3Fitting Model
From the scatter diagram, it appears that there exists a linear association between the cubic feet moved and the labor hours required. Using JMPIN to conduct a simple regression fit of Hours by Feet (refer to Appendix A for details), we’ve obtained a linear model for the data as:
Hours = 2.36966 + 0.0500803 Feet
with the correlation coefficient r = 0.942998 and r^{2} = 0.889246. As r^{2} = 0.889246, it indicates that the fitted model can explain a large proportion of the total variation: approximately 88.9% of the variation in the labor hours is explained by the model.
Now, let’s conduct a hypothesis testing for zero slope (β_{1} = 0) to verify that a straightline model in cubic feet is better than a model that does not include cubic feet at all. For the full hypothesis testing procedure, refer to Appendix D. Since we reject the null hypothesis of zero slope for the straight line, the choice of the linear model is convincingly reasonable. More over, the Analysis of Variance section in the JMPIN output (refer to Appendix B) further shows that the null hypothesis of zero slope should be rejected as the pvalue < 0.001 < 0.05 (the significance level).
By plotting the residuals of Hours and graphing the histogram for the residuals (refer to Appendix B), we can see no violation to model assumptions.
4Application of the Fitted Model
By obtaining the relationship between the labor hours required and cubic feet to be moved with the linear model:
Hours = 2.36966 + 0.0500803 Feet
the company now can make more reliable predictions on the labor hours easily and accurately. For example, to estimate the labor hours needed to move X_{0} = 800 cubic feet, all the company needs to do is to do a simple calculation by substitute 800 into the model to get an estimated point as:
Predicted Hours = 2.36966 + 0.0500803 * 800 ≈ 37.69
Then by doing the following calculation, the company can get a 95% prediction interval (PI):
Predicted Hours + b1 * (X_{0} – Mean Feet) ± t_{n2, 1α/2} * S_{YX} * sqrt(1 + 1/n + (X_{0} – Meat Feet)^{2}/((n – 1) * S_{X}^{2}))
≈ 37.69 + 0.0500803 * (800  625.555556) ± 2.030 * 5.031427 * sqrt(1 + 1/36 + (800 – 625.555556)^{2}/((36 – 1) * 78726.654))
≈ (36.02, 56.84)
Therefore, if the labor hour required to move 800 cubic feet is within 36 hours and 57 hours, then it is normal. If the labor hour is below 36 hours, then the move is more efficient than expected. If the labor hour is above 57 hours, then the company needs to investigate to verify what has been wrong with the move. There might have some other factors that affect the move as it has an extraordinary result.
Appendix A: Bivariate Fit of Hours By Feet
Figure 2 Mean and Regression Fit of Hours By Feet
Table 1: Fit Mean
Mean

28.95833

Std Dev [RMSE]

14.90104

Std Error

2.483507

SSE

7771.438

Table 2: Summary of Fit
RSquare

0.889246

RSquare Adj

0.885988

Root Mean Square Error

5.031427

Mean of Response

28.95833

Observations (or Sum Wgts)

36

Table 3: Analysis of Variance (ANOVA Table)
Source

DF

Sum of Squares

Mean Square

F Ratio

Model

1

6910.7189

6910.72

272.9864

Error

34

860.7186

25.32

Prob > F

C. Total

35

7771.4375


<.0001

Table 4: Parameter Estimates
Term


Estimate

Std Error

t Ratio

Prob>t

Intercept


2.36966

2.073261

1.14

0.2610

Feet


0.0500803

0.003031

16.52

<.0001

Table 5: Data Summary
N Rows

Sum(Hours)

Sum(Feet)

Mean(Hours)

Mean(Feet)

Std Dev(Hours)

Std Dev(Feet)

36

1042.5

22520

28.9583333

625.555556

14.9010426

280.582704

Figure 3 Residual Plot
Figure 4 Distributions Residuals Hours
Appendix C: Hypothesis Testing for Zero Slope: β_{1} = 0
Testing Procedure:

Assumptions: The variable β_{1} has a normal distribution, from which a random sample has been selected.

Hypotheses: H_{0}: β_{1} = 0
H_{A}: β_{1} ≠ 0

Use 95% significant level: α = 0.05

Test Statistic: T = (b_{1} – β_{1}) / S_{b1}, where

S_{b1} = S_{YX} / (S_{X} * sqrt(n – 1))

S^{2}_{YX} = (1 / (n – 2)) * ∑(Y_{i} – Ŷ_{i})^{2}

S^{2}_{X} = (1 / (n – 2)) * ∑(X_{i} – X_{i})^{2}

Sample size n = 36

Rejection regions: reject H_{0} if  T  ≥ t_{n2, 1α/2} = t_{34, 0.975} ≈ 2.030; do not reject H_{0} otherwise.

Calculation of T:
From the JMPIN output in Appendix B, we get
b_{1} = 0.0500803
S^{2}_{YX} = 25.32 => S_{YX} ≈ 5.032
S^{2}_{X} = 78726.654 => S_{X} ≈ 280.583
S_{b1} = 5.032 / (280.583 * sqrt(36 – 1)) ≈ 0.00303
T = (0.0500803 – 0) / 0.00303 ≈ 16.528

Since T ≈ 16.528 > t_{34, 0.975} ≈ 2.030, we reject H_{0} at significance level 0.05 and conclude that there is evidence that the cubic feet to be moved indeed provides significant information for predicting the labor hours needed, that is, a straightline model in cubic feet is better than a model that does not include cubic feet at all.