Download 1998 Exam – Question 5 - Spokane Public Schools

AP STATS REVIEW – INFERENCE PROCEDURES
(2009 #5)
For many years, the medically accepted practice of giving aid to a person experiencing a heart attack was to have the
person who placed the emergency call administer chest compression (CC) plus standard mouth-to-mouth
resuscitation (MMR) to the heart attack patient until the emergency response team arrived. However, some
researchers believed that CC alone would be a more effective approach.
In the 1990s a study was conducted in Seattle in which 518 cases were randomly assigned to treatments: 278 to CC
plus standard MMR and 240 to CC alone. A total of 64 patients survived the heart attack: 29 in the group receiving
CC plus standard MMR, and 35 in the group receiving CC alone. A test of significance was conducted on the
following hypotheses.
Ho: The survival rates for the two treatments are equal.
Ha: The treatment that uses CC alone produces a higher survival rate.
This test resulted in a p-value of 0.0761.
a)
Interpret what this p-value measures in the context of this study.
b) Based on this p-value and study design, what conclusion should be drawn in the context of this study? Use
a significance level of α = 0.05.
c)
Based on your conclusion in part (b), which type of error, Type I or Type II, could have been made? What
is one potential consequence of this error?
(2010 #3)
A humane society wanted to estimate with 95 percent confidence the proportion of households in its county that own
at least one dog.
a)
Interpret the 95 percent confidence level in this context.
The humane society selected a random sample of households in its county and used the sample to estimate the
proportion of all households that own at least one dog. The conditions for calculating a 95 percent confidence
interval for the proportion of households in this county that own at least one dog were checked and verified, and the
resulting confidence interval was 0.417 ± 0.119.
b) A national pet products association claimed that 39 percent of all American households owned at least one
dog. Does the humane society’s interval estimate provide evidence that the proportion of dog owners in its
county is different from the claimed national proportion? Explain.
c)
How many households were selected in the humane society’s sample? Show how you obtained your
answer.
(2010 #5)
A large pet store buys the identical species of adult tropical fish from two different suppliers—Buy-Rite Pets and
Fish Friends. Several of the managers at the pet store suspect that the lengths of the fish from Fish Friends are
consistently greater than the lengths of the fish from Buy-Rite Pets. Random samples of 8 adult fish of the species
from Buy-Rite Pets and 10 adult fish of the same species from Fish Friends were selected and the lengths of the fish,
in inches, were recorded, as shown in the table below.
Do the data provide convincing evidence that the mean length of the adult fish of the species from Fish Friends is
greater than the mean length of the adult fish of the same species from Buy-Rite Pets?
(2011 #4)
High cholesterol levels in people can be reduced by exercise, diet, and medication. Twenty middle-aged males with
cholesterol readings between 220 and 240 milligrams per deciliter (mg/dL) of blood were randomly selected from
the population of such male patients at a large local hospital. Ten of the 20 males were randomly assigned to group
A, advised on appropriate exercise and diet, and also received a placebo. The other 10 males were assigned to group
B, received the same advice on appropriate exercise and diet, but received a drug intended to reduce cholesterol
instead of a placebo. After three months, posttreatment cholesterol readings were taken for all 20 males and
compared to pretreatment cholesterol readings. The tables below give the reduction in cholesterol level
(pretreatment reading minus posttreatment reading) for each male in the study.
Do the data provide convincing evidence, at the α = 0.01 level, that the cholesterol drug is effective in producing a
reduction in mean cholesterol level beyond that produced by exercise and diet?
(2011B #4)
A parent advisory board for a certain university was concerned about the effect of part-time jobs on the academic
achievement of students attending the university. To obtain some information, the advisory board surveyed a simple
random sample of 200 of the more than 20,000 students attending the university. Each student reported the average
number of hours spent working part-time each week and his or her perception of the effect of part-time work on
academic achievement. The data in the table below summarize the students’ responses by average number of hours
worked per week (less than 11, 11 to 20, more than 20) and perception of the effect of part-time work on academic
achievement (positive, no effect, negative).
A chi-square test was used to determine if there is an association between the effect of part-time work on academic
achievement and the average number of hours per week that students work. Computer output that resulted from
performing this test is shown below.
a)
State the null and alternative hypotheses for this test.
b) Discuss whether the conditions for a chi-square inference procedure are met for these data.
c)
Given the results from the chi-square test, what should the advisory board conclude?
d) Based on your conclusion in part (c), which type of error (Type I or Type II) might the advisory board have
made? Describe this error in the context of the question.
