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Unformatted text preview: Chapter 25 l’aired Samples and Blocks (Review) Name MULTEPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Indicate the correct test procedure and reasoning. i) A teacher is interested in performing a hypothesis test to compare the mean math score of the girls 1)
and the mean math score of the boys. She randomly selects 10 girls from the class and then
randomly selects 10 boys. She arranges the girls' names alphabetically and uses this list to assign
each girl a number between 1 and 10. She does the same thing for the boys. A) l—sample t—test. The teacher should compare the sample mean for the girls against the
population mean for the boys. 8) Paired tutes’c Since there are 10 boys and 10 girls, we can link the two samples.
C) Twowsarnple tutest. There is no natural pairing between the two populations. D) Paired twtest. Since the boys and girls are in the same class, and are hence dependent samples,
they are can be linked. E) Either twosample or paired ttest Wiil work. 2) A researcher Wishes to determine Whether listening to music affects students‘ performance on 2)
memory test. He randomly selects 50 students and has each student perform a memory test once
while listening to music and once without listening to music. He obtains the mean and standard
deviation of the 50 "with music" scores and obtains the mean and standard deviation of the 50
"Without music scores". A) Two—sample t—test, since the researcher has two samples. B) Pooled t—test, since the standard deviation is likely to be the same in both populations
C) Paired t—test, since there is a "with music” and "without music" score for each person.
D) Not enough information given to determine correct type of test. E) z—test for each sample, since the researcher has the sample standard deviation. 3) A researcher wishes to compare how students at two different schools perform on a math test. He 3)
randomly selects 40 students from each school and obtains their test scores. A) Pooled tutest, since the standard deviations of the two populations are likely to he the same.
B) Either twousample or paired tutests wiii work equally well. (I) Paired tmtest, since the students at the two schools can be paired together. D) Twowsarnple intest, since the samples are independent. E) Not enough information is given to determine the correct type of test. SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Decide whether or not the conditions and assumptions for inference with a paired 1—test are satisfied. Explain your
answer. 4) The owner of a large apartment complex had an in—ground swimming pool installed in an 4)
effort to increase tenant satisfaction. Nine tenants were randomly selected to complete a
questionnaire that asSeSSed their level of satisfaction with the apartment complex. Their
scores before and after the installation of the pool are shown. The owner wants to assess
the effectiveness of the pool in increasing tenant satisfaction. Tenant Level of Satisfaction Level of Satisfaction
Before Pool After Pool (.0 WWNWNAWM \om‘qoxmsroam
nmnmnmmn 5) An agricultural company wanted to know if a new insecticide increases corn yields. Eight 5)
test plots showed an average increase of 3.125 bushels per acre when the insecticide was
used compared to the previous year. The standard deviation of the increases was 2.911 bushels per acre. The company wants to use a paired t—test to determine the mean increase
in yield. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Use the paired tainterval procedure to obtain the required conﬁdence interval for the mean difference. Assume that the
conditions and assumptions for inference are satisﬁed. 6) An agricultural company wanted to know if a new insecticide would increase corn yields. Eight 6)
test plots showed an average increase of 3.125 bushels per acre. The standard deviation of the increases was 2.911 bushels per acre. Determine a 99% conﬁdence interval for the mean increase in
yield. A) (—0.476, 4.399)
B) (—0.476, 6.726)
C) (6.726, 4.399)
D) (1.851, 4.399)
a) (1.851, 6.726) 7) Ten different families are tested for the number of gallons of water a day they use before and after 7)
viewing a conservation video. Construct a 90% conﬁdence interval for the mean of the difference of the "before" minus the "after" times if d—(after—before) : —4.8 and sd=52451 Before 33 33 38 33 35 35 40 40 40 31
After 34 28 25 28 35 33 31 28 35 33 A) (3.8.5.3) B) (2.13.5) C) (15,831) D) (2.5.7.1) s) (1.87.8) 8) A test of writing ability is given to a random sample of students before and after they compieted a 8)
formal writing course. The results are given below. Construct a 99% conﬁdence interval for the mean difference between the before and after scores ifd(after—before) e —2.0 and ed =2.6457 Before 70 80 92 99 93 97 76 63 68 71 ‘74
After 69 79 90 96 91 95 75 64: 62 64 76 A) (—0.2, 4.2)
B) (—0.1, 4.1)
C) (—05.45)
