Question

A taxi company is trying to decide whether to purchase brand A or brand B tires for its fleet of taxis. To estimate the difference in the two brands, an experiment is conducted using 8 of each brand, assigned at random to the left and right rear wheels of 8 taxis. The tires are run until they wear out and the distances, in kilometers, are recorded in the accompanying data set. Find a 90% confidence interval for y- H2. Assume that the differences of the distances are approximately normally distributed Click here to view the data set. Click here to view page 1 of the table of critical values of the t-distribution. Click here to view page 2 of the table of critical values of the t-distribution. Let Hy be the population mean for brand A and let uz be the population mean for brand B. The confidence interval is <H-H2 (Round to one decimal place as needed.)

Answer 1

	Brand A ( X )	Σ ( Xi- X̅ )2	Brand B ( Y )	Σ ( Yi- Y̅ )2
	34300	14726406.25	36100	7770156.25
	41400	10643906.25	41800	8482656.25
	37200	878906.25	37700	1410156.25
	33100	25376406.25	32000	47437656.25
	41500	11306406.25	40100	1470156.25
	44600	41763906.25	48300	88595156.25
	37000	1293906.25	38100	620156.25
	36000	4568906.25	37000	3562656.25
Total	305100	110558750	311100	159348750

Mean X̅ = Σ Xi / n
X̅ = 305100 / 8 = 38137.5
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 110558750 / 8 -1 ) = 3974.1801

Mean Y̅ = ΣYi / n
Y̅ = 311100 / 8 = 38887.5
Sample Standard deviation S_Y = √ ( (Yi - Y̅ )² / n - 1 )
S_Y = √ ( 159348750 / 8 -1) = 4771.1746

Confidence interval :-
$\( \bar{X}_{1} - \bar{X}_{2}) \pm t_{\alpha /2 , DF}\sqrt{ (S_{1}^{2}/n1) &plus; (S_{2}^{2}/n2)}$

$DF = ( ( S_{1}^{2} / n1 &plus; S_{2}^{2} / n2))^{2} / ( (S_{1}^{2} / n1)^{2}/n1-1) &plus; (S_{2}^{2} / n2)^{2}/n2-1) )$
$DF = ( ( 3974.1801^{2} / 8 &plus; 4771.1746^{2} / 8 ))^{2} / ( ( 3974.1801^{2} / 8 )^{2}/ 8 -1) &plus; ( 4771.1746^{2} / 8 )^{2}/ 8 -1) )$
DF = 13

t(α/2, DF) = t(0.1 /2, 13 ) = 1.771
$( 38137.5 - 38887.5 ) \pm t_{ 0.1/2 , 13 }\sqrt{ ( 3974.1801^{2}/ 8 ) &plus; ( 4771.1746^{2}/ 8 )}$
Lower Limit = $( 38137.5 - 38887.5 ) - t_{ 0.1/2 , 13 }\sqrt{ ( 3974.1801^{2}/ 8 ) &plus; ( 4771.1746^{2}/ 8 )}$
Lower Limit = -4637.9
Upper Limit = $( 38137.5 - 38887.5 ) &plus; t_{ 0.1/2 , 13 }\sqrt{ ( 3974.1801^{2}/ 8 ) &plus; ( 4771.1746^{2}/ 8 )}$
Upper Limit = 3137.90
90% Confidence interval is ( -4637.9 , 3137.9 )