(2011B #5)
During a flu vaccine shortage in the United States, it was believed that 45 percent of vaccine-eligible people
received flu vaccine. The results of a survey given to a random sample of 2,350 vaccine-eligible people indicated
that 978 of the 2,350 people had received flu vaccine.
a) Construct a 99 percent confidence interval for the proportion of vaccine-eligible people who had received
flu vaccine. Use your confidence interval to comment on the belief that 45 percent of the vaccine-eligible
people had received flu vaccine.
b) Suppose a similar survey will be given to vaccine-eligible people in Canada by Canadian health officials. A
99 percent confidence interval for the proportion of people who will have received flu vaccine is to be
constructed. What is the smallest sample size that can be used to guarantee that the margin of error will be
less than or equal to 0.02 ?
MULTIPLE CHOICE:
CHAPTER 10
1.
You want to compute a 96% confidence interval for a population mean. Assume that the population standard
deviation is known to be 10 and the sample size is 50. The value of z* to be used in this calculation is
(a)1.960
(b) 1.645
(c) 1.7507
(d) 2.0537
(e) None. The answer is
2.
You want to estimate the mean SAT score for a population of students with a 90% confidence interval. Assume
that the population standard deviation is  = 100. If you want the margin of error to be approximately 10, you
will need a sample size of
(a) 16
(b) 271
(c) 38
(d) 1476
(e) None. The answer is
.
3.
A significance test gives a P-value of 0.04. From this we can
(a) Reject H0 at the 1% significance level
(b) Reject H0 at the 5% significance level
(c) Say that the probability that H0 is false is 0.04
(d) Say that the probability that H 0 is true is 0.04
(e) None of the above. The answer is
.
4.
A significance test was performed to test the null hypothesis H 0: µ = 2 versus the alternative Ha: µ  2. The
test statistic is z = 1.40. The P-value for this test is approximately
(a) 0.16
(b) 0.08
(c) 0.003
(d) 0.92
(e) 0.70
(f) None. The answer is
.
5.
You have measured the systolic blood pressure of a random sample of 25 employees of a
company located near you. A 95% confidence interval for the mean systolic blood pressure for the employees
of this company is (122, 138). Which of the following statements gives a valid interpretation of this interval?
(a) Ninety-five percent of the sample of employees has a systolic blood pressure between 122 and 138.
(b) Ninety-five percent of the population of employees has a systolic blood pressure between 122 and 138.
(c) If the procedure were repeated many times, 95% of the resulting confidence intervals would contain the
population mean systolic blood pressure.
(d) The probability that the population mean blood pressure is between 122 and 138 is .95.
(e) If the procedure were repeated many times, 95% of the sample means would be between 122 and 138.
(f) None of the above. The answer is
.
6.
An analyst, using a random sample of n = 500 families, obtained a 90% confidence interval for mean monthly
family income for a large population: ($600, $800). If the analyst had used a 99% confidence coefficient
instead, the confidence interval would be:
(a)
(b)
(c)
(d)
(e)
7.
Narrower and would involve a larger risk of being incorrect
Wider and would involve a smaller risk of being incorrect
Narrower and would involve a smaller risk of being incorrect
Wider and would involve a larger risk of being incorrect
Wider but it cannot be determined whether the risk of being incorrect would be larger or smaller
To assess the accuracy of a laboratory scale, a standard weight that is known to weigh 1 gram is repeatedly
weighed a total of n times and the mean x of the weighings is computed. Suppose the scale readings are
normally distributed with unknown mean  and standard deviation
 = 0.01 g. How large should n be so that a 95% confidence interval for  has a margin of error of ± 0.0001?
(a) 100
(b) 196
(c) 27061
(d) 10000
(e) 38416
.
CHAPTER 11
1.
You want to compute a 90% confidence interval for the mean of a population with unknown population standard
deviation. The sample size is 30. The value of t* you would use for this interval is
(a) 1.96
(b) 1.645
(c) 1.699
(d) 0.90
(e) 1.311
(f) None of the above
2.
A 95% confidence interval for the mean reading achievement score for a population of third-grade students is
(44.2, 54.2). The margin of error of this interval is
(a) 95% (b) 5
(c) 2.5 (d) 10 (e)The answer cannot be determined from the information given.
3.
