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Unformatted text preview: Practice Test 11 AP Statistics Name: Directions: Work on these Sheets. Tables and formulas appear on a separate sheet. Part 1: Multiple Choice. Circle the letter corresponding to the best answer. 1. The heightséfin inches) of males in the United States are believed to be normally distributed with mean it. The average height of a random sample of 25 American adult males is found to be i = 69.72
inches and the standard deviation of the 25 heights is found to be s = 4.15. The standard error of it: is
(a) 0.17 "
(b) 0.69 ﬂ
@083 at?
(d) 1.856 r 1 “M
(e) 2.04 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 1 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. 2. The appropriate null and alternative hypotheses are:
(a) H0: u = 5.39 vs Ha: p $5.39
(b) H0: 11 = 5.39 vs Ha: u < 5.00
(c) H0: u = 5 vs Ha: u = 5.39 (d H0:u=5vsHa:u<5
Hozu=5vsHazu>5
3. The value of the test statistic, the rejection region (0. = 0.05), and the Pvalue (computed by a
computer) are: 
(a) z" = 2.24; reject if 2* > 1.960; Pvalue = 0.0125
(15) z" = 2.24; reject if z" > 1.645; Pvalue = 0.0125 _
( t“ = 2.24 with 25 df; reject tft“ > 1.708; P—value = 0.0171 .t * = 2.24 with 24 (if; reject ift’t > 1.711; Pvalue = 0.0173?
(c) t = 2.24 with 24 df; reject if t > 2.064; P—value = 0.0173,?“ ,1 a) A teacher compares the pretest and posttest scores of students.
A teacher compares the scores of students using a computer based method of instruction with the scores of other students using a traditional method of instruction.
. {c}...A. teacher. compares the scores of students inher class on astandardized t...th udtlathenatienal
average score.
((1) A teacher calculates the average of scores of students on a pair of tests and wishes to see if this
average is larger than 80%. " I (e) None of these. 4. ich of the following is an example of a matched pairs design?
(
) Chapter ] 1 l 5. Popular wisdom is that eating presweetened cereal tends to increase the number of dental caries
(cavities) in children. A sample of children was (with parental consent) entered into a study and
followed for several years. Each child was classiﬁed as a sweetenedcereal lover or a
nonsweetened—cereal lover. At the end of the study, the amount of tooth damage was measured. Here is the summary data: ‘ . Grou :1 mean std. dev
Sugar Bombed 10 6.41 5.0
No sugar ' ’ "'15 " 5.20 15.0 An approximate 95% conﬁdence interval for the difference in the mean tooth damage is: /"’
Is—Ts // , 1
(@ ﬁﬁleiﬂﬂi226——+—m (% a. ‘3; /://»’
yr / \lm“ 1L @(6.41—5.20)i2.261/E+F 9Q / \tli \
25 225
(9) ﬁssizaszoltisa;¥%%;;:;:: (d)(a41—520}t226 3§1+EE§
100 225 (@ @41—5ﬂﬂ:196.31+2£§
.__W,.... . . . .. so.. .____.,. 6. You are thinking of using a t procedure to test hypotheses about the mean of a population using a
signiﬁcance level of 0.05. You suspect that the distribution of the population is not normal and may be moderately skewed. Which of the following statements is correct?
(a) You should not use the tprocedure because the population does not have a normal distribution. _ You may use the I procedure provided your sample size is large, say at least
c ’ou may use the rprocedure, but you should probably claim only that the signiﬁcance level is 0.10.
((1) You may not use the a: procedure. I procedures are robust to nonnorrnality for conﬁdence intervals NE, but not for tests of hypotheses.
s) (6) You may use the r procedure provided that there are no outliers. 1/, ,
/ Chapter I 1 2 Part2: Free Response
Answer completely, but be concise. Write sequentially and Show all steps. 7. Mutual fund performance. Many mutual funds compare their performance with that of a
benchmark, an index of the returns on all securities of the kind the fund buys'.‘ The Vanguard
International Growth Fund, for example, takes as its benchmark the Morgan Stanley EAFE ' (Europe, Australasia, Far East) index of overseas stock market performance. Here are the percent
returns for the fund and for the EAFE from 1982 (the ﬁrst full year of the fund’s existence) to 2000. "a nail RAF}: .1 sea 4 1 e32 '11 mes Sans are: lama tat:4 :21; 2.2;:
15:54 m2 12am: ' was 55.1113
193?» £36.94 5:35;: tea; use i 1 lest; zen 3:12.94 £1396; 14,55. we
19$? was 224.93 rec: 4.12 2m
1935 11.6] salsa was 1633 2:333
1959 24.?6 Just; 1999 26.34 emu
ism 12.115 m 23.21?) seam wash 7 13.96 199] at .74 1233!} Does the fund signiﬁcantly outperform its benchmark? (a) Explain clearly whether the matchedpairs t test or the two~sample t test is the proper
choice to answer this question. Matt1M6 {paow (b) Carry out the appropriate test and state your conclusion about the fund’s performance. Follow
the Inference Toolbox! j .mu’prk {$6 3‘3 (PD (/LL : ' urn/airs m U) G 2,‘ ’05? ,_ . . gig)2.“ L q 0+ “Sub; a Vﬂ ﬂakL, ﬁzﬂ‘yabgf' 10“ Chi, elc Fl UY'r'KLJL‘fy I
(piCLbLWVVVe‘DLCEﬁLQ to I 0 mc‘fhhw l I. ﬂigccj/ H) u/L_Lb.__4,m bauxite gate : 0 Chapter 1 l 3 8. Researchers studying the learning of speech often compare measurements made on the recorded
speech of adults and children. One variable of interest is called the voice onset time (VOT). Here
are the results for 6yearold children and adults asked to pronounce the word “bees.” The VOT is measured in milliseconds and can be either positive or negative. Group a T .2:
(I’laiktmu 31'! ~3J’ST 33.6.59 Adults in 4;?:3.17 stat25 (a) What is thistandard error of the difference between the mean VOT for children and adults? IJr Lili— '—': 15.96? [<3 2C (b) The researchers were investigating whether VOT distinguishes adults from children. Construct
a 95% confidence interval for the difference in mean VOTs when pronouncing the word “bees.” Follow the Inference Toolbox. Mm W aka mm r. a:
(uL..m.i”3 CQLLU’L I , x ,I I 3" L l Salt/Wt 3Q 9’0 Jr is at I beet (Limit4 [Limm C in _/L.§;3.Md;_¢7
l/ﬂ/QL 41,ch  Sat;an Jew.. ,(rtrhltt‘ll. ' _ . , _ . : (if; 2537? J Gilczﬁt. (02a)
6R air: q {72.291 I (_15‘5:)51L.3) (c) Based on your work in part (b), how would you answer the researchers’ question? Explain. «hart w ct (Ltﬁzamza bacon/12 (in/41 Mater, (d) The researchers in the study looked at VOTs for adults and children pronouncing several
different words. Explain why they should not perform a separate twosample t test for each
word and conclude that the words with a signiﬁcant difference (say, P < 0.05) distinguish
children from adults. (The researchers did not make this mistake.) (MAJEij of.) provideUte D Luau.
rm ‘LL L4, LLJ‘QLL' :43. wapw (.3, J9? .. fact by rattling a.t,m___ Chapter I] 4 ...
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This note was uploaded on 11/03/2011 for the course STATS 101 taught by Professor Smith during the Spring '11 term at University of Colorado Denver.
 Spring '11
 Smith
 Statistics

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