The manager of a tire store in Minneapolis has been concerned with the high cost
ID: 3019939 • Letter: T
Question
The manager of a tire store in Minneapolis has been concerned with the high cost of inventory. The current policy is to stock all the snow tires that are predicted to sell over the entire winter at the beginning of the season (end of October). The manager can reduce inventory costs by having suppliers deliver snow tires regularly from October to February. However, he needs to be able to predict weekly sales to avoid stockouts that will ultimately lose sales. To help develop a forecasting model, he records the number of snow tires sold weekly during the last winter and the amount of snowfall (in inches) in each week. a. Develop a regression model and use a software package to produce the statistics. b. Perform a complete diagnostic analysis to determine whether the required conditions are satisfied. c. If one or more conditions are unsatisfied, attempt to remedy the problem. d. Use whatever procedures you wish to assess how well the new model fits the data. e. Interpret and test each of the coefficients.
Explanation / Answer
Let, y = the number of snow tires sold weekly
x1 = snowball per week
x2 = time
a. Develop a regression model and use a software package to produce the statistics.
This analysis we can done by using MINITAB.
First plug the data in minitab file.
steps are,
START --> STAT --> Regression --> Regression --> Response : y --> Predictors --> x1,x2 --> Results : select second option --> ok --> ok
Output is as follows :
Regression Analysis: y versus x1, x2
The regression equation is
y = 59 + 216 x1 + 3.4 x2
Predictor Coef SE Coef T P
Constant 58.5 245.2 0.24 0.814
x1 216.39 39.66 5.46 0.000
x2 3.39 21.20 0.16 0.875
S = 510.420 R-Sq = 67.3% R-Sq(adj) = 63.4%
Analysis of Variance
Source DF SS MS F P
Regression 2 9095488 4547744 17.46 0.000
Residual Error 17 4428990 260529
Total 19 13524478
Interpretation :
Here hypothesis for the test is,
H0 : Bj = 0 Vs H1 : Bj 0
where Bj is the population slope.
We can see that the coefficients of x1 and x2 are positive they are known as slopes.
From these two slopes we can say that there is positive correlation between y and x1 x2.
S is known both as the standard error of the regression and as the standard error of the estimate.
Standard error of the regression is 510.420.
We see that P-value for x1 is 0.000.
Assume that alpha = level of significance = 0.05
A low p-value (< 0.05) indicates that you can reject the null hypothesis. In other words, a predictor that has a low p-value is likely to be a meaningful addition to your model because changes in the predictor's value are related to changes in the response variable.
Conclusion : x1 ( snowball per week) is statistically significant.
Similarly P-value for x2 is 0.875 which is larger than 0.05 (alpha).
Fail to reject H0.
A larger (insignificant) p-value suggests that changes in the predictor are not associated with changes in the response.
x2 (time) is not statistically significant.
R-sq is 67.3%.
It indicates that the proportion of the variation in y which is explained by variation in x1 and x2.
In general, the higher the R-squared, the better the model fits your data
But here R-sq is 67.3% which is less hence not better the model fits your data.
ANOVA table :
Interpretation: In this case ANOVA tests the hypothesis that beta=0. In fact F is nothing but T-square. A low p-value suggest that beta plays a significant role in the model, this is just reassurance of the t-test.
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