D) (1.2.2.8) s) (—13.6, 17.6) Interpret the given conﬁdence interval. 9) A high school coach uses a new technique in training middle distance runners. He records the 9)
times for 4 different athletes to run 800 meters before and after this training. A9096 confidence
interval for the difference of the means before and after the training, [a3 — )1 A, was determined to be (2.6, 4.8). A) Based on this sample, with 90% conﬁdence, the average time for the BOG—meter run for middle
distance runners at this high school is between 2.6 and 4.8 seconds ionger after the new
training. B) Based on this sample, with 90% conﬁdence, the average time for the BOO«meter run for middle distance runners at this high school is between 2.6 and 4.8 seconds shorter after the new
training. (I) We are 90% conﬁdent that a randomly selected middle distance rurmer at this high schooi
will have a time for the BOO—meter run that is between 2.6 and 4.8 seconds shorter after the
training than before the training. D) We know that 90% of all random sampies done on runners at this high school will show that
the mean time difference before and after the training is between 2.6 and 4.8 seconds. E) We know that 90% of the middle distance runners shortened their times between 2.6 and 4.8
seconds after the training. 10) A study was conducted to determine the average number of words children learn during their third 10)
year. For a random sample of one hundred children, parents reported the number of words spoken
at 24 months and then again at 36 months of age. A9898 conﬁdence interval for the difference of
the means at 36 months and 24 months, i136 — 1,94, was determined to be (209, 290). A) Based on this sample, we are 98% conﬁdent the average number of words spoken by children
is between 209 and 290 words greater at 36 months than at 24 months of age. 8) We know that 98% of all random samples done on the population wiii show that the average
number of words spoken by children is between 209 and 290 words greater at 36 months than
at 24 months of age. C) We know that 98% of children say between 209 and 290 more words at 36 months of age than
at 24 months of age. D) We are 98% conﬁdent that the children in this sample said, on average, between 209 and 290 I
more words at 36 months than at 24 months of age. E) We are 98% confident that a randomly selected child will say between209 and 290 more
words at 36 months than at 24 months of age. SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Use a paired t—test to perform the required hypothesis test for two population means. Assume that the conditions and
assumptions for inference are satisﬁed. 11) Five students took a math test before and after tutoring. Their scores were as follows) 11) Subject A B C D E
Before 67 75 79 69 75
After 71 84 '77 72 87 Do the data suggest that the tutoring has an effect on the math scores? Perform a paired
tutest at the 5% significance ievel. 12) The table below shows the weights of seven subjects before and after following a particular 12)
diet for two months. Subject A B C D E F (3
Before 193 173 185 158 196 199 157
After 186 164 183 163 182 201 145 Do the data suggest that the diet is effective in reducing weight? Perform a paired t—test at
the 1% significance level. 13) A test of abstract reasoning is given to a random sample of students before and after they 13)
completed a formal logic course. The results are given below. Do the data suggest that the
mean score after the course differs from the mean score before the course? Perform a
paired twtest at the 5% signiﬁcance level. Before 74 83 75 88 84 63 93 84 91 77
After 73 7'7 70 77 ’74 67 95 83 84 75 TRUE/FALSE. Write 'T' if the statement is true and 'F' if the statement is false. Pravide an appropriate response. 14) Suppose x is a variable on each of two populations whose members can be paired and that a paired 14)
difference is the difference between the values of the variable x on the members of a pair. True or
false? The mean of the paired differences equals the difference between the two population means. SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 15) A researcher wants to investigate whether people can concentrate as effectively in the 15)
morning it they do not eat breakfast. He will score some volunteers on reading
comprehension tests. Some of the testing will be done after volunteers have eaten
breakfast, and some testing will be done after voiunteers have skipped breakfast. Design
an experiment that will require a two—sampie t—test to analyze the results. Design an
experiment that will require a matched pairs tutest to analyze the results. Which
experiment would you consider to be the stronger design? 16) A researcher was interested in examining the difference in the number of hours of sleep 16)
married women and single women get each night. Five married women and five single
women reported their average number of hours of sleep per night for the last week.