The effect of acid rain upon the yield of crops is of concern in many places. In order to determine baseline
yields, a sample of 13 fields was selected, and the yield of barley (g/400m2) was determined. The output from
SAS appears below:
QUANTILES(DEF=4)
EXTREMES
N
13 SUM WGTS
13
100% MAX 392
99% 392
LOW
HIGH
MEAN
220.231 SUM
2863
75% Q3
234
95% 392
161
225
STD DEV 58.5721 VAR
3430.69
50% MED 221
90% 330
168
232
SKEW
2.21591 KURT
6.61979
25% Q1
174
10% 163
169
236
USS
671689 CSS
41168.3
0% MIN 161
5% 161
179
239
CV
26.5958 STD MEAN 16.245
1% 161
205
392
A 95% confidence interval for the mean yield is:
(a) 220.2 ± 1.96(58.6)
(b) 220.2 ± 1.96(16.2)
(c) 220.2 ± 2.18(58.6)
(d) 220.2 ± 2.18(16.2)
(e) 220.2 ± 2.16(16.2)
4.
To use the two-sample t procedure to perform a significance test on the difference between two means, we assume
(a) The populations’ standard deviations are known
(b) The samples from each population are independent
(c) The distributions are exactly normal in each population
(d) The sample sizes are large
(e) All of the above
5. We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the
experiment it was found that the new feed gave a 10 kg bigger gain than the old feed. A two-sample t-test with
the proper one-sided alternative was done and the resulting P-value was 0.082. This means:
(a) There is an 8.2% chance the null hypothesis is true.
(b) There was only a 8.2% chance of observing an increase greater than 10 kg (assuming the null hypothesis
was true).
(c) There was only an 8.2% chance of observing an increase greater than 10 kg (assuming the null hypothesis
was false).
(d) There is an 8.2% chance the alternate hypothesis is true.
(e) There is only an 8.2% chance of getting a 10 kg. increase.
6.
The water diet requires one to drink two cups of water every half 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
1
180
170
2
125
130
3
240
215
4__
150
152
For the population of all adults, assume that the weight loss after six weeks on the diet (weight before beginning
the diet – weight after six weeks on the diet) is normally distributed with mean µ. To determine if the diet leads to
weight loss, we test the hypotheses
H0: = 0, Ha:  > 0.
Based on these data we conclude that
(a) We would not reject H0 at significance level 0.10.
(b) We would reject H0 at significance level 0.10 but not at 0.05.
(c) We would reject H0 at significance level 0.05 but not at 0.01.
(d) We would reject H0 at significance level 0.01.
(e) The sample size is too small to allow use of the t procedures.
The next two questions refer to the following situation: In some mining operations, a byproduct of the
processing is mildly radioactive. Of prime concern is the possibility that release of these byproducts into the
environment may contaminate the freshwater supply. There are strict regulations for the maximum allowable
radioactivity in supplies of drinking water, namely an average of 5 picocuries per liter (pCi/L) or less. However,
it is well known that even safe water has occasional hot spots that eventually get diluted, so samples of water
are assumed safe unless there is evidence to the contrary. A random sample of 25 specimens of water from a
city’s water supply gave a mean of 5.39 pCi/L and a standard deviation of 0.87 pCi/L.
7. The appropriate null and alternative hypotheses are:
(a) H0: µ = 5.39 vs Ha: µ 5.39
(b) H0: µ = 5.39 vs Ha: µ < 5.00
(c) H0: µ = 5 vs Ha: µ = 5.39
(d) H0: µ = 5 vs Ha: µ < 5
(e) H0: µ = 5 vs Ha: µ > 5
8.
The value of the test statistic, the rejection region ( = 0.05), and the P-value (computed by a computer) are:
(a) z* = 2.24; reject if z* > 1.960; P-value = 0.0125
(b) z* = 2.24; reject if z* > 1.645; P-value = 0.0125
(c) t* = 2.24 with 25 df; reject if t* > 1.708; P-value = 0.0171
(d) t* = 2.24 with 24 df; reject if t* > 1.711; P-value = 0.0173
(e) t* = 2.24 with 24 df; reject if t* > 2.064; P-value = 0.0173
CHAPTER 12
1.
Some scientists believe that a new drug would benefit about half of all people with a certain blood disorder. To
estimate the proportion of patients who would benefit from taking the drug, the scientists will administer it to a
random sample of patients who have the blood disorder. What sample size is needed so that the 95%
confidence interval will have a width of 0.06?
(a) 748
(b) 1,068
(c) 1,503
(d) 2,056
(e) 2,401
2. A random sample of 900 individuals has been selected from a large population. It was found that 180 are regular
users of vitamins. Thus, the proportion of the regular users of vitamins in the population is estimated to be 0.20.
