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Unformatted text preview: MATH 2300
Homework # 8 Answers Provide an appropriate response.
1) Many people think that a national lobby's successful fight against gun control legislation is reflecting the 1) C will of a minority of Americans. A random sample of 4,000 citizens yielded 2290 who are in favor of gun control legislation. Find the point estimate for estimating the prop 'on of all Americans W112 are in favor gfgun controlwle‘gis‘lation.
A) 0.4275 B) 4000 @5725 D) 2290 E) 0.8588 {)0} 0% Q3 two“? Sift)“: Pepi/«)9 \‘0 V‘ PW $07 Vow 15’ go‘w ? w lb? km” A
r  2200
55.. it vmtwt'oe = P' ——~— : 05 725
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Select the most appropriate answer.
2) L 2) In an effort to monitor the level of lead in the air after an explosion at a battery factory, the following lead readings were taken for 6 days following the explosion (in ug/m3). What is the point estimate for the
population mean lead level in the air over the 6 days following the explosion? Wednesday Thursday 1.54 B) 0.50 C) 2.91 D) 0.73 E) 0.42 Pop/y)“ QSk'tMﬁkz. (Evy M vogwi‘akﬁow Mean :5 we?!" 2 x 5. to 1. t. m + ,9_,_5ez,:3j’:.:j:f€§13 ‘0 ___’— 56m?\£ Mean : 5t. 1‘ s
rw b
: #54» Provide an appropriate response.
3) In practice a
A) population standard deviation
B) sample standard deviation
C) none of these
D) margin of error @standard error. is an estimated standard deviation of a sampling distribution. Find the standard error
4) In a survey of 3200 T.V. viewers, 20% said they watch network news programs. Find the standard error 4) B
for the sample proportion. W A) 00835“ 0.0071 C) 0.0649 D) 00142 E) 0.0721 A A H .............. M»
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e 3200 : Fur?) : r w.
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A
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: o ' o 0:} l Use the given degree of confidence and sample data to construct a confidence interval for the population proportion.
5) Of 369 randomly selected medical students) 23 said that they planned to work in a rural community. __C_z 5) Construct a 9§fézgeaﬁdence intern} forﬁhsﬁrcemage of, all medical Sweeneyth plan to workiaawral community. A) (2.99%, 9.47%) B) (4.16%, 8.30%) , 23 7 0,0623 3L9 hr ind/“pie, FWY‘" ~°"‘ 7 F *
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:3 2:202; (yo
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: ocoL‘ZS'E 6) Of 346 items tested, 12 are found to be defective. proportion of all suchritems that are defective.
B) (0.093, 0.600) A) (0.015, 0.054) I 35% .z
./ Provide an appropriate response 7) You are planning to use a sample proportion p to estimate a population proportion, p. A sample size of 0003825 /\ @3.77%, 8.70%) (CM Mm *1” MW‘ D) (3.77%, 9.47%) 39/. 5‘” (>on\0\‘~0“
‘9 i (gram) (eae)
, M.Q. 0.0623: (0' 01258) @0623 f O'Ol‘tg
0:092; +0102 49) VWYQY 3‘3")“ :5 2) (0'0673 ovoztt ,
:45 (005??) 00863) :5 (5'7?'/) siéa’ﬂ)
‘ 6);; Construct a 98% confidence intervalVtoﬁestimﬁlﬁlhﬁ @(0012, 0.058)
a8 f 01' i (grSCov‘eB D) (0014,0055) é‘w Povmlaxs on vao‘f 05C)“. .....—» ,2»? 003468 ’1' (2‘53)(otooﬂ8£$) 0~0346$ﬁ O‘ szgﬂ ’2'33 :3 > (ooaorfas—oozzas) o~oaef>8+crozzeg) \
(O, 0H8 j 0‘ 0535 (crow) o~ 0.583 \ r 3) 7>
99 7) 100 and a confidence level of 95% yielded a margin of error of 0.025. Which of the following will result aglarg’erﬂ argin ofﬂerror? —x I:
/ II: Increasing the sample size while keeping the same confidence level
Decreasing the sample size while keeping the same confidence level \/ III: Increasing the confidence level while keeping the same sample size )1 IV: Decreasing the
A) IV A" A ‘
