In: Math
Suppose you are interested in buying a new Toyota Corolla. You are standing on the sales lot looking at a model with different options. The list price is on the vehicle. As a salesperson approaches, you wonder what the dealer invoice price is for this model with its options. The following data are based on a random selection of Toyota Corollas of different models and options. Let y be the dealer invoice (in thousands of dollars) for the given vehicle.
x | 12.9 | 13.0 | 12.8 | 13.6 | 13.4 | 14.2 |
y | 11.6 | 11.9 | 11.5 | 12.2 | 12.0 | 12.8 |
(a) Verify that Σx = 79.9, Σy = 72, Σx2 = 1065.41, Σy2 = 865.1, Σxy = 960.02, and r ≈ 0.980.
Σx | = |
Σy | = |
Σx2 | = |
Σy2 | = |
Σxy | = |
r | = |
(b) Use a 1% level of significance to test the claim that
ρ > 0. (Use 2 decimal places.)
t | = |
critical t | = |
Conclusion
Reject the null hypothesis, there is sufficient evidence that ρ > 0.
Reject the null hypothesis, there is insufficient evidence that ρ > 0.
Fail to reject the null hypothesis, there is insufficient evidence that ρ > 0.
Fail to reject the null hypothesis, there is sufficient evidence that ρ > 0.
(c) Verify that Se ≈ 0.1039, a ≈
0.464, and b ≈ 0.866.
Se | = |
a | = |
b | = |
(d) Find the predicted dealer invoice when the list price is
x = 13.6 (thousand dollars). (Use 2 decimal places.)
(e) Find a 90% confidence interval for y when x =
13.6 (thousand dollars). (Use 2 decimal places.)
lower limit | = |
upper limit | = |
(f) Use a 1% level of significance to test the claim that
β > 0. (Use 2 decimal places.)
t | = |
critical t | = |
Conclusion
Reject the null hypothesis, there is sufficient evidence that β > 0.
Reject the null hypothesis, there is insufficient evidence that β > 0.
Fail to reject the null hypothesis, there is insufficient evidence that β > 0.
Fail to reject the null hypothesis, there is sufficient evidence that β > 0.
(g) Find a 90% confidence interval for β and interpret its
meaning. (Use 2 decimal places.)
lower limit | = |
upper limit | = |
Interpretation
For every $1,000 increase in list price, the dealer price increases by an amount that falls within the confidence interval.
For every $1,000 increase in list price, the dealer price increases by an amount that falls outside the confidence interval.
For every $1,000 increase in list price, the dealer price decreases by an amount that falls within the confidence interval.
For every $1,000 increase in list price, the dealer price decreases by an amount that falls outside the confidence interval.
a)
X | Y | XY | X² | Y² |
12.9 | 11.6 | 149.64 | 166.41 | 134.56 |
13.0 | 11.9 | 154.7 | 169 | 141.61 |
12.8 | 11.5 | 147.2 | 163.84 | 132.25 |
13.6 | 12.2 | 165.92 | 184.96 | 148.84 |
13.4 | 12.0 | 160.8 | 179.56 | 144 |
14.2 | 12.8 | 181.76 | 201.64 | 163.84 |
Ʃx = | Ʃy = | Ʃxy = | Ʃx² = | Ʃy² = |
79.9 | 72 | 960.02 | 1065.41 | 865.1 |
Sample size, n = | 6 |
x̅ = Ʃx/n = | 13.3167 |
y̅ = Ʃy/n = | 12 |
SSxx = Ʃx² - (Ʃx)²/n = | 1.40833 |
SSyy = Ʃy² - (Ʃy)²/n = | 1.1 |
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = | 1.22 |
Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 0.9802 =0.980
-------------------------------------
b) Null and alternative hypothesis:
Ho: ρ = 0, Ha: ρ > 0
α = 0.01
Test statistic :
t = r*√(n-2)/√(1-r²) = 0.9802 *√(6 - 2)/√(1 - 0.9802²) = 9.90
df = n-2 = 4
Critical value, t_c = T.INV(0.01, 4) = 3.75
Conclusion:
Reject the null hypothesis, there is sufficient evidence that ρ > 0.
-------------------------------------
(c) Sum of Square error, SSE = SSyy -SSxy²/SSxx = 0.043147929
Standard error, se = √(SSE/(n-2)) = 0.1039
Slope, b = SSxy/SSxx = 0.866
y-intercept, a = y̅ -b* x̅ = 0.464
-------------------------------------
(d) ŷ = 0.464 + (0.866) * 13.6 = 12.25
-------------------------------------
(e) Significance level, α = 1 - 0.90 = 0.10
df = n-2 = 4
Critical value, t_c = T.INV.2T(0.10, 4) = 2.1318
90% confidence interval for y when x = 13.6
Lower limit = 12.00
Upper limit = 12.49
-------------------------------------
(f) Null and alternative hypothesis:
Ho: β₁ = 0, Ha: β₁ > 0
Test statistic:
t = b /(se/√SSxx) = 9.90
df = n-2 = 4
Critical value, t_c = T.INV(0.01, 4) = 3.75
Conclusion
Reject the null hypothesis, there is sufficient evidence that β > 0.
(g) Significance level, α = 0.10
Critical value, t_c = T.INV.2T(0.1, 4) = 2.1318
90% confidence interval for β :
Interpretation
For every $1,000 increase in list price, the dealer price increases by an amount that falls within the confidence interval.