Mathematics, 06.05.2020 04:33 ConstanceBhoo301

The manager of Colonial Furniture has been reviewing weekly advertising expenditures. During the past 6 months, all advertisements for the store have appeared in the local newspaper. The number of ads per week has varied from one to seven. The store’s sales staff has been track- ing the number of customers who enter the store each week. The number of ads and the number of customers per week for the past 26 weeks were recorded.

a. Determine the sample regression line.
b. Interpret the coefficients.
c. Can the manager infer that the larger the number of ads, the larger the number of customers?
d. Find and interpret the coefficient of determination.
e. In your opinion, is it a worthwhile exercise to use the regression equation to predict the number of customers who will enter the store, given that Colonial intends to advertise five times in the newspaper? If so, find a 95% prediction interval. If not, explain why not.

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Answer from: izabelllreyes

Step-by-step explanation:

Hello!

The objective is to estimate the linear regression between the number of customers in the store per week and the number of ads put in the local newspaper.

The owner of the store is interested in finding out if the advertisement that appears in the newspapers affects the number of clients he has per week so the dependent and independent variables are:

Y: Number of customers in the Colonial Furniture store per week.

X: Number of adds put in the local newspaper per week.

The linear regression model is:

E(Y)= α + βX

α= intercept

β= slope

To estimate the equation you have to calculate the point estimations of the slope and the intercept

^α= a= Y[bar] -bX[bar]

^β= b= $\frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }$

∑X= 107 ∑X²= 527 ∑Y= 10005 ∑Y²= 4313803 ∑XY= 43025 n=26weeks

X[bar]= ∑X/n= 107/26= 4.12

Y[bar]= ∑Y/n= 10005/26= 384.81

b= $\frac{43025-\frac{107*10005}{26} }{527-\frac{(107)^2}{26} }= 21.36$

a= 384.81 - 21.36*4.12= 296.92

The estimated equation is ^Y= 296.92 + 21.36X

The general interpretations of the regression coefficients are:

Intercept

α: Is the value of the population mean of Y when X equals zero

a: Is the value of the estimated mean of T when X equal zero

Slope

β: Is the modification of the population mean of Y when X increases in one unit.

b: Is the modification of the estimated mean of Y when X increases in one unit.

In this example

α: Is the value of the population average number of weekly customers in the store when no adds where published in the local newspaper.

a: 296.92 is the estimated average number of weekly customers in the store when no adds appeared in the local newspaper.

β: Is the modification of the population average number of weekly customers in the store when the number of ads published in the local newspaper increases one unit.

b: The estimated average number of weekly customers in the store increases 21.36 when the number of ads published in the local newspaper increases one unit.

If the number of ads increases the number of weekly customers in the store, then the slope of the regression, β, should be positive, symbolically: β > 0. If not then the slope will be at most zero, β ≤ 0. These are the hypotheses to test to answer the question:

H₀: β ≤ 0

H₁: β > 0

I'll set the level of significance α: 0.05

For the simple linear regression analysis the statistic to use is a Students t:

$t= \frac{b-\beta }{Sb} ~t_{n-2}$

$t_{H_0}= \frac{21.36-0}{14.28} = 1.495 = 1.50$

Using the critical value approach, the rejection region is one-tailed to the right:

$t_{n-2;1-\alpha }= t_{24;0.95}= 1.711$

Thre decision rule is:

If $t_{H_0}$ ≥ 1.711, the decision is to reject the null hypothesis.

If $t_{H_0}$ < 1.711, the decision is to not reject the null hypothesis.

The value of $t_{H_0}$ =1.50 is less than the critical value, the decision is to not reject the null hypothesis.

Using a significance level of 5%, you can conclude that the slope of the regression is at most equal to zero, i.e. the population means of weekly customers in the store is at most zero when the number of ads is increased in one unit. Then the number of ads in the local newspaper does not change the weekly number of customers in the store.

The coefficient of determination is:

$R^2= \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{sumY^2-\frac{(sumY)^2}{n} }$ .

$R^2= \frac{21.36^2[527-\frac{(107)^2}{26} ]}{4313803-\frac{(10005)^2}{26} }= 0.09$

R²= 9%

Only 9% of the variability of the weekly customers of the furniture store ins explained by the number of ads that appeared in the local newspaper under the estimated regression ^Y= 296.92 + 21.36X

No.

In item c) the one-tailed hypothesis test wasn't significant, leading to think that there may not be a functional association between both variables.

Strong evidence of this supposition is the coefficient of determination.

The calculated coefficient is only 9%, meaning that no matter how many ads the shop publishes in the newspaper, it will have little to none effect on the weekly number of customers the store receives.

The independent variable chosen for this regression isn't good to explain the variability of the dependent variable. Therefore a CI using this model would be not reliable to predict the variability of weekly customers of the store.

I hope this helps!