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#5.48 ACT scores of high school seniors. The scores of high school seniors on the ACT college entrance e
(a) What is the approximate probability that a single student randomly chosen from all those tak
(b) Now taken as SRS of 25 students who took the test. What are the mean and standard deviatio
(c) What is the approximate probability that the mean score
of these students is 23 or higher?
(d) Which of your two Normal probability calculations in (a) and (c) is more accurate? Why?
Solutions: Given that, =20.8 and SD =4.8
(a)
Required Prob: P( X >= 23 )
P( X >= 23 ) = 1 - P( X < 23 )
P( X < 23 ) = 0.6766 (by using excel normdist() function, click on the respectiv
P( X >= 23 ) = 1 - P( X < 23 ) = 1-0.6766 = 0.3234
(b)
For this problem lets recollect that
By the properties of means and variances of random variables, the mean and variance of the sa
Note: Here x-bar also refer as M
Now, when n=25
Expected value of M = 20.8
Standard deviation of M = / sqrt(n) = 4.8 / sqrt(25) = 4.8 / 5 = 0.96
( c)
Required Prob: P( sample mean(M) >= 23 )
P( M >= 23 ) = 1 - P( M < 23 )
P( M < 23 ) = 0.9890 (by using eMcel normdist() function, click on the respecti
P( M >= 23 ) = 1 - P( M < 23 ) = 1-0.9890 = 0.0110
(d)
Normal probability calculation in (c) is more accurate, because it is capturing the current sample
whereas (a) captures overall population's variability
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amination in 2003 had mean =20.8 and SD =4.8. The distribution of scores is only roughly Normal.
ing the test scores 23 or higher?
n of the sample means score x of these 25 students?
value for how it is calculated)
mple mean are the following:
e value for how it is calculated)
's variability,
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#5.52 A Lottery payoff. A $1 bet in a state lotterys Pick 3 game pays $500 if the three-digit number you cho
the winning number, which is drawn at random. Here is the distribution of the payoff X:
Payoff X $0 $500
Probability 0.999 0.001
Each days drawing is independent of other drawings.(a) What are the mean and SD of x?
(b) Joe buys a Pick 3 ticket twice a week. What does the law of large numbers say about the average
(c) What does the central limit theorem say about the distribution of Joes average payoff after 104
(d) Joe comes out ahead for the year if his average payoff is greater than $1 (the amount he spent e
What is the probability that Joe ends the year ahead?
Solutions:
(a)
x $0 $500 Imp note: Generally there should be loss f
P(x) 0.999 0.001 but considering as it is to avoid
mean = xp(x) = 0*0.999 + 500*0.001 = 0.5
SD = sqrt ( V(x) )
V(x) = x*x*p(x) - (mean*mean) = 0*0*0.999 + 500*500*0.001 - 0.5*0.5 = 249.75
SD = sqrt ( 249.75 ) = 15.803
(b)
From the law of large numbers, the average payoff joe receives from his bets will be close populati
(c)
The central limit theorem (CLT) states conditions under which the mean of a sufficiently
large number of independent random variables, each with finite mean and variance, will be approxi
(d)
Required prob: P( X > 1)
As here, n = 104
Mean of sample mean = 0.5
SD of sample mean = / sqrt(n) = 15.803 / sqrt(104) = 1.5496
therefore X follows normal dist with mean = 0.5 and SD=1.5496
P(X>1) = 1 - P(X
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ose exactly matches
payoff Joe receives from his bets?
bets in a year?
ch day on a ticket).
ctor, here it should be -1 in place of 0
e the confusion
n's average payoff
ately normally distributed
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#5.60 Advertisements and brand image. Many companies place advertisements to improve the image of
(a) There are 28 students in each group. Although individual scores are discrete, the mean score
(b) What are the means and SD of the sample mean scoresfor theJournalgroup and
for the Enquirer group?
(c) We can take all 56 scores to be independent because students are not told each others score
between the mean scores in the two groups?
