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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


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


Confidence Interval


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is known to be 0.0002 gram.

or the mean of repeated measurements of the weight.




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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



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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


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



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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

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