Question

A random sample of 50 customer orders for wash down motors used in the food processing industry. The order size refers to the number of motors ordered and the total cost is the total production cost for that order.

1. Define the Independent and Dependent Variables
2. Create and interpret the scatter plot
3. Find and evaluate r
4. Conduct the regression
5. Meet the assumptions for the regression
6. Hypothesis test the regression
7. Create the regression model
8. Create the 95% confidence interval model
9. Predict the cost for batches of 10, 20 and 30

Order Size	Total Cost
28	$         7,531
27	$         6,329
34	$         8,413
30	$         7,793
19	$         5,360
21	$         4,838
11	$         2,551
17	$         3,899
33	$         8,326
23	$         5,465
7	$         2,283
23	$         5,413
18	$         4,238
26	$         6,911
20	$         6,315
29	$         8,243
10	$         2,866
22	$         6,775
14	$         4,289
6	$         1,475
15	$         3,590
35	$         9,439
23	$         6,760
18	$         5,170
31	$         7,780
21	$         4,896
35	$         8,816
29	$         8,116
21	$         6,212
24	$         5,551
27	$         7,080
36	$         9,826
24	$         6,129
20	$         5,094
13	$         3,568
14	$         3,738
19	$         5,332
14	$         3,286
28	$         6,664
24	$         5,990
27	$         7,093
16	$         3,975
36	$         9,046
20	$         4,906
19	$         5,324
7	$         2,734
32	$         8,138
21	$         5,376
29	$         7,763
23	$         5,964

Answer 1

Independent Variable : Order Size

Dependent Variables : Total cost

----

Scatter plot:

The scatter plot indicates a positive and strong linear relationship between order size and Total cost.

---

Ʃx =	1119
Ʃy =	292669
Ʃxy =	7297616
Ʃx² =	28031
Ʃy² =	1911167665
Sample size, n =	50
x̅ = Ʃx/n = 1119/50 =	22.38
y̅ = Ʃy/n = 292669/50 =	5853.38
SSxx = Ʃx² - (Ʃx)²/n = 28031 - (1119)²/50 =	2987.78
SSyy = Ʃy² - (Ʃy)²/n = 1911167665 - (292669)²/50 =	198064793.8
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 7297616 - (1119)(292669)/50 =	747683.78

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 747683.78/√(2987.78*198064793.78) = 0.9719

There is strong and positive linear relationship.

----

Slope, b = SSxy/SSxx = 747683.78/2987.78 = 250.24727

y-intercept, a = y̅ -b* x̅ = 5853.38 - (250.24727)*22.38 = 252.84616

Regression equation :

ŷ = 252.8462 + (250.2473) x

----

Slope Hypothesis test:

Null and alternative hypothesis:

Ho: β₁ = 0

Ha: β₁ ≠ 0

Slope, b = 250.2473

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 198064793.78 - (747683.78)²/2987.78 = 10958971.1

Standard error, se = √(SSE/(n-2)) = √(10958971.10394/(50-2)) = 477.81994

Test statistic:

t = b/(se/√SSxx) = 250.2473 /(477.8199/√2987.78) = 28.6272

df = n-2 = 48

p-value = T.DIST.2T(ABS(28.6272), 48) = 0.0000

Conclusion:

p-value < α Reject the null hypothesis.

The model is significant.

-----

Critical value, t_c = T.INV.2T(0.05, 48) = 2.0106

95% Confidence interval for slope:

Lower limit = β₁ - tc*se/√SSxx = 250.2473 - 2.0106*477.8199/√2987.78 = 232.6711

Upper limit = β₁ + tc*se/√SSxx = 250.2473 + 2.0106*477.8199/√2987.78 = 267.8234

-----

Predicted value of y at x = 10

ŷ = 252.8462 + (250.2473) * 10 = 2755.3188

Predicted value of y at x = 20

ŷ = 252.8462 + (250.2473) * 20 = 5257.7915

Predicted value of y at x = 30

ŷ = 252.8462 + (250.2473) * 30 = 7760.2642