Question

Please use R program to solve with explanation.

Enterprise Industries produces Fresh, a liquid landry detergent. The company wishes to study the relationship between price and demand for the large size bottle of Fresh in its sales regions. The company has gathered data (see Table) concerning demand for Fresh in 30 sales regions of equal sales potential.

i = 1,2,.,30 ythe demand for the large size bottle of Fresh (in hundreds of thousand) in sales region i Xithe price (in dollars) of Fresh as offered by x2 the average industry price (in dollars) of competitors' similar detergents in sales region. x4 Xi2-Xi1 the "price difference" in sales region i Sales Price For Average Industry Price Difference Demand ,Xi2 xi4X12X1 Region fer Fresh Fresh Price. У. 7.38 (dellars) -0.05 3.85 3.80 3.75 4.00 0.25 8.51 3 3.70 4.30 9.52 0.60 3.70 3.70 7.50 4 0,00 3.60 3.85 0.25 9.33 Build a simple linear regression relating demand (y) to price difference (x) 3.60 3.80 0.20 8.28 3.75 3.60 0.15 8.75 7 3.80 3.85 0.05 7.87 Find and report the least squares point estimates bo and b,. Use them to write the least squares prediction equation Find and report the total variation, explained variation, SSE, and r2. Interpret the value of a. 7.10 3.80 3.65 -0.15 3.85 4.00 0.15 10 8.00 7.89 b. 1 3.90 4.10 0.20 3.90 0.10 8.15 12 4.00 13 3.70 4,10 0.40 9.10 c. Find and report the t statistic for testing Ho: B 0and the associated prob-value. Use the prob-value to test the significance of the regression relationship d. Find and report F(model) and the associated 14 3.75 4.20 0.45 8.86 0.35 15 3.75 4.10 8.90 3.80 4,10 8.87 16 0.30 17 3.70 4,20 0.50 9.26 3.80 4.30 18 0.50 9,00 prob-value. Use the prob-value to test the significance of the regression relationship Find and report the t statistic for testing Ho: Bo 0 a-d the associat-d prob-value Use the prob-value to test Ho: Po = 0 . f. Find and report s2 and s g. Calculate a 95% confidence interval for B. Interpret this interval h. Find and report the 95% prediction interval for an individual demand when the price difference is S.60. Interpret this interval 8.70 19 3.70 4,10 0.40 3.75 3.80 7.95 20 -0.05 e. 7.65 3.80 3.75 -0.05 21 3.65 22 3.75 -0.10 7.27 23 3,70 3,90 0.20 8.00 3.65 3.65 8.50 24 0.10 3.60 8.75 25 4.10 0.50 3.65 4.25 60 26 9.21 3.70 3.65 27 -0.05 8.27 28 3.75 3.75 0.00 7.67 3.85 0.05 7.93 29 3.80 3.7 30 4,25 0.65 9.26

Answer 1

I used R software to solve this problem.

R code:

> x=c(-0.05,0.25,0.6,0,0.25,0.2,0.15,0.05,-0.15,0.15,0.2,0.1,0.4,0.45,0.35,0.3,0.5,0.5,0.4,-0.05,-0.05,-0.1,0.2,0.1,0.5,0.6,-0.05,0,0.05,0.65)
> y=c(7.38,8.51,9.52,7.5,9.33,8.28,8.75,7.87,7.1,8,7.89,8.15,9.1,8.86,8.9,8.87,9.26,9,8.7,7.95,7.65,7.27,8,8.5,8.75,9.21,8.27,7.67,7.93,9.26)
> fit=lm(y~x)
> summary(fit)

Call:
lm(formula = y ~ x)

Residuals:
Min 1Q Median 3Q Max
-0.44792 -0.23301 -0.06905 0.16779 0.86284

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 7.8209 0.0807 96.91 < 2e-16 ***
x 2.5849 0.2558 10.10 7.66e-11 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.3212 on 28 degrees of freedom
Multiple R-squared: 0.7847, Adjusted R-squared: 0.7771
F-statistic: 102.1 on 1 and 28 DF, p-value: 7.664e-11

a) Least square regression equation:

Y = 7.8209 + 2.5849 X

b)

> anova(fit)
Analysis of Variance Table

Response: y
Df Sum Sq Mean Sq F value Pr(>F)
x 1 10.5347 10.5347 102.08 7.664e-11 ***
Residuals 28 2.8896 0.1032
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Total variation = 10.5347 + 2.8896 = 13.4243

Explained variation = 10.5347

SSE = 2.8896

r² = 0.7847

It means that price difference explains 78.47 % of variation in demand.

c)

t statistic = 10.10 p value = 7.66e-11

Since p value is less than 0.05 we reject H₀.

d)

F statistic = 102.08 p value = 7.664e-11

Since p value is less than 0.05 we reject H₀ and conclude that relationship is statistically significant.

e)

t statistic = 96.91 p value < 2e-16

Hence coefficient of intercept (beta) is statistically significant.

f)

s = residual's standard error = 0.3212

s² = Residual variance = 0.1032