Question

QUESTION 1

1. A consumer advocacy group compares the amount of nicotine in two brands of cigarettes. A sample of 64 cigarettes from Brand A has an average of 1.95 mg with a standard deviation of 0.24 mg. A sample of 81 cigarettes from Brand B has an average of 1.85 mg and a standard deviation of 0.22.  

   
   Suppose you want to estimate how much more nicotine is in Brand A cigarettes in general using a 90% confidence interval.
   
   1. The facts of this problem are:

   		mu(A)=1.95;mu(B)=1.85;sigma(A)=0.24;sigma(B)=0.22;n(A)=64;n(B)=81
   		x-bar(A)=1.95;x-bar(B)=1.85;s(A)=0.24;s(B)=0.22;n(A)=64;n(B)=81
   		x-bar(A)=0.24;x-bar(B)=0.22;s(A)=1.95;s(B)=1.85;n(A)=64;n(B)=81

10 points   

QUESTION 2

1. What is the correct table to use for this problem?

   		Table F because it is a chi-square problem
   		Table A because this is a proportion problem
   		Table D because sigma is not known

10 points   

QUESTION 3

1. The degrees of freedom based directly on the facts of the problem are

   		64
   		81
   		63
   		80

10 points   

QUESTION 4

1. The degrees of freedom that you end up using from table D is

   		64
   		81
   		63
   		60

10 points   

QUESTION 5

1. The correct formula to use is

   		(xbar-mu)/[s/square root(n)]
   		(p-hat minus p)/square-root of [p*(1-p)/n]
   		[xbar(suv)-xbar(small)] +/- t * square root of [square(s(suv))/n(suv)+square(s(small))/n(small)]
   		(p-hat minus p)/[p*(1-p)/n]

10 points   

QUESTION 6

1. The t value for the problem is

   		2.000
   		2.660
   		1.671
   		-1.671

10 points   

QUESTION 7

1. The margin of error for this problem is (Remember that the margin of error is the value to the right of the +/- sign)

   		7.9
   		17.9
   		0.065
   		-0.065

10 points   

QUESTION 8

1. The confidence interval for this problem is [remember a confidence interval has the form (lower,upper)]

   		(29.09,64.91)
   		(0.035,0.165)
   		(0.165,0.035)
   		(2.6%,47%)

10 points   

QUESTION 9

1. A spokeperson for the Ministry of Public Health says "Based on the results of this study, the additional amount of nicotine in Brand A could be 0.12 mg" Based on the confidence interval your business conclusion would be...

   		The spokesperson is wrong
   		The spokesperson is right

Answer 1

QUESTION 1:

Answer: x-bar(A)=1.95;x-bar(B)=1.85;s(A)=0.24;s(B)=0.22;n(A)=64;n(B)=81

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QUESTION 2

Answer: Table D because sigma is not known

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QUESTION 3

Degrees of freedom based directly on the facts of the problem : min(n1-1, n2-1) = 63

Answer: 63

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QUESTION 4

Degrees of freedom that you end up using from table D :

Answer: 60

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QUESTION 5

The correct formula to use is

Answer: [xbar(suv)-xbar(small)] +/- t * square root of [square(s(suv))/n(suv)+square(s(small))/n(small)]

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QUESTION 6

At α = 0.1 and df = 60, two tailed critical value, t_c = T.INV.2T(0.1, 60) = 1.671

Answer: 1.671

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QUESTION 7

Margin of error, E = t_c*√(s1²/n1 +s2²/n2) = 1.671*√(0.24²/64 + 0.22²/81) = 0.065

Answer: 0.065

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QUESTION 8

90% Confidence interval for the difference :

Lower Bound = (x̅1 - x̅2) - t_c*√(s1²/n1 +s2²/n2)

= (1.95 - 1.85) - 1.671*√(0.24²/64 + 0.22²/81) = 0.035

Upper Bound = (x̅1 - x̅2) + t_c*√(s1²/n1 +s2²/n2)

= (1.95 - 1.85) + 1.671*√(0.24²/64 + 0.22²/81) = 0.165

Answer: (0.035,0.165)

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QUESTION 9

Answer: The spokesperson is right