I can\'t afford to be wrong. Can anyone help me find the missing piece. I have a

Question

I can't afford to be wrong. Can anyone help me find the missing piece. I have already tried -0.11 and -0.12. One more and I'm out. Please help.

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These data are observations collected using a completely randomized design. Sample 1 Sample 2 Sample 3 (a) Find a 90% confidence interval for u1. Round your answers to two decimal places 1.93 to 4.16 (b) Find a 90% confidence interval for the difference (u1 u3). Round your answers to two decimal places x to 3.12 Need Help? Read It

Explanation / Answer

a) mean = 3 , std,deviation = 1.224 , n = 5

t value for 90% confidence interval = 2.132

CI = mean + / - t * ( s / sqrt(n))

= 3 + / - 2.132 * ( 1.224 / sqrt(5))

= (1.93 , 4.16)

b) mean 1 = 3 , s1 = 1.224 , mean 3 = 1.5 , s3 = 0.577 , n1 = 5 , n3 = 4

t value for 90% confidence interval with df = 7 = 1.895

CI = (mean1 - mean3 ) + / - t * sqrt(s1^2 / n1 + s3^2/n3)

= ( 3 - 1.5) + / -1.895 * sqrt(1.224^2 / 5 + 0.577^2/4)

= (0.33 , 2.67)

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