Shown below is software output for two possible tests for the data. Paired retest of p(l~—2) TestHo:u(SM)=Ovsi~la: u(S—M)=t0 Mean of Paired Differences = 1.6 t—Statistic = 4 WM; df
p x 0.0161 2Sarnple t—Test of pi  u2 Ho:uS—uM=D Ha:uS—pM¢O Test Ho: MS)  MM) a 0 vs Ha: MS)  MM) 1 0 Difference Between Means = 1.6 t~Statistic = 1.8353 w]? .27 cit
p = 0.1075 Which of thesa tests is appropriate for these data? Explain. Using the test you select, state
your conclusion. 17) Ten different families are tested for the number of gaiions of water they use each day 17)
before and after viewing a conservation video. We want to see if there is strong evidence
that the conservation video changes people‘s water consumption rate. A tutest of the null
hypothesis of no difference has a t—statistic of 2.894 with a P—vaiue of 0.0178. Interpret this
result by explaining the meaning of the P—value. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question 18) At one SAT test site students taking the test for a second time volunteered to inhaie supplementa? 18)
oxygen for 10 minutes before the test. In fact, some received oxygen, but others (randomly
assigned) were given just normai air. Test resuits showed that 42 of 66 students who breathed oxygen improved their SAT scores, compared to only 35 of 63 students who did not get the oxygen.
Which procedure should we use to see if there is evidence that breathing extra oxygen can help
test—takers think more clearly? A) 2—sarnple tmtest I B) 2—proporti0n zwtest
C) 1—sample t—test
D) 1—proportion z—test E) matched pairs t—test SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question 19) Many states mandate tests that have to be passed in order for the smdents to graduate Witt 19)
a high school diploma. A locai schooi superintendent believes that afteruschool tutoring
will improve the scores of students in his district on the state's graduation test. A tutor
agrees to work with 15 students for a month before the superintendent will approach the
school board about implementing an after«school tutoring program. The after—school
tutoring program will be implemented it student scores increase by more than 20 points. The superintendent will test a hypothesis using on 2 0.02. Write an appropriate hypotheses
(in words and in symbols). 20) Many states mandate tests that have to be passed in order for the students to graduate witi’ 20)
a high school diploma. A locai school superintendent believes that afterwschool tutoring
will improve the scores of students in his district on the state's graduation test. A tutor
agrees to work with 15 students for a month before the superintendent will approach the
school board about implementing an afterwschool tutoring program. The after—school
tutoring program will be implemented if student scores increase by more than 20 points.
The superintendent will test a hypothesis using or = 0.02. After this trial produced
inconclusive results, the superintendent decided to test the after«school tutoring program
again with another group of students. Describe two changes he could make in the trial to
increase the power of the test, and explain the disadvantages of each. 21) You need to find a new hair stylist and know that there are two terriﬁc salons in your area, 21)
Hair by Charles and Curl Up (is Dye. You want a really good haircut, but you do not want
to pay too much for the cut. A random sample of costs for 10 different stylists was taken at
each salon (each salon employs over 100 stylists). A friend tells you that he has heard that
Curl Up :32 Dye is the more expensive salon.