(d) FindP ( - x-bar 1 ).
Solutions:
(a)
Because the central limit theorem (CLT) states conditions under which the mean of a sufficientl
large number of independent random variables, each with finite mean and variance, will be appr
So based on this CLT,we could say that mean score for a group of 28 will be close to Normal
(b)
Given that, Journal's mean 4.8 and SD 1.5, and Enquirer's mean 2.4 and SD 1.6
here sample number, n=28
By the properties of means and variances of random variables, the mean and variance of the sa
Now, for jounral when n=28
Expected value of of journal = 4.8
Standard deviation of = / sqrt(n) = 1.5 / sqrt(28) = 4.8 / 5 0.283
Now, for Enquirer when n=28
Expected value of M of Enquirer = 2.4 Note: Her
Standard deviation of M = / sqrt(n) = 1.6 / sqrt(28) = 4.8 / 0.302
(c)
The distribution of y-bar - x-bar would be normal based central limit theorem and normal dist pro
From CLT,we already know that y-bar and x-bar will be normal and form normal dist properties,
if x1 and x2 follow normal dist then x1+x2 and x1-x2 also follows normal dist
(d)
for required prob, first we will need to calculate mean and SD of y-bar - x-bar
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using (b) and (c ), y-bar mean = 4.8 and var = square of 0.283 = 0.0804
x-bar mean = 2.4 and var = square of 0.302 = 0.0914
y-bar - x-bar follows normal dist with mean = 4.8-2.4 = 2.4 and var =
SD = sqrt(0.1718) =
P ( - x-bar 1 ) = 1 - P ( - x-bar
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their brand rather than to promote specific products. In a randomized comparative experiment, busi
or a group of 28 will be close to Normal. Why?
s. What is the distribution of the difference-
oximately normally distributed
mple mean are the following:
x-bar refers as M
perties
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.0804 + 0.0914 = 0.1718
0.4145
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ess students read ads that cited either the Wall Street Journal or the National Enquirer for importan
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t facts about a fictitious company. The students then rated the trustworthiness of the source on a 7-p
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oint scale. Suppose that in the population of all students scores for the Journal have mean 4.8 and S
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1.5, while scores for the Enquirer have mean 2.4 and SD 1.6
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#6.18 Mean OC in young women. Refer to the previous exercises. A biomarket for bone formations mea
in the same study was osteocalcin (OC), measured in the blood. The units are nanograms per milliliter (ng/
For the 31 subjects in the study the mean was 33.4 ng/ml. Assume that the SD is known to be 19.6 ng/ml.
Confidence Interval Estimate for the Mean
DataPopulation Standard Deviation 19.6
Sample Mean 33.4
Sample Size 31
Confidence Level 95%
Standard Error of the Mean 3.5203 Note: Click on corresponding cell, to kno
Z Value -1.9600
Interval Half Width 6.8996
Interval Lower Limit 26.5004Interval Upper Limit 40.2996
Intermediate Calculations
Confidence Interval
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ured
l).
eport the 95% confidence interval.
how it is calculated
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#6.32 Accuracy of a laboratory scale. To assess the accuracy of a laboratory scale,
a standard weight known to weigh 10 grams is weighed repeatedly. The scale readings are Normally
distributed with unknown mean (this mean is 10 grams if the scale has no bias.) the SD of the scale reading
(a) The weight is measured five times. The main result is 10.0023 grams. Give a 98% confidence interval f
(b) How many measurements must be averaged to get a margin of error of 0.0001 with 98% confidence
(a)
Confidence Interval Estimate for the Mean
Data
Population Standard Deviation 0.0002
Sample Mean 10.0023
Sample Size 5
Confidence Level 98%
Standard Error of the Mean 0.0001 Note: Click on corresponding cell, to kno
Z Value -2.3263Interval Half Width 0.0002
Interval Lower Limit 10.0021
Interval Upper Limit 10.0025
(b)
Sample Size Determination
Data
Population Standard Deviation 0.0002
Sampling Error 0.0001Confidence Level 98%
Z Value -2.32635 Note: Click on corresponding cell, to kno
Calculated Sample Size 21.64758
Sample Size Needed 22
Intemediate Calculations
Result
Intermediate Calculations
Confidence Interval
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is known to be 0.0002 gram.
or the mean of repeated measurements of the weight.
how it is calculated
how it is calculated
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#6.58 A two-sided test and the confidence interval. The P-value for a two-sided test of the null hypo
(a) Does the 95% confidence interval include the value 30? Why?
