Question
In: Statistics and Probability
Determine the 95% confidence interval for the difference between two population means where sample 1 has...
Determine the 95% confidence interval for the difference between two population means where sample 1 has data: 16, 14, 19, 18, 19, 20, 15, 18, 17, 18, and sample 2 has data: 13, 19, 14, 17, 21, 14, 15, 10, 13, 15. (Assume equal population variances)
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Expert Solution
| sample 1 ( X ) | Σ ( Xi- X̅ )2 | Sample 2 ( Y ) | Σ ( Yi- Y̅ )2 | |
| 16 | 1.96 | 13 | 4.41 | |
| 14 | 11.56 | 19 | 15.21 | |
| 19 | 2.56 | 14 | 1.21 | |
| 18 | 0.36 | 17 | 3.61 | |
| 19 | 2.56 | 21 | 34.81 | |
| 20 | 6.76 | 14 | 1.21 | |
| 15 | 5.76 | 15 | 0.01 | |
| 18 | 0.36 | 10 | 26.01 | |
| 17 | 0.16 | 13 | 4.41 | |
| 18 | 0.36 | 15 | 0.01 | |
| Total | 174 | 32.4 | 151 | 90.9 |
Mean X̅ = Σ Xi / n
X̅ = 174 / 10 = 17.4
Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1
)
SX = √ ( 32.4 / 10 -1 ) = 1.8974
Mean Y̅ = ΣYi / n
Y̅ = 151 / 10 = 15.1
Sample Standard deviation SY = √ ( (Yi - Y̅ )2 / n - 1
)
SY = √ ( 90.9 / 10 -1) = 3.178
Confidence interval is :-
( X̅1 - X̅2 ) ± t( α/2 , n1+n2-2) SP √( (1/n1) + (1/n2))
t(α/2, n1 + n1 - 2) = t( 0.05/2, 10 + 10 - 2) = 2.101
( 17.4 - 15.1 ) ± t(0.05/2 , 10 + 10 -2) 2.6172 √ ( (1/10) +
(1/10))
Lower Limit = ( 17.4 - 15.1 ) - t(0.05/2 , 10 + 10 -2) 2.6172 √(
(1/10) + (1/10))
Lower Limit = -0.159
Upper Limit = ( 17.4 - 15.1 ) + t(0.05/2 , 10 + 10 -2) 2.6172 √(
(1/10) + (1/10))
Upper Limit = 4.759
95% Confidence Interval is ( -0.159 , 4.759 )
orchestra answered 1 year agoRelated Solutions
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