Question

The general manager of a chain of pharmaceutical stores reported the results of a regression analysis,...

The general manager of a chain of pharmaceutical stores reported the results of a regression analysis, designed to predict the annual sales for all the stores in the chain (Y) – measured in millions of dollars. One independent variable used to predict annual sales of stores is the size of the store (X) – measured in thousands of square feet. Data for 14 pharmaceutical stores were used to fit a linear model. The results of the simple linear regression are provided below.

        Y = 0.964 + 1.670X;  SYX =$0.9664 million; 2 – tailed p value = 0.00004 (for testing ß1);                          

                        Sb1=0.157;    X = 2.9124; SSX=Σ( Xi –X )2=37.924;   n=14 ;

1. Which of the following is a correct conclusion?

a. There is insufficient evidence to draw a correct conclusion

b. We cannot reject the null hypothesis at a level of significance 0.05 and, therefore conclude that the size of the store is not a useful linear predictor for the annual sales of the store

c. We can reject the null hypothesis at a level of significance 0.05 and, therefore conclude that there is significant evidence to show that the size of a store is not a useful linear predictor for the annual sales of the store

d. We can reject the null hypothesis at a level of significance 0.05 and, therefore conclude that there is sufficient evidence to show that the size of a store is a useful linear predictor for the annual sales of the store.  

2. Predict the average annual sales for pharmaceutical stores of 4000 sq. ft.

    a)  $7.6440 mill

    b) $67.7640 mill

    c) $668.9640 mill

    d) $6680.9640 mill

3.  Suppose the manager obtains a 95% confidence interval estimate for the population slope of annual sales of pharmaceutical stores and the square footage size of the stores of (1.328, 2.012). Which of the following will be a correct conclusion?

a. We can reject the null hypothesis at α = 0.10 and, therefore, conclude that there is sufficient evidence to show that the size of pharmaceutical stores are a useful linear predictor for the annual sales.

b. We can reject the null hypothesis at α = 0.05 and, therefore, conclude at this level of significance that there is evidence to show that the size of pharmaceutical stores are a useful linear predictor for the annual sales of the store

c. We cannot reject the null hypothesis at α = 0.05 and, therefore, conclude that there is sufficient evidence to show that the size of pharmaceutical stores are a useful linear predictor for the annual sales of the stores.

d. We can reject the null hypothesis at α = 0.01 and, therefore, conclude that there is sufficient evidence to show that the size of pharmaceutical stores are a useful linear predictor for the annual sales of the stores

4.  Interpret the estimate for the standard error of the estimate in the model:

a. For every 1.0 thousand sq. foot increase in the size of the pharmaceutical store, we expect the annual sales to increase by 0.9664 million dollars.

b. About 95% of the observed annual sales will fall within 2 x 0.9664 million dollars of the least squares line (the line of regression).

c. About 95% of the observed annual sales fall within 0.9664 million dollars of the least squares line

d. About 95% of the observed annual sales fall within the predicted values

Homework Answers

Answer #1

1) We can reject the null hypothesis at a level of significance 0.05 and, therefore conclude that there is sufficient evidence to show that the size of a store is a useful linear predictor for the annual sales of the store.

2) The predicted average annual sales for pharmaceutical stores of 4000 sq. ft.

= 0.964 + 4*1.670

= $7.6440 mill

3) We can reject the null hypothesis at α = 0.05 and, therefore, conclude at this level of significance that there is evidence to show that the size of pharmaceutical stores are a useful linear predictor for the annual sales of the store

4) About 95% of the observed annual sales will fall within 2 x 0.9664 million dollars of the least squares line (the line of regression).

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