Question

Vehicle Name	Weight
Jaguar X-Type 2.5 4dr	3428
Cadillac CTS VVT 4dr	3694
Jaguar X-Type 3.0 4dr	3516
Cadillac Deville 4dr	3984
Cadillac Deville DTS 4dr	4044
Cadillac Seville SLS 4dr	3992
Jaguar S-Type 3.0 4dr	3777
Jaguar S-Type 4.2 4dr	3874
Jaguar S-Type R 4dr	4046
Jaguar Vanden Plas 4dr	3803
Jaguar XJ8 4dr	3803
Jaguar XJR 4dr	3948

In what follows use any of the following tests/procedures: Regression, multiple regressions, confidence intervals, one-sided t-test or two-sided t-test. All the procedures should be done with 5% P-value or 95% confidence interval

Upload CARS data. SETUP: It is believed that Lincolns and Cadillac have different weigths. Given the data your job is to confirm or disprove this belief. (CAREFULL: sort the data in order to extract the needed information).

9. What test/procedure did you perform?

• a. One-sided t-test
• b. Two-sided t-test
• c. Regression
• d. ​​Confidence interval

10. What is the P-value/margin of error?

• a. 339.06234
• b. 0.598587029
• c. 0.299293514
• d. 8.065421
• e. ​​None of these

11. Statistical Interpretation

• a. Due to small margin of error we are confident that the average mpg is above 3500.
• b. Since P-value is large we cannot claim that the averages are different.
• c. Since P-value is very small we are confident that the slope of regression line is not zero.
• d. None of these

12. Conclusion

• a. Yes, I am confident that the above assertion is correct.
• b. No, we cannot claim that the above assertion is correct.

Task 2

Answer 1

Data:

Sample 1	Weight		Sample 2	Weight
Cadillac CTS VVT 4dr	3694		Jaguar S-Type 3.0 4dr	3777
Cadillac Deville 4dr	3984		Jaguar S-Type 4.2 4dr	3874
Cadillac Deville DTS 4dr	4044		Jaguar S-Type R 4dr	4046
Cadillac Seville SLS 4dr	3992		Jaguar Vanden Plas 4dr	3803
			Jaguar XJ8 4dr	3803
			Jaguar XJR 4dr	3948
			Jaguar X-Type 2.5 4dr	3428
			Jaguar X-Type 3.0 4dr	3516

A.

Two-sided t-test

i.e Option B is correct.

B.

Difference Scores Calculations

Treatment 1

N₁: 4
df₁ = N - 1 = 4 - 1 = 3
M₁: 3928.5
SS₁: 75443
s²₁ = SS₁/(N - 1) = 75443/(4-1) = 25147.67


Treatment 2

N₂: 8
df₂ = N - 1 = 8 - 1 = 7
M₂: 3774.38
SS₂: 302229.88
s²₂ = SS₂/(N - 1) = 302229.88/(8-1) = 43175.7


T-value Calculation

s²_p = ((df₁/(df₁ + df₂)) * s²₁) + ((df₂/(df₂ + df₂)) * s²₂) = ((3/10) * 25147.67) + ((7/10) * 43175.7) = 37767.29

s²_M1 = s²_p/N₁ = 37767.29/4 = 9441.82
s²_M2 = s²_p/N₂ = 37767.29/8 = 4720.91

t = (M₁ - M₂)/√(s²_M1 + s²_M2) = 154.12/√14162.73 = 1.3

df = 10

p-value = 0.22439 > 0.05 (The result is not significant )

i.e Option C is correct (Closest answer)

C.

Since P-value is large we cannot claim that the averages are different.

i.e Option B is correct.

D.

No, we cannot claim that the above assertion is correct.

i.e Option B is correct.

Please upvote if you have liked my answer would be of great help. Thank you.