5) A company that manufactures baseball bats believes that its new bat will allow players to hit the ball for 30 feet farther than its current model. The owner hires a professional baseball player known for hitting home runs to hit ten balls with each bat and he measures the distance each ball is hit to test the company’s claim. The results of the batting experiment are shown in the following table. Construct and interpret a 90% confidence interval for the true difference between the mean distance hit with the new model and the mean distance hit with the older model. Assume that the variances of the two populations are the same. Round your CI limits to 1 decimal place.
HItting distance in feet
New model: 235 240 253 267 243 237 250 241 251 248
Old model : 200 210 231 218 210 209 210 229 234 231
For New Model :
∑x = 2465
∑x² = 608427
n1 = 10
Mean , x̅1 = Ʃx/n = 2465/10 = 246.5
Standard deviation, s1 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(608427-(2465)²/10)/(10-1)] = 9.4546
For Old Model :
∑x = 2182
∑x² = 477424
n2 = 10
Mean , x̅2 = Ʃx/n = 2182/10 = 218.2
Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(477424-(2182)²/10)/(10-1)] = 12.0720
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90% Confidence interval :
At α = 0.1 and df = n1+n2-2 = 18, two tailed critical value, t-crit = T.INV.2T(0.1, 18) = 1.734
S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((10-1)*9.4546² + (10-1)*12.072²) / (10+10-2) = 117.561
Lower Bound = (x̅1 - x̅2) - t-crit*√(S²p*(1/n1 +1/n2))
= (246.5 - 218.2) - 1.734*√(117.5611*(1/10 + 1/10)) = 19.9
Upper Bound = (x̅1 - x̅2) + t-crit*√(S²p*(1/n1 +1/n2))
= (246.5 - 218.2) + 1.734*√(117.5611*(1/10 + 1/10)) = 36.7
19.9 < µ1 - µ2 < 36.7
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