Question

1. The data in the following MINITAB output refer to an automobile’s stopping distance (in feet) at different speeds.

Data Display Row     Speed StopDist 1          25        63 2          25        56 3          30        84 4          35        107 5          45        153 6          45        164 7          55        204 8          55        220 9          65        285 10        65        303 Descriptive Statistics Variable          N          Mean               Median            TrMean            StDev              SEMean Speed              10        44.50               45.00               44.38               15.36               4.86 StopDist          10        163.9               158.5               160.0               88.4                 28.0 Variable          Min      Max     Q1       Q3 Speed              25.00 65.00 28.75 57.50 StopDist          56.0     303.0 78.7     236.2 Regression Analysis The regression equation is StopDist = __________________________________ Predictor         Coef     Stdev t-ratio p Constant          −89.99             12.68 −7.10 0.000 Speed              5.7053             0.2707   ___________ s = 12.47         R-sq = 98.2% R-sq(adj) = 98.0%

1. What’s the equation of the regression line you’d use to predict stopping distance based on speed?

Answer:

2. Interpret the slope of the regression line.

Answer:

3. Find a 95% confidence interval for the slope of the regression line.

Answer:

4. Test the hypothesis H₀: β₁ = 0 and H_a: β₁ ≠ 0.

Answer:

5. Suppose, instead of testing H₀: β₁ = 0, you decided to test H₀: β₁ = 5 against H_a: β₁ ≠ 5. What P-value would be associated with this test?

Answer:

Answer 1

a)

stopdist = -89.99 + 5.7053*spped

b)

for every unit increase in speed, value of stop distance get increase by 5.7053

c)

n =   10
alpha,α =    0.05
estimated slope=   5.7053
std error =    0.2707

Df =    n - 2 =   8
critical t-value =    2.3060 [excel function: =t.inv.2t(0.05,8)

margin of error ,E =    t*Std error =   0.6242
      
confidence interval is      
lower bound =    ß1 - E=   5.0811
upper bound =    ß1 + E=   6.3295

d)

Slope hypothesis test      
Ho:   ß = 0  
Ha:   ß ╪ 0  
      
n =   10  
alpha,α =    0.05  
estimated slope=   5.7053  
std error =    0.2707  
      
t-test statistic =    t = (estimated slope -ß)/ std error = (5.7053-0)/0.2707 = 21.0761
      
Df =    n - 2 =   8
critical t-value =    2.3060  

since, t-stat = 21.0761 >t-critical, reject Ho

so, slope is significant at α=0.05

e)

t-test statistic =    t = (estimated slope -ß)/ std error = (5.7053-2)/0.2707 =   13.688

df=8

p-value=0.0000 [excel function: =t.dist.2t(13.688,8) ]