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In: Math

Your scores from an afternoon of bowling were 128, 189, 156, 143, 117, and 152. Your...

Your scores from an afternoon of bowling were 128, 189, 156, 143, 117, and 152. Your sisters bowling scores were 204, 176, 189, 198, 218, 177, 190,and 203. Find the(two-sided) 98% confidence interval for (sigma^2*you) /(sigma^2*Sister). Assume bowling is normally distributed.

Solutions

Expert Solution

Your scores bowling  ( X ) Σ ( Xi- X̅ )2 sisters bowling ( Y ) Σ ( Yi- Y̅ )2
128 380.25 204 92.6406
189 1722.25 176 337.6406
156 72.25 189 28.8906
143 20.25 198 13.1406
117 930.25 218 558.1406
152 20.25 177 301.8906
190 19.1406
203 74.3906
Total 885 3145.5 1555 1425.8748

Mean X̅ = Σ Xi / n
X̅ = 885 / 6 = 147.5
Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 3145.5 / 6 -1 ) = 25.0819

Mean Y̅ = ΣYi / n
Y̅ = 1555 / 8 = 194.375
Sample Standard deviation SY = √ ( (Yi - Y̅ )2 / n - 1 )
SY = √ ( 1425.8748 / 8 -1) = 14.2722

Confidence Interval



F(0.02/2, n1,n2) = F(0.02/2, 5,7) = 7.46
F(0.02/2, n1,n2) = F(0.02/2, 7,5) = 10.46
Lower Limit = ( 629.1017 / 203.6957 ) * ( 1 / F(0.02/2, 5,7)) = 0.4140
Upper Limit = ( 629.1017 / 203.6957 ) * ( F(0.02/2, 7,5)) = 32.3051
98% Confidence interval is ( 0.414 , 32.3051 )
( 0.4140 <      < 32.3051 )


milcah answered 1 hour ago
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