-value less than 0.05. CAUTION: The next two questions refer to the following information.
Studies conducted in Australia indicate that there may be a diffnce between the pain thresholds of
blonds and brunettes. An experiment was conducted to determine whether hair colour is related to the amount of
pain produced by common types of mishaps. Nineteen randomly-sampled men were divided into four categories
of hair colour: light blond (A), dark blond (B), light brunette (C), or dark brunette (D); there were five men of
each hair colour except for light brunette, which had four men. Each man in the experiment was given a pain
threshold sensitivity score based on his performance in a pain sensitivity test (the higher the score, the higher the
person’s pain tolerance). The following output comes from an S-Plus analysis of the data.
*** Analysis of Variance Model ***
Df Sum of Sq Mean Sq F Value Pr(F)
colour 3 1360.726 453.5754 6.791407 0.004114227
Residuals 15 1001.800 66.7867
Tables of means
Grand mean
47.842
colour
A B C D
59.2 51.2 42.5 37.4
rep 5.0 5.0 4.0 5.0
95 % non-simultaneous confidence intervals for specified
linear combinations, by the Fisher LSD method
critical point: 2.1314
response variable: sensitivity
intervals excluding 0 are flagged by '****'
Estimate Std.Error Lower Bound Upper Bound
A-B 8.0 5.17 -3.02 19.0
A-C 16.7 5.48 5.02 28.4 ****
A-D 21.8 5.17 10.80 32.8 ****
B-C 8.7 5.48 -2.98 20.4
B-D 13.8 5.17 2.78 24.8 ****
C-D 5.1 5.48 -6.58 16.8
2 2
50. Calculating 5(59.2 ! 47.842) + ... + 5(37.4 ! 47.842) would give you:
(A) Sum of Squares for Treatments
(B) Sum of Squares for Error
(C) Total Sum of Squares
(D) Mean Square for Treatments
(E) Mean Square for Error
51. How many pairs of means were found to differ at the 5% level of significance by the Fisher least significant
difference procedure?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6 52. A researcher wishes to estimate the difference between two population means within 3 units of the true
value with a 95% confidence interval, using independent random samples of equal size. If the
population standard deviations are known and both are equal to 10 units, what minimum sample size is
required for each sample?
(A) 43
(B) 86
(C) 171
(D) 342
(E) 684
CAUTION: The next two questions refer to the following information.
A major beverage company (Brand A) and its closest competitor (Brand B) each had 40% of the cola
market, while another competitor (Brand C) had 20% of the market. After an aggressive marketing
campaign, the company sets out to evaluate the effect of the campaign. The company randomly selected
a panel of 100 people and asked them to taste and state their preference among three glasses of the three
competing colas. 50 preferred Brand A, 35 preferred Brand B, and 15 preferred Brand C.
53. If one tested the hypothesis that the probabilities of preferring Brand A, B, and C were 2/5, 2/5, and 1/5,
respectively, then what is the value of the chi-square test statistic?
(A) < 3.0
(B) between 3.0 and 4.5
(C) between 4.5 and 6.0
(D) between 6.0 and 9.0
(E) > 9.0
54. What are the degrees of freedom for the appropriate test statistic of the associated null hypothesis?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
55. A day's production of automobile radiators was randomly sampled for quality control to estimate the
proportion that leaked. The sample proportion was 0.08, and the corresponding 95% confidence
interval was (0.05, 0.11). Suppose the production process doesn't change, so that the population
proportion that are leaking remains constant from day to day. What can we say about similar random
samples of the same size to be taken over the next 10 years (2400 working days)?
(A) About 95% of the sample proportions will be equal to the population proportion.
(B) About 95% of the sample proportions will fall in the interval (0.05, 0.11)
(C) About 95% of the confidence intervals (similarly constructed) will cover the value 0.08.
(D) About 95% of the confidence intervals (similarly constructed) will cover the population proportion.
(E) About 95% of the radiators will last at least 10 years without springing a leak. CAUTION: The next two questions refer to the following information.
To test if two bee species differ in their preferences for two species of flowers, 250 visits by different
individual bees were recorded (assume these 250 visits constitute a random sample from the population
of interest). The data are summarized in the following table.
