8/8/2019 Ch8 Confidence Interval Estimation Part1 - Suggested Problems Solutions

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STATISTICS FOR SOCIAL SCIENCESBUS 152

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CONFIDENCE INTERVAL ESTIMATIONPart1

8.1 Confidence Interval Estimate for the Mean ( Known)

Problem 1 (8.1)

SOLUTION:For 95% Confidence level,1 - = 0.95 = 0.05 /2 = 0.025 ---> Standard Normal Table ---> Z= -1.96

XZ

n =851.96

8

64 83.04

86.96_________________________________________________________________________________

Problem 2 (8.3)

SOLUTION:If all possible samples of the same size n are taken, 95% of them include the truepopulation average monthly sales of the product within the interval developed. Thus youare 95 percent confident that this sample is one that does correctly estimate the trueaverage amount.

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Problem 3 (8.7)

SOLUTION:a) For 99% Confidence level,

1 - = 0.99 = 0.01 /2 = 0.005 ---> Standard Normal Table ---> Z= -2.58

XZ

n= 0.995 2.58

0.02

500.9877 1.0023

b) Since the value of 1.0 is included in the interval, there is no reason to believethat the mean is different from 1.0 gallon.

c) No. Since is known and n = 50, from the Central Limit Theorem, we may assume

that the sampling distribution of X is approximately normal.d) For 95% Confidence level,

1 - = 0.95 = 0.05 /2 = 0.025 ---> Standard Normal Table ---> Z= -1.96

The reduced confidence level narrows the width of the confidence interval.

a) XZ

n= 0.9951.96

0.02

500.9895 1.0005

b) Since the value of 1.0 is still included in the interval, there is no reason tobelieve that the mean is different from 1.0 gallon.

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Problem 4 (8.9)

SOLUTION:a) For 95% Confidence level,

1 - = 0.95 = 0.05

/2 = 0.025 ---> Standard Normal Table ---> Z= -1.96XZ

n=1.99 1.96

0.05

1001.9802 1.9998

b) No. Since is known and n = 100, from the central limit theorem, we may assumethat the sampling distribution of X is approximately normal.

c) An individual value of 2.02 is only 0.60 standard deviation above the sample meanof 1.99. The confidence interval represents bounds on the estimate of theaverage of a sample of 100, not an individual value.

d) A shift of 0.02 units in the sample average shifts the confidence interval by thesame distance without affecting the width of the resulting interval.

a)XZ

n=1.97 1.96

0.05

100 1.9602

1.9798_________________________________________________________________________________

8.2 Confidence Interval Estimation for the Mean ( Unknown)

Problem 5 (8.11)

SOLUTION:For 95% Confidence level,1 - = 0.95 = 0.05

/2 = 0.025d.f. = 36-1 = 35 ---> Table of Critical Value of t ---> t= 2.0301

24

75 2.030136

SX t

n = 66.8796 83.1204

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Problem 6 (8.15)

SOLUTION:a) For 95% Confidence level,

1 - = 0.95 = 0.05

/2 = 0.025d.f. = 20-1 = 19 ---> Table of Critical Value of t ---> t= 2.0930

0.321.67 2.0930

20

SX t

n = $1.52 $1.82

b) The store owner can be 95% confident that the population mean retail value ofgreeting cards that the store has in its inventory is somewhere in between $1.52and $1.82. The store owner could multiply the ends of the confidence interval bythe number of cards to estimate the total value of her inventory.

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Problem 7 (8.19)

SOLUTION:a) For 95% Confidence level,

1 - = 0.95, = 0.05, /2 = 0.025d.f. = 26-1 = 25 ---> Table of Critical Value of t ---> t= 2.0595

3.05547.1538 2.059526

SX t

n = 5.92 $8.39

b) You can be 95% confident that the population mean monthly service fee, if a customersaccount falls below the minimum required balance, is somewhere between $5.92 and $8.39.

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Problem 7 (8.22)

SOLUTION:

a)44.2700

182.4 2.093020

SX t

n = $161.68 $203.12

b)10.0263

45 2.0930 20

S

X tn

= $40.31 $49.69

c) The population distribution needs to be normally distributed.d)

Norma

300

Hotel

Box-a

8/8/2019 Ch8 Confidence Interval Estimation Part1 - Suggested Problems Solutions

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Both the normal probability plot and the box-and-whisker show that thepopulation distribution for hotel cost is not normally distributed and is skewed tothe right.

8.22 d)

cont.

Nor

80

C

ars

Bo

Both the normal probability plot and the box-and-whisker show that thepopulation distribution for rental car cost is not normally distributed and isskewed to the right.

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