(b) Does the 90% confidence interval include the value 30? Why?
Solutions:
(a)No, it does not include the value 30 at 95% confidence interval
Because generally, one rejects the null hypothesis if the p-value is smaller than or equal to t
Here, P-value 0.04 < 0.05 (significance level), so it is rejecting the null hypothesis.
Means, it doesn't inlcude 30
(b)
No, it does not include the value 30 at 90% confidence interval
Because generally, one rejects the null hypothesis if the p-value is smaller than or equal to t
Here, P-value 0.04 < 0.1 (significance level), so it is rejecting the null hypothesis.
Means, it doesn't inlcude 30
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hesis H 0 :=30 is 0.0
he significance level
he significance level
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#6.66 Are the pine trees randomly distributed north to south? In example 6.1 we looked at the distrib
One way to formulate hypotheses about whether or not the trees are randomly distributed in the tract is to
north-south direction. The values range from 0 to 200, so if the trees are uniformly distributed in this direct
values (100) should be due to chance variation. The sample means for the 584 trees in the tract is 99.74.
assumption that the trees are uniformly distributed gives a SD of 58. Carefully state the null and alternativ
Note that this requires that you translate the research question about the random distribution of the trees i
of a probability distribution. Test your hypotheses, report your results, and write a short summary of what
Solution:
Null hypothesis: H0 : The pine trees are randomly distributed north to south ( = 100)
Alternative hypothesis H1 : The pine trees are not randomly distributed north to south ( 100)
Z Test of Hypothesis for the Mean
Null Hypothesis m= 100Level of Significance 0.05
Population Standard Deviation 58Sample Size 584
Sample Mean 99.74
Standard Error of the Mean 2.400057
Z Test Statistic -0.10833
Two-Tail Test
Lower Critical Value -1.95996
Upper Critical Value 1.959964
p -Value 0.913733
Do not reject the null hypothesis
As p-value is grater than 0.05 (alpha), we do not reject the null hypothesis. Means, The pine trees
Data
Intermediate Calculations
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tion of longleaf pine trees in the Wade Tract.
examine the average location in the
ion, any difference from the middle
theoretical calculation based on the
hypotheses in terms of this variable.
to specific statements about the mean
ou have found.
re randomly distributed north to south
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6.68 Who is the author? Statistics can help decide the authorship of literary works.
Sonnets by a certain Elizabethan poet are known to contain an average of =8.9 new words (words are n
other works). The SD of the number of new word is =2.5. Now a manuscript with debating whether it is t
=10.2 words not used in the poets known works. We expect poems by another author to contain mor
H 0 : = 8.
H a : > 8.9Give the z test statistics and its P-value. What do you conclude about the authorship of the new poem
Solution:
Z Test of Hypothesis for the Mean
Null Hypothesis m= 8.9Level of Significance 0.05
Population Standard Deviation 2.5
Sample Size 1
Sample Mean 10.2
Standard Error of the Mean 2.5
Z Test Statistic 0.52
Upper-Tail Test
Upper Critical Value 1.6449
p -Value 0.3015
Do not reject the null hypothesis
As we are not rejecting the null hypothesis, it is saying that =8.9, so the authorship of the new p
Data
Intermediate Calculations
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ot used in the poets
he poets work. The new sonnets contain an average of
new words, so to see if we have evidence that the new sonnets are not by our poet we test
?
ems is Elizabethan poet