Question

Please I want someone help me to solve this question a,b,c,d,e

I’m not sure about my solution

A food processor was receiving complaints from its customers about the firmness of its canned swe potatoes. The firm's research scientist decided to gather data on their product to determine if addin pectin to the sweet potatoes might result in a product with a more desirable firmness. The scientist measured firmness on canned sweet potatoes at various pectin concentrations. Before testing, the ca were sealed and placed in a 25 C environment for 30 days. Pectin Concentration Firmness Readin 0% 1% 2% 3% 6.90, 50.20 1.30 56.48, 59.34, 62.9767.91, 70.78, 73.67 68.13, 70.85, 72.34 (a) Let x denote the pectin concentration of the sweet potatoes in a can and y denote the firmness reading following the 30 days of storage at 25 C. Plot the sample data in a scatterplot. Does firmness appear to have a linear relationship with pectin? Justify your answer (b) Using the model y-B+BX+E estimate parameters, test appropriate hypotheses for the slope c Make a residual plot to see whether the model appears to be appropriately specified. Justify you (d) Predict the firm.ness of a can of sweet potatoes treated with a 1.5% concentration of pectin (by (e) Make a new scatterplot of the sample data and include the estimated regression line. Comment parameter, and write a short paragraph stating your findings. answer weight) after 30 days of storage at 25 C. Comment on your prediction. this graph.

This is the data

# Set directory to data folder

setwd("C:data")

# getwd()

# Read data from csv file

data <- read.csv("SweetPotatoFirmness.csv",header=TRUE, sep=",")

head(data)

str(data)

# scatterplot of independent and dependent variables

plot(data$pectin,data$firmness,xlab="Pectin, %",ylab="Firmness")

par(mfrow = c(2, 2)) # Split the plotting panel into a 2 x 2 grid

model <- lm(firmness ~ pectin , data=data)

summary(model)

plot(model)

par(mfrow=c(1,1))

# Residual Plot

data$residuals <- resid(model)

data$predict <- predict(model)

plot(data$predict,data$residuals,xlab="Fitted Values",ylab="Residuals")

# Estimated regression line and scatterplot of data

plot(data$pectin,data$firmness,xlab="Pectin, %", ylab="Firmness",

ylim=c(45,75),xlim=c(0,3),main="Simple Linear Regression",

pch=19,cex=1.5)

lines(sort(data$pectin),fitted(model)[order(data$pectin)], col="blue", type="l")

Answer 1

(a) following scatter plot shows there is strong relationship between pectine and firmness

(b) the regression line is given as , y=51.456+7.411x

the p-value =0.000 of the statistic t of the slope variable x, is less than typical value of alpha=0.05, so slope is significant

following regression analysis information has been generated using ms-excel

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.913470961
R Square	0.834429196
Adjusted R Square	0.817872116
Standard Error	4.043147413
Observations	12
ANOVA
	df	SS	MS	F	Significance F
Regression	1	823.8438	823.8438	50.39712	3.3E-05
Residual	10	163.4704	16.34704
Total	11	987.3142
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	51.456	1.953026	26.34681	1.43E-10	47.10439	55.80761
X	7.411	1.043936	7.099093	3.3E-05	5.084965	9.737035

(c) residual plot, showed model is appropriate as it is scattered

(d) for x=1.5, y=51.456+7.411*1.5=62.57

(e)please find the required scatter plot