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low carb n=77 average weight loss 4.7kg standard deviation7.2kg low fat n=79 average weight loss 2.6kg...

low carb n=77 average weight loss 4.7kg standard deviation7.2kg
low fat n=79 average weight loss 2.6kg standard deviation5.9kg
  1. For this problem, the degrees of freedom is 147, which gives the critical t-value for a 95% confidence interval to be TC = 1.992. Use this value and the values in the table to calculate the margin of error for the difference in population means: E=Tc s1 2/n1+s22/ n2.
  2. Calculate the difference in sample means, x1-x2, using the low-carbohydrate group as group 1 and the low-fat group as group 2. And calculate the 95% confidence interval for the difference in population means, μ1-μ2.
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Answer #1

For Sample 1 : x̅1 = 4.7, s1 = 7.2, n1 = 77

For Sample 2 : x̅2 = 2.6, s2 = 5.9, n2 = 79

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 146.7768 = 147

95% Confidence interval for the difference :

At α = 0.05 and df = 147, two tailed critical value, t_c = T.INV.2T(0.05, 147) = 1.976

Lower Bound = (x̅1 - x̅2) - t_c*√(s1²/n1 +s2²/n2) = (4.7 - 2.6) - 1.976*√(7.2²/77 + 5.9²/79) = 0.0143

Upper Bound = (x̅1 - x̅2) + t_c*√(s1²/n1 +s2²/n2) = (4.7 - 2.6) + 1.976*√(7.2²/77 + 5.9²/79) = 4.1857

0.0143 < µ1 - µ2 < 4.1857

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Note: critical value with df = 147 and α = 0.05 using excel is 1.976. with TC = 1.992 answer is:

Lower Bound = (x̅1 - x̅2) - t_c*√(s1²/n1 +s2²/n2) = (4.7 - 2.6) - 1.992*√(7.2²/77 + 5.9²/79) = -0.0024

Upper Bound = (x̅1 - x̅2) + t_c*√(s1²/n1 +s2²/n2) = (4.7 - 2.6) + 1.992*√(7.2²/77 + 5.9²/79) = 4.2024

-0.0024 < µ1 - µ2 < 4.2024

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