Statistics question a.4

Statistics
Tutor: None Selected Time limit: 1 Day

Apr 29th, 2015

s1² =√ (8250 - 8179.6)/10 = 7.04

s2² = √(8496 - 8000.9/10) = 8.71

s² = (7.04 + 8.71)/2 = 7.875

and s = 2.806

Now critical t value (18 df) at 90% confidence is 1.734

difference in  means = 28.6 - 28.3 = 0.3

Hence confidence interval =( 0.3 -1.734* 2.806, 0.3 + 1.734*2.806)

= (-0.187,  0 .7.866)

Apr 29th, 2015

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Apr 29th, 2015
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Apr 29th, 2015
Dec 5th, 2016
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