Question

This is the third time I've posted this question. I'm asking that a computer program is not used and that work is shown. This isn't for a grade but I'm likely to be tested on this soon. The information is from 11.7 and the questions I need to answer are from 11.13. Included in the post are the formulas I need to use. Please help me.

Ag. 11.7 canned sweet potatoes. The firm's research scientist decided to conduct an experiment to determine if adding pectin to the sweet potatoes might result in a product with a more desira ble firmness. The experiment was designed using three concentrations of pectin (by weight)- 1%, 2%, and 396-and a control with 0%. The processor packed 12 cans with sweet potatoes with a 25% (by weight) sugar solution. Three cans were randomly assigned to each of the pectin concentrations with the appropriate percentage of pectin added to the sugar syrup. The cans were sealed and placed in a 25°C environment for 30 days. At the end of the storage time the cans were opened, and a firmness determination was made for the contents of each can These appear below: A food processor was receiving complaints from its customers about the firmness of its Pectin concentration 0% 1% 2% 3% Firmness reading 46.90, 50.20,51.30 56.48, 59.34, 62.97 67.91,70.78, 73.67 68.13, 70.85, 72.34 Ag. 11.13 Refer to the data of Exercise 11.7 a. Calculate a 95% confidence interval for A). b. What is the interpretation of Ho: Bo0 for the problem in Exercise 11.7? c. Test the hypotheses Ho: A)-0 versus Ha: 0 d. Determine the p-value of the test of Ho: Bo 0. Key Formulas 1. Least-squares estimates of slope 4, Confidence interval for β1 and intercept 5, F test for A Ho: β1-0 (two-tailed ) TS: F MSRegression 6. Confidence interval for E(y+1 MS(Error) wher 1+1 In 7. Prediction interval for y+1 2. Estimate of ơ 8. Test for lack of fit in linear regression = SS(Error) where 3. Statistical test for p1 SSP Ho: Bi0 (two-tailed) T.S: Σ(n-1 ) and 11. Confidence interval for pyx SS(Error)- SSP 12. Statistical test for pyx Ho, pyr-0 (two-tailed) 9. Correlation coefficient 13. Spearman rank correlation 10. Coefficient of determination coefficient SS(Total) SS(Error) SS(Total) where x and y are ranks

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
0	46.9	2.25	245.627	23.51
0	50.2	2.25	153.079	18.56
0	51.3	2.25	127.069	16.91
1	56.48	0.25	37.119	3.05
1	59.34	0.25	10.449	1.62
1	62.97	0.25	0.158	-0.20
2	67.91	0.25	28.489	2.67
2	70.78	0.25	67.363	4.104
2	73.67	0.250	123.155	5.549
3	68.13	2.250	30.886	8.336
3	70.85	2.250	68.517	12.416
3	72.34	2.250	95.404	14.651

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	18	750.87	15.000	987.3	111.165
mean	1.50	62.57	SSxx	SSyy	SSxy

sample size ,   n =   12          
here, x̅ = Σx / n=   1.50   ,     ȳ = Σy/n =   62.57  
                  
SSxx =    Σ(x-x̅)² =    15.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =   111.2          
                  
estimated slope , ß1 = SSxy/SSxx =   111.2   /   15.000   =   7.4110

intercept,   ß0 = y̅-ß1* x̄ =   51.4560          

SSE=   (SSxx * SSyy - SS²xy)/SSxx =    163.470   
std error ,Se =    √(SSE/(n-2)) =    4.04315

a)

confidence interval for intercept
Degree of freedom ,df = n-2=   10

alpha =   0.05

t critical value=t α/2 =    2.2281   [excel function: =t.inv.2t(α/2,df) ]          
estimated std error of intercept =Se(ßo) =    Se*√(1/n+x̅²/Sxx)=       1.9530      
margin of error ,E=   t*std error =        4.3516      
                  
lower confidence limit = ß̂o - E =   51.4560   -   4.3516   =   47.1044
upper confidence limit= ß̂o + E =   51.4560   +   4.3516   =   55.8076

b)

when firmness reading is zero, then predicted Pectin concentration is also zero

c)

Ho:   ßo=   0
H1:   ßo ╪   0

alpha =   0.05
Degree of freedom ,df = n-2=   10

estimated Intercept = 51.4560

estimated std error of intercept =Se(ßo) =    Se*√(1/n+x̅²/Sxx)=       1.9530

test stat= estimated intercept/ estimated std error of intercept = 51.4560/1.9530 = 26.3468

critical value=2.228

since, test stat > critical value, reject Ho

hence, intercept is significantly different from zero

d)

p value=0.0000 [excel function: =t.dist.2t(26.3468,10)]