# SOLUTION: I am having difficulty with linear rgression. I could sure use some help with my problems. At the very end is what I have so far. Thanks in advance for all your help. The dat

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: I am having difficulty with linear rgression. I could sure use some help with my problems. At the very end is what I have so far. Thanks in advance for all your help. The dat      Log On

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 Algebra: Probability and statistics Solvers Lessons Answers archive Quiz In Depth

 Question 192061: I am having difficulty with linear rgression. I could sure use some help with my problems. At the very end is what I have so far. Thanks in advance for all your help. The data below are the gestation periods, in months, of randomly selected animals and their corresponding life spans, in years. Gestation, x: 8, 2.1, 1.3, 1, 11.5, 5.3, 3.8, 24.3 Life span, y: 30, 12, 6, 3, 25, 12, 10, 40 n= 8 A. find the equation of the regression line for the given data. round the line values to the nearest two decimal points. B. using the equation found in part a, predict the life span when the gestation is 10 months. Round to the nearest absence. Guidlines -- Hypothesis testing steps: 1. State Ho and Ha 2. Specify the level of significance alpha a 3. Find the standarized test statistic t = r / sqrt [1 - r squared] [n-2] 4. Find the criticalvalue(s) to. Use the method specified in the problem statement. 5. Define the rejection region using critical value(s). 6. Make a decision to reject or fail to rejct the null hypothesis. 7. Interpret the decision in the context of the original claim. I think Ho and Ha are: Ho: p = 0 and Ha: p does not equal 0, which would make this a 2 tail This is what I have: X Y X * Y X² Y² 8 30 240 64 900 2.1 12 25.2 4.41 144 1.3 6 7.8 1.69 36 1 3 3 1 9 11.5 25 287.5 132.25 625 5.3 12 63.6 28.09 144 3.8 10 38 14.44 100 24.3 40 972 590.49 1600 ∑x═ ∑y═ ∑xy═ ∑x²═ ∑y²═ 57.3 138 1637.1 836.37 3558 n = 8 d.f = 8 - 2 = 6 (-3.707, 3.707) {two-tailed test} cu = ±3.707; Reject Ho if t < -3.707 or t > 3.707 r = 0.915931364 cv = 0.917 Strong positive correlation because "r" is close to 1 8 (1637.1) - (57.3) (138) = 5189.4 {13096.8 - 7907.4} 8 (836.37) - (57.3)² = 3407.67 {6690.96 - 3283.29} 8 (3558) - (138)² = 9420 {28464 - 19044} 5189.4 / sqrt [3407.67] [9420] = 0.9159313639 or 0.916 (rounded to 3 decimal places) t= 0.916 / sqrt 1-(0.916)² / 8-2 sqrt (1-(.916)²) = 0.4011782646 / sqrt (8-2) = 0.1637803407 .916 / 0.1637803407 = 5.59286 or 5.593 (rounded to 3 decimal places) Because t is in the rejection region, you should decide to reject the null hypothesis. There is enough evidence to conclude that a significant linear correlation exists. Answer by stanbon(57203)   (Show Source): You can put this solution on YOUR website!I am having difficulty with linear rgression. I could sure use some help with my problems. At the very end is what I have so far. Thanks in advance for all your help. The data below are the gestation periods, in months, of randomly selected animals and their corresponding life spans, in years. Gestation, x: 8, 2.1, 1.3, 1, 11.5, 5.3, 3.8, 24.3 Life span, y: 30, 12, 6, 3, 25, 12, 10, 40 n= 8 A. find the equation of the regression line for the given data. round the line values to the nearest two decimal points. y = 1.52x + 6.34 --------------------------- B. using the equation found in part a, predict the life span when the gestation is 10 months. Round to the nearest absence. f(10) = 1.52*10 + 6.34 f(10) = 15.2 + 6.34 f(10) = 21.54 ------------------------------ Guidlines -- Hypothesis testing steps: 1. State Ho and Ha 2. Specify the level of significance alpha a 3. Find the standarized test statistic t = r / sqrt [1 - r squared] [n-2] 4. Find the criticalvalue(s) to. Use the method specified in the problem statement. 5. Define the rejection region using critical value(s). 6. Make a decision to reject or fail to rejct the null hypothesis. 7. Interpret the decision in the context of the original claim. I think Ho and Ha are: Ho: p = 0 and Ha: p does not equal 0, which would make this a 2 tail This is what I have: X Y X * Y X² Y² 8 30 240 64 900 2.1 12 25.2 4.41 144 1.3 6 7.8 1.69 36 1 3 3 1 9 11.5 25 287.5 132.25 625 5.3 12 63.6 28.09 144 3.8 10 38 14.44 100 24.3 40 972 590.49 1600 ∑x═ ∑y═ ∑xy═ ∑x²═ ∑y²═ 57.3 138 1637.1 836.37 3558 n = 8 d.f = 8 - 2 = 6 (-3.707, 3.707) {two-tailed test} cu = ±3.707; Reject Ho if t < -3.707 or t > 3.707 r = 0.915931364 cv = 0.917 Strong positive correlation because "r" is close to 1 8 (1637.1) - (57.3) (138) = 5189.4 {13096.8 - 7907.4} 8 (836.37) - (57.3)² = 3407.67 {6690.96 - 3283.29} 8 (3558) - (138)² = 9420 {28464 - 19044} 5189.4 / sqrt [3407.67] [9420] = 0.9159313639 or 0.916 (rounded to 3 decimal places) t= 0.916 / sqrt 1-(0.916)² / 8-2 sqrt (1-(.916)²) = 0.4011782646 / sqrt (8-2) = 0.1637803407 .916 / 0.1637803407 = 5.59286 or 5.593 (rounded to 3 decimal places) Because t is in the rejection region, you should decide to reject the null hypothesis. There is enough evidence to conclude that a significant linear correlation exists. ------------------------------------------------ I have not checked your work below # 7. Cheers, Stan H.