Bee Species 1 Bee Species 2
Flower Species 1 52 83
Flower Species 2 68 47
56. The null hypothesis that bee species do not differ in their choice of flower species can be tested either
with a chi-square test or with a test that uses the standard normal distribution. If the latter test was used,
the absolute value of the “z” test statistic is
(A) 1.46
(B) 2.14
(C) 2.52
(D) 2.68
(E) 3.25
57. Suppose it had been hypothesized, prior to gathering the data, that Bee Species 2 will have a stronger
preference than Bee Species 1 for Flower Species 1. Let1p and 2 be the true proportion of bees that
visit Flower Species 1 for Bee Species 1 and 2, respectively. Against the null hypothesis of no
difference, the alternative hypothesis would be:
(A) p 1 p >20
(B) p ! p < 0
1 2
(C) p 1 p
20
(D) p 1 p $20
(E) p 1 p #20
58. Retired athletes in three professional sports were examined for unusually high levels of knee damage.
The following data set was obtained.
Hockey Soccer Football Total
Level of Minimal 43 16 21 80
Knee
Moderate 26 17 32 75
Damage?
High 22 8 42 72
Total 91 41 95 227
To test if the level of knee damage was independent of the sport played, a chi-square contingency table
analysis could be used. Under the appropriate null hypothesis, the expected frequency for the category
“soccer player and minimal damage” would be:
(A) 6.37
(B) 12.61
(C) 14.45
(D) 17.24
(E) 18.42 Note: The next two questions refer to the following information.
Lea (1965) conducted a study in certain regions of Great Britain, Norway, and Sweden to investigate the
relationship between mean annual temperature and the mortality rate for a type of breast cancer in
women. The graph of the data includes the fitted least squares regression line for regressing the
mortality index on temperature. Only part of the S-Plus analysis is provided.
Coefficients:
Value Std. Error t value Pr(>|t|)
(Intercetemp-212.3577 10.3489 -6.7578 0.0000
Analysis of Variance Table
Response: mortality
Terms added sequentially (first to last)
Df Sum of Sq Mean Sq F Value Pr(F)
temp 1 2599.534 2599.534 45.66847 9.201838e-006
Residuals 14 796.906 56.922
59. What is the value of the residual for the point (34.0, 52.5)?
(A) !5.87
(B) !4.68
(C) 3.65
(D) 4.68
(E) 5.87
60. A 95% confidence interval for the true slope (1 ) has the limits:
(A) 2.36 ± 0.35
(B) 2.36 ± 0.75
(C) 2.36 ± 0.93
(D) 2.36 ± 2.15
(E) 2.36 ± 2.36
61. The following four statistics are often calculated in simple linear regression and correlation analysis: y-
intercept, slope, correlation coefficient, and coefficient of determination. How many of these statistics
will always be unitless?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 62. A researcher calculated the sample size required to estimate a mean within a certain bound using a 95%
confidence interval. However, the sample size required was very large, in fact in excess of n = 400. Two options
were then considered:
I. Relax the bound, such that the length of the confidence interval would be increased by 50%.
II. Use a 90% confidence interval rather than a 95% confidence interval, thus doubling the chance of not
capturing the mean in the confidence interval.
Of these two options, which one will give the greatest decrease in the required sample size?
(A) I
(B) II
(C) Insufficient information to decide
(D) I if ì/ó < 0.10, but II if ì/ó > 0.10
(E) I if the population mean is positive, but II if the population mean is negative
63. Krefting and Roe (1949) conducted an experiment to investigate the rates of seed survival when the seed
passes through the digestive system of two species of birds, ring-necked pheasant and bob-white quail.
Recovery rates of whole grey dogwood seeds from the guano (bird shit) were 35% for pheasants and
46% for quail. Suppose that 500 seeds were fed to each species, and we can assume that the usual
conditions for binomial proportions apply. While we could test for a difference of binomial proportions
using a “z” test, we could also use a 2×2 contingency table. If we use the latter, the value of the chi-
square test statistic would be closest to:
(A) 2.51
(B) 5.08
(C) 6.28
(D) 7.47
(E) 12.55
64. Two groups of rats [(a) normal and (b) adrenalectomized] were tested for blood viscosity. The summary
statistics are:
(a) (b)
sample size 11 9
sample mean 3.921 4.111
sample s.d. 0.527 0.601
If it was hypothesized, prior to obtaining the data, that adrenalectomized rats would have higher mean
blood viscosity, the appropriate alternative hypothesis for comparing the population mean of normal rats
(ì1) with the population mean of adrenalectomized rats (ì ) wo2ld be:
(A) ì !1ì > 2
(B) ì !1ì < 2
(C) ì !1ì
2
(D) ì !1ì $ 2
(E) ì !1ì # 2
65. In simple linear regression and correlation analysis, the rate at which Y changes as X increases is
measured by:
(A) the Y intercept
(B) the slope
(C) the correlation coefficient
(D) the variance about the regression line
(E) the coefficient of determination Note: The next two questions refer to the following information.
A track coach has learned of two new training techniques that are designed to reduce the time required to
run the mile. Three random samples of

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