m. e . s rsror¢DC$réii§> : (VZ«3(BY¢> [6???) co fidence level while keeping the same sample size
‘ C) 11 and iv II and III D) I and IV E) I and III N 2 Using the t—tables, report the t—score for the given confidence interval and degrees of freedom. 8) A 99% confidence interval from a sample of size 19 8)
(ﬁgcévaﬁ 0E Xf'aeéow: V)" :‘ ig~i : ‘8
9) 95% confidence interval with df = 25
A) 1.960 B) 2.145 @2060 D) 1.753 E) 2.120 Provide an appropriate response. 10) In a survey of 1,000 television viewers, 40% said they watch network news programs. For a 99%
confidence level, the margin of error for this estimate is 3.99%. If we only want to be 90% confident, how
will the margin of error change? A) Since less confidence allows a wider interval, the margin of error will be larger.
B) Since less confidence allows a wider interval, the margin of error will be smaller.
C) the margin of error will remain the same.
@Since less confidence allows a narrower interval, the margin of error will be smaller.
E) Since less confidence allows a narrower interval, the margin of error will be larger. 11) Suppose that you wish to obtain a confidence intervalgforgaﬂpﬂgpulationgmean. Under the conditions described below, should you use the z—interval, the tinterval, or neither?  The population standard deviation is unknown.
' The population is normally distributed. ~ The sample size is small. @ t—interval A) Z—interval B) neither Find the requested value
12) A researcher for a car insurance company wishes to estimate the mean annual premium that women aged 25—30 pay for their car insurance. A random sample of 16 women aged between 25 and 30 yields the
following annual premiums, in dollars. 582 658 466 941 748 662 777 704 594 723 580 725 856 610 720 985 Use the data to obtain a point estimate of the mean annual premium for all women aged between 25 and
r 30. Round your answer to the nearest dollar. A) $727 $709 Q5 \'~ Max 6, C) $705 D) $718 E) $721 POW/xii
582+ 6515+ 44,54».. + ?Zo+ M60» 3 (M VOEM\D‘X’"6V‘ V‘AQOM '38 “A; meow) 2838 10) CD 11) 12) 38$
71: :M
M. H:
= 708‘ [8‘ RI
2" a 773A:—
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wlfiﬁ l2“ AVSLrcv g pvxguev Determine the margin of error in estimating the population parameter. 8 13) How tall is your average English classmate? To determine this, you measure the height of a random 13)
sample of 15 of your 100 fellow students, ﬁnding a 95% confidence interval for the mean height of 67.25 to
69.75 inches.
A) 0.75 inches 1.25 inches C) 1.06 inches D) 1.5 inches
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Construct the requested confidence interval from the supplied information. 8
14) How tall is your average English classmate? To determine this, you measure the height of a random 14) sample of 15 of your 200 fellow students, finding a mean height of 68 inches and a standard deviation of
2.3 inches. Construct a 90% confidence interval for the mean height of your classmates. A) (67.023, 68.977) (66.954, 69.046) C) (67.730, 68.270) D) (65.908, 70.092)
Mrs a Mswxl‘t 80% C3, 2}»: Wash.“ mean
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A a) (68, wow? 3 66* "Mg?> 15) Among a sample of 65 students selected at random from one high school, the mean number of siblings is 15) B
1.3 with a standard deviation of 1.1. Find a 95% confidence interval for the mean number of siblings for
all students at this high school. A) (1.16, 1.44) (103,157) C) (127,133) D) (63.07, 66.93)
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 Spring '08
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 Math, Statistics

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