Given two dependent random samples with the following results:
Population 1 | 19 | 26 | 26 | 30 | 48 | 33 | 31 |
---|---|---|---|---|---|---|---|
Population 2 | 24 | 38 | 40 | 38 | 39 | 41 | 29 |
Use this data to find the 98% confidence interval for the true difference between the population means.
Let d=(Population 1 entry)−(Population 2 entry)d=(Population 1 entry)−(Population 2 entry). Assume that both populations are normally distributed.
Step 1 of 4 : Find the mean of the paired differences, x‾d. Round your answer to one decimal place.
Step 2 of 4: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
Step 3 of 4: Find the standard deviation of the paired differences to be used in constructing the confidence interval. Round your answer to one decimal place.
Step 4 of 4: Construct the 98% confidence interval. Round your answers to one decimal place.
Population 1 | Population 2 | Difference |
19 | 24 | -5 |
26 | 38 | -12 |
26 | 40 | -14 |
30 | 38 | -8 |
48 | 39 | 9 |
33 | 41 | -8 |
31 | 29 | 2 |
1)
∑d = -36
∑d² = 578
n = 7
Mean , x̅d = Ʃd/n = -36/7 = -5.1429 = -5.1
2)
At α = 0.02 and df = n-1 = 6, two tailed critical value, t-crit = T.INV.2T(0.02, 6) = 3.143
3)
Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] = √[(578-(-36)²/7)/(7-1)] = 8.0917 = 8.1
4)
98% Confidence interval :
At α = 0.02 and df = n-1 = 6, two tailed critical value, t-crit = T.INV.2T(0.02, 6) = 3.143
Lower Bound = x̅d - t-crit*sd/√n = -5.1429 - 3.143 * 8.0917/√7 = -14.8
Upper Bound = x̅d + t-crit*sd/√n = -5.1429 + 3.143 * 8.0917/√7 = 4.5
-14.8 < µd < 4.5
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