1. Write hypotheses for your friend’s claim.
ii. The following are computer outputs. Which output is the correct one to use for this
test? Explain. Output A: Twowsarnple T for Hair by Charles vs Curl Up (in Dye
N Mean StDev SE Mean Hair by Charles 10 22.10 6.33 2.0 Curl Up «S: Dye 10 26.00 4.81 1.5 Difference m mu (Hair by Charles) — mu (Curl Up 8: Dye) Estimate for difference: —3.90000 95% CI for difference: 0922983, 1.42983) T—Test of difference = 0 (vs not =): T—Value = —1.55 P~Value : 0.140 D? = 16 Output B:
Paired T for Hair by Charles —— Curl Up 8: Dye N Mean StDev SE Mean
Hair by Charles 10 22.1000 6.3325 2.0025
Curl Up&rDye 10 26.0000 4.8074 1.5202
Difference 10 —3.90000 7.37036 2.33071 95% CI for mean difference: («—9.17244, 1.37244)
T—Test of mean difference = 0 (vs not = 0): T—Value = ~16? PwValue m 0.129 iii. Use the appropriate computer output to make a conclusion about the hypothesis test
based on the data. Make sure to state your conclusion in context. 22) A packing company considers hiring a national training consultant in hopes of improving 22)
productivity on the packing line. The national consultant agrees to work with 18
employees for one week as part of a trial before the packing company makes a decision
about the training program. The training program will be implemented if the average
product packed increases by more than 10 cases per day per employee. The packing
company manager will test a hypothesis using or 2 0.05. In this context, which do you
consider to be more serious — a Type I or a Type 11 error? Explain briefly. 23) Researchers developing new drugs must be concerned about possible side effects. They 23)
must check a new medication for arthritis to be sure that it does not cause an unsafe
increase in blood pressure. They measure the blood pressures of a group of 12 subjects,
then administer the drug and recheck the blood pressures one hour later. The drug will be
approved for use unless there is evidence that blood pressure has increased an average of
more than 20 points. They will test a hypothesis using a. = 0.05. Write appropriate
hypotheses (in words and in symbols). 24) Researchers developing new drugs must be concerned about possible side effects. They 24)
must check a new medication for arthritis to be sure that it does not cause an unsafe
increase in blood pressure. They measure the blood pressures of a group of 12 subjects,
then administer the drug and recheck the blood pressures one hour later. The drug will be
approved for use uniess there is evidence that biood pressure has increased an average of
more than 20 points. They will test a hypothesis using or u 0.05. After this experiment
produced inconclusive results the researchers decided to test the drug again another group
of patients. Describe two changes they could make in their experiment to increase the
power of their test, and explain the disadvantages of each. Suppose you were asked to analyze each of the situations desaibed below. (NOTE: Do not do these problems!) For each
indicate which procedure you would use (pick the appropriate test from the list below), the test statistic (z or t), and, if t,
the number of degrees of freedom. proportion — 1 sample difference of proportions — 2 samples mean  1 sample difference of means — independent samples
mean of differences — matched pairs none of these 25) Are there more broken bones in summer or winter? We get records about the number oi 25)
fractures treated in )anuary and Juiy at a random sample of 25 emergency rooms. 26) The school newspaper wants a 95% confidence intervai for the road test failure rate. In a 26)
random sample of 65 student drivers, 37 said they failed their driver's test at least once. 27) Tags piaced on garbage cans aliow the disposai of up to 30 pounds of garbage. A random 2'7)
sample of 22 cans averaged 33.2 pounds with a standard deviation of 3.2 pounds. Is this
strong evidence that residents overload their garbage cans? 28) An oral surgeon is interested in estimating how long it takes to extract all four wisdom 28) teem. The doctor records the times for 24 randomly chosen surgeries. Estimate the time it
takes to perform the surgery with a 95% conﬁdence interval. 29) The owner of a construction company would Iike to know if his current work teams can 29)
build room additions quicker than the time allotted for by the contract. A random sample
of 15 room additions completed recently revealed an average completion time of 0.32 days
faster than contracted. is this strong evidence that the teams can cornplete room additions
in less than the contract times? 30) In a study to determine whether there is a difference between the average jail time 30)
convicted bank robbers and car thieves are sentenced to, the few students randomly Selected 20 cases of each type that resulted in jail sentences during the previous year. A
90% confidence interval was created from the results. Answer Key
Testnarne: CHAPTER 25 PAIRED SAMPLES AND BLOCKS (REVIEW) 1) C 2) C 3) D 4) Since the data consist of before and after satisfaction scores for individual tenants, the data are clearly paired. The
tenants in the sample were chosen randomiy and are independent of each other. If we assume that this is a large
apartment complex, then the sample should be no more than 10% of the population of tenants. The boxplot of the
differences, shown below, is symmetric with no outliers. ﬂ 5) The data are paired since the data consist of before and after yields for the same test plots. We can assume the test
plots are independent if they were not geographicaiiy clustered or farmed by the same operator. The test plots include
fewer than 10% of all farms using the insecticide. We can assume the population of differences has a Normal model,
although we would need the actual data in order to make a boxplot or normal probability plot of the differences. It
would be preferable to have a control group in this study since yield differences could be due to other factors, such as
weather. 6) B
7) E
8) C
9) B
10) A
11) H O : it d :0
H A : fad a 0
Test statistic: t = ~2.134
P—value a 0.0998
Do not reject H 0 _ At the 5% signiﬁcance level, the data do not provide sufficient evidence to conclude that the
tutoring has an effect on the math scores.
