JustAnswer
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|(a) b0 and b1 alone are parameters but s is not a parameter in simple linear regression. This is because the distribution |
|of y given a value of x has equal standard deviation for all x values and is centered about the least squares regression line. |
|(b) Here Ho is not b1 = 0 because we are testing for the population slope and not sample slope. Ho should be β1 = 0 |
|(c) The prediction interval is wider than the confidence interval. This is because it is much easier to predict what will happen on|
|average, in the long run, than it is to predict what will happen in any particular measurement. |
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|(a) |
|IBI: x-bar = 65.94 and s = 18.280 |
|Area: x-bar = 28.29 and s = 17.714 |
|(b) |
|y = 52.923 + 0.460x |
|F = 11.67 |
|P = 0.001 |
|r = 0.446 |
|Option A |
|Option B |
|Option B |
|Option C |
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|(a) |
|y = 53.40 + (-0.1959)x |
|Option A |
|(b) |
|Option D |
|t = -1.265 |
|df = 14 |
|P = 0.2264 |
|Option B |
|(c) |
|Option A |
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|(a) Option A |
|(b) Option C |
|(c) Option C |
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|(5) |
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|(b) |
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|(c) |
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|The 95% confidence interval is [0.186, 0.202] |
| |
|(6) |
|n = 20 |
|R = 0.5 |
|Hypotheses: |
|Ho: There is no significant correlation, that is ρ = 0 |
|Ha: There is significant correlation, that is ρ ≠ 0 |
|Decision Rule: |
|t (Two-tailed), α = 0.05 |
|Degrees of freedom = 20 - 2 = 18 |
|Lower Critical t- score = -2.100922037 |
|Upper Critical t- score = 2.100922037 |
|Reject Ho if |t| > 2.100922037 |
|Test Statistic: |
|SE = √{(1 - R^2)/DOF} = √((1 - 0.5^2)/18) = 0.204124145 |
|t = R/SE = 0.5/0.204124145231932 = 2.449489743 |
|p- value = 0.024769559 |
|Decision (in terms of the hypotheses): |
|Since 2.449489743 > 2.100922037 we reject Ho |
|Conclusion (in terms of the problem): |
|It appears that the variables are correlated |
| |
|n = 10 |
|R = 0.5 |
|Hypotheses: |
|Ho: There is no significant correlation, that is ρ = 0 |
|Ha: There is significant correlation, that is ρ ≠ 0 |
|Decision Rule: |
|t (Two-tailed), α = 0.05 |
|Degrees of freedom = 10 - 2 = 8 |
|Lower Critical t- score = -2.306004133 |
|Upper Critical t- score = 2.306004133 |
|Reject Ho if |t| > 2.306004133 |
|Test Statistic: |
|SE = √{(1 - R^2)/DOF} = √((1 - 0.5^2)/8) = 0.306186218 |
|t = R/SE = 0.5/0.306186217847897 = 1.632993162 |
|p- value = 0.14111328 |
|Decision (in terms of the hypotheses): |
|Since 1.632993162 < 2.306004133 we do not reject Ho |
|Conclusion (in terms of the problem): |
|There is no sufficient evidence that the variables are correlated |
|The conclusions are different because of different sample sizes. In the first case, sample size is larger and we found evidence of |
|correlation, but in the second case, sample size is smaller and we found no evidence of correlation. |
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|(a) |
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|The regression equation is y = -115.102 + 3.1019x |
|r = 0.993 implies a strong correlation between x and y |
|r^2 = 0.986 means about 98.6% of the variation in y is explained by the variation in x |
|The regression is significant since the p- value is 0. Length * Width is a strong predictor of Weight. |
|(b) The other regression analyses that can be done are the simple linear regression of Weight vs Length alone and the multiple |
|linear regression of Weight vs Length and Length * Width |
|Here are the results for the multiple regression model … |
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|The regression equation is y = 10.2085 - 9.2137*x1 + 4.0365*x2 |
|r = 0.995 implies a strong correlation between the variables |
|r^2 = 0.991 means about 99.1% of the variation in y is explained by the variation in x1 and x2 |
|The p- value is > 0.05 for length. This means Length is not a significant predictor. But the p- value is < 0.05 for Length * Width.|
|This means Length * Width is a strong predictor of Weight. |
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