The standard error of this estimate is approximately:
(a) 0.1600
(b) 0.0002
(c) 0.4000
(d) 0.0133
(e) 0.0267
3. The college newspaper of a large Midwestern university periodically conducts a survey of students on campus to
determine the attitude on campus concerning issues of interest. Pictures of the students interviewed along with
quotes of their responses are printed in the paper. Students are interviewed by a reporter “roaming” the campus
selecting students to interview “haphazardly.” On a particular day the reporter interviews five students and asks
them if they feel there is adequate student parking on campus. Four of the students say no.
Which of the following conditions for inference about a proportion using a confidence interval are violated in
this example?
(a) The data are an SRS from the population of interest.
(b) The population is at least ten times as large as the sample.
ˆ )  10 .
(c) npˆ  10 and n(1  p
(d) We are interested in inference about a proportion.
(e) More than one condition is violated.
4.
A 95% confidence interval for p, the proportion of Canadian beer drinkers who prefer Lion Red was found to be
(0.236,0.282). Which of the following is correct?
(a) About 95% of beer drinkers have between a 23.6% and a 28.2% chance of drinking Lion Red.
(b) There is a 95% probability that the sample proportion lies between 0.236 and 0.282.
(c) If a second sample was taken, there is a 95% chance that its confidence interval would contain 0.25.
(d) This confidence interval indicates that we would likely reject the hypothesis H0: p = 0.25.
(e) We are reasonably certain that the true proportion of beer drinkers who prefer Lion Red is between 24%
and 28%.
5. Refer to the previous question. Suppose that the same poll was repeated in the United States (whose population is
10 times larger than Canada), but in this new pool, four times the number of people were interviewed. The
resulting 95% confidence intervals will be:
(a) about 1/2 as wide as the Canadian interval
(b) about 1/4 as wide as the Canadian interval
(c) about 1/10 as wide as the Canadian interval
(d) about 4/10 times as wide as the Canadian interval
(e) the same size as the Canadian interval
6.
In a large Midwestern university (the class of entering freshmen being on the order of 6000 or more students),
an SRS of 100 entering freshmen in 1993 found that 20 finished in the bottom third of their high school class.
Admission standards at the university were tightened in 1995. In 1997 an SRS of 100 entering freshmen found
that 10 finished in the bottom third of their high school class. Let p1 and p2 be the proportion of all entering
freshmen in 1993 and 1997, respectively, who graduated in the bottom third of their high school class.
What conclusion should we draw?
(a) We are 95% confident that the admissions standards have been tightened.
(b) Reject H0 at the  = 0.01 significance level.
(c) Fail to reject H0 at the  = 0.05 significance level.
(d) There is significant evidence of a decrease in the proportion of freshmen who graduated in the bottom third
of their high school class that were admitted by the university.
(e) If we reject H0 at the  = .05 significance level based on these results, we have a 5% chance of being
wrong.
7.
In a test of H0: p = 0.4 against Ha: p  0.4, a sample of size 100 produces z = 1.28 for the value of the test
statistic. Thus the P-value (or observed level of significance) of the test is approximately equal to:
(a) 0.90
(b) 0.40
(c) 0.05
(d) 0.20
(e) 0.10
CHAPTER 13
The next set of questions refer to the following situation
In the paper "Color Association of Male and Female Fourth-Grade School Children" (J. Psych., 1988, 383–8),
children were asked to indicate what emotion they associated with the color red. The response and the sex of the
child are noted and summarized below. The first number in each cell is the count, the second number is the row
percent.
Frequency|
Row Pct |anger
|happy
|love
|pain
|
---------+--------+--------+--------+--------+
f
|
27 |
19 |
39 |
17 |
---------+--------+--------+--------+--------+
m
|
34 |
12 |
38 |
28 |
---------+--------+--------+--------+--------+
Total
61
31
77
45
Total
102
112
214
Statistic
DF
Value
Prob
-----------------------------------------------------Pearson Chi-Square
*
4.629
*****
1.
The expected frequency for the cell corresponding to Anger and Males is:
(a) 15.9
(b) 55.7
(c) 30.4
(d) 31.9
(e) 29.1
2. The null hypothesis is:
(a) Emotional association with red is independent of gender.
(b) Gender is dependent upon the emotional association with red.
(c) The probability of selecting an emotion with red is related to gender.
(d) The number of children in each cell does not depend upon gender or upon emotion.
(e) The color red is independent of the emotion associated with it and with gender.
3. The null hypothesis will be rejected at =0.05 if the test statistic exceeds:
(a) 3.84
(b) 5.99
(c) 7.81
(d) 9.49
(e) 14.07
4. The approximate P-value is:
(a) Between 0.100 and 0.900
(b) Between 0.050 and 0.100
(c) Between 0.025 and 0.050
(d) Between 0.010 and 0.025
(e) Between 0.005 and 0.010