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Unformatted text preview: STAT 31111 Homework 9 Solutions
Instructor: Alicia Johnson 9.11 9.16 9.22 9.31} 9.39 Get Pvaloe from r: a) tits
b} use
e) {1.85
d} None et‘these Pvalues gives strong evidence against llu. All ol'them indicate that the null hypothesis is plausible. Another test of astrology a} p — proportion of adults who guess correctly
H..: p H3 and H.: p e “3 b} p — 23I33' 0.33?
se = so.“ puy'n}  lgesssu
z = to??? 0.333jiﬂﬂ52 HUB c} The Pvalue is 0.4T. “the null hypothesis were true, the probability would be 0.4? of
getting a test statistic at least as extreme as the value observed. d} At the o ll.tl1 level we do not have strong evidence to reject the null hypothesis. Hence
I would not conclude that people are more liker to select their "correct" horoscope than
il‘they were randomly guessing. esssyssi ease Exit poll predictions: b] liu'. p 0.50: Ii”: p r [1.50
The lirst assumption is that the data are categorical. which is true for voting status.
Second. the sample must he random, which we will assume is true. Third. the sample
must be sulliciently large to assume that the distribution nt‘the sample proportion is
approximately normal: np {EDI  H05) 3 15 and n[l p} (2m Milli} e 15. c} Pavalue  [1.92 {alternative hypothesis is twovsided]. AssumingerI  [1.511). the probability
of obtaining a sample promotion where 50. I‘ll; or more of the voters voted for Webb or
the other extreme. less than 49.9% voted for Webb. is about 92%. d} Since the PvaltJe ‘= CLUE, we cannot reject the null hypothesis. There is insufﬁcient
evidence to predict who won lhc election. Low carbohydrate diet: a} Let ,u =' the mean weight change
H”: a  t} against H": .e s: it b) se — 5N” — 3.4mm — [1.531
1* r (x... some — peso eyesai —ts.air c} Ltmking at Table B. for 40 degrees of freedom we see that the 4! critical values for every
given right tail probability is less than [3.21 Hence the Pvalue for this test will be less
than out“. So if we assume that the null hypothesis is true, ifp = t], the probability
would be less than (Will of getting a test statistic less than or equal to 1Ei.2'i'. d] At the o = [H]! level we actually do have strong evidence to reject the null hypothesis
{Pvalue < [Will { Dill). Hence we can conclude that subjects on such a diet lost weight [In EVCIHEE. Anorexia in teenage girls:
a) See attached Rweb output. We see that the data is approximately normally distributed. b} See attached Ftweb output. Dur assumptions are that we have a quantitative variable
with population a deﬁned. that data are obtained using randomization, and that the
population is approximately normal, which we can assume to be true. Let a = mean weight change. Then He: a — i] and H”: a is G. Our P—value is {1.ﬂtiﬂ3433'. which means that il~ the null hypothesis were true. the
probability would be 13.013335 of getting a test statistic greater than or equal to the one
observed. We have strong evidence to reject the null hypothesis, hence we can conclude
that the mean weight change was greater than Cl. 9.42 Dr. Dog: a} For the significance level ol'tlﬂS. we would reject the null hypothesis. We have strung
evidence that dogs can detect urine from bladder cancer patients at a rate higher than
would be expected by chance. b} ll'wc made an error. it was a Type 1 error. which would indicate that we
concluded that dogs could detect urine from bladder cancer patients. but they really were
not able to do so any better than by chance. 9.47 Errors in the courtroom:
a} A Type i error would result in finding the defendant guilty when hershe is actually innocent.
h} AType II error wculd result in Failing triconvict a defendant who is actually guilty.
c) if conviction results in the death penalty, a Type l error would be considered more serious because it would mean putting an innocent person to death. Rweh l[C‘llutpuuts Hweh:s pnslscripttliln “rrmpfﬂdut.?4?iﬁtps"i Rweb:h x < read.tablet”rtmpKRdata.2423a.data". headar=Tt
Rweb:> attaehtxl Rweb:> namearxt [1] “Change” Hweb:> Hueb:> Rweb:} stemtthanget The decimal point is l digittst to the right of the I D  543 0 I 046$?59
1  111334
2  2
Rweb:> t.testtx=Change. conf.level=o.95. alternative=ugreater”. nu=Dl
One Sample ttest data: Change t = 4.1859, df = 15. p—value = D.DDD34BT
alternative hypothesis: true mean is greater than 0
95 percent confidence interval: 4.252552 Inf sample estimates:
mean of x 7.294118 ...
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 Spring '09
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