12) H G : u d = O
H A : n d > 0
Test statistic: t n 1,954
Pvalue = 0.049
Do not reject H 0 . At the 1% significance levei, the data do not provide sufficient evidence to conclude that the diet is
effective in reducing weight.
13) HO : fJ. d z 0
H A : p. d a 0
Test statistic: t m 2.366
P—value = 0.0422
Reject H 0 . At the 5% significance level, the data provide sufﬁcient evidence to conclude that the mean score after the course differs from the mean score before the course.
14) TRUE Answer Key
Testname: CHAPTER 25 PAIRED SAMPLES AND BLOCKS (REVIEW) 15) Twowsarnple tutest: Randomly assign half the volunteers to do the reading comprehension test after eating breakfast
Assign the other half to do the reading comprehension test with no breakfast. Compare the scores for the two groups. Matched pairs t«test: Randomly assign haif the volunteers to take a reading comprehension test after eating breakfast,
and half to take it with no breakfast. Then, the next day, have each group take a reading comprehension test under the
other conditions. Look at the differences in test scores for each individuai with and without breakfast. The matched pairs design is stronger since people vary in their reading comprehension ability. The matched pairs
design wiii remove this extra variation. 16) The 2wsampie t«test is appropriate since the married and single women are independent of each other. There is no
natural pairing between the two groups. According to the 2~sampie tutest, there does not appear to be a signiﬁcant
difference (P—value 0.1075) in the amount of sleep between married and single women. 17) if there is no difference in water consumption rates after viewing the conservation video, the chance of seeing a
difference as large or larger than the one observed is about 1.78%. 18) B 19) Ho: it = 20; The difference between the mean number of points before and after the tutoring program is not more than
20. H A: u > 20; The difference between the mean number of points before and after the tutoring program is more than 20. 2.0) To increase the power of the test, we couid increase the level of significance (or) or increase the sample size. By
increasing the level of significance, it could lead to adopting a tutoring program that actually doesn't help. By
increasing the sampie size, the trial cost would increase. 21) i.Let H = Hair by Charles and C = Curl Up & Dye. Ho: pH — tic (There is no difference in the mean cost of haircuts at the two saions.)
HA: pH < uc (The mean cost of haircuts is higher at Curl Up & Dye.) We would use Output A, since we are doing a two—sample t—test. (Output B is for a paired—t test.)
iii. The Pwvalue of 0.070 is high, so I faii to reject the null hypothesis. There is no evidence that Curl Up ti: Dye is any
more expensive on average than Hair by Charles. 22) A Type I error would be very expensive for the packing company. A Type 1 error would mean that the manager
rejected the null hypothesis when in fact the null hypothesis is true. In this situation, by rejecting the null hypothesis
the company thought the training improved productivity, so they paid for the consuitant to train all empioyees. In
reality, the training did not improve productivity so the company wasted money on training that did not help. 23) H0: pd = 20 The mean increase in biood pressure is safe. HA: pd > 20 The mean increase in hiood pressure exceeds the safe iimit. 24) Increase alpha; could lead to rejecting a medication that's actually okay. Increase n; more costly and difficult, and could
endanger more subjects. 25) mean of differences — matched pairs; t; 24 degrees of freedom 26) proportion  1 sample; 2 27) mean  1 sample; t; 21 degrees of freedom 28) mean — 1 sampie; t; 23 degrees of freedom 29) mean of differences  matched pairs; t; 14 degrees of freedom 30) difference of means  independent samples; t; 19 degrees of freedom 10 ...
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This note was uploaded on 11/10/2009 for the course CAL 3452 taught by Professor Mr.sun during the Spring '09 term at Sungkyunkwan.
 Spring '09
 Mr.Sun

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