Question

part b

Jaaman une Regression Analysis Problem #L The following table shows the sales (in $100,000) of a certain product ma fnction of the pat 10 months and the level of advertisement (in $1,000) for the coresponding month 10 I 107 110 118 120 117 1221 120125 100 Sales 7 6 2.1 2.6 2,8 3.4 2.9 4 2 Month 10 3.4 3.5 3.7 36 Advertisement 1.5 Part A: Considering a simple linear regression of sales vs. time (months), perform ths folloming analysis manually, showing your detailed work. What is the trend equation? What is the degree for the goodness of fin b. Forecast the expected sales for periods 11, 12 and 15. /c. Determine the 95% C.I for the values obtained above. d. Determine the 95 % prediction interval for the next value of sales comresponding to the periods indicated above. Plot the regression line along with all the results obtained in parts b, c, and d. Form the confidence interval and prediction interval bands. c. Part B: Considering a simple linear regression of sales vs. advertisement, perform the following analysis manually, showing your detailed work. What is the trend equation? What % of times can advertisement reduce the sales variability? b. Is the relation between sales and advertisement statistically significant? Perform the analysis by two different methods, t-test, and F-test. What is the expected sales if the current period will spend only $3,000 on advertisement? What is a 99% C.I. for it? d. Verify your manual computations (for parts a-e) by using Minitab. e. Using Minitab, find the coefficient of correlation between sales and the advertisement. Is the true coefficient of correlation between these two variables significant? a. c. Check the assumptions of the regression model f we can be 951 conhdund that Cor cury of incrase the Inche oanela and Is beturcun

Answer 1

Part B

Sl.No	Advertisement (Xi)	Sales (Yi)	Xi2	Yi2	Xi * Yi
1	1.5	100	2.25	10000	150
2	2.1	110	4.41	12100	231
3	2.6	107	6.76	11449	278.2
4	2.8	110	7.84	12100	308
5	3.4	118	11.56	13924	401.2
6	2.9	120	8.41	14400	348
7	3.4	117	11.56	13689	397.8
8	3.5	122	12.25	14884	427
9	3.7	120	13.69	14400	444
10	3.6	125	12.96	15625	450
Sum $\sum$	29.5	1149	91.69	132571	3435.2
Average	$\bar{X}$ = 2.95	$\bar{Y}$ = 114.9

n =10,

= 4.665

-Σ+-5 Σ 2 π. i-1 i i-1

= 550.9

2 n i-1 n -1

= 45.65

Σ) (Σ: Sy ΣJi i-1 yi i=1 i-1

simple linear regression equation Y = b₀ + b₁X

b₁ = S_xy / S_xx = 45.65/4.665 = 9.78

b₀ = = 114.9 - 9.78*2.95 = 86.05

Y- b1X

Y = 86.05 + 9.78X

b) Hypothesis

H₀ : b₀ = 0

H₁ : b₀$\neq$ 0

SS_E = S_yy - b₁S_xy = 550.9 - 9.78*45.65 = 104.443

t-test

MS_E = SS_E/(n-2) = 104.443/8 = 13.055

test statistic t₀ = = 5.846

1 MSp Sz

critical value = = 2.306

to.025,&

since critical value is less than test statistic we reject null hypothesis, and conclude that there is significant linear relation associated in between sales and advertisement.

F- Test

ANOVA
	df	SS	MS	F	critical F
Regression	1	446.7144	446.7144	34.30142	5.12
Residual	8	104.1856	13.0232
Total	9	550.9

since critical value is less than test statistic we reject null hypothesis, and conclude that there is significant linear relation associated in between sales and advertisement.

c) expected sales when advertising = $3000

y = 86.05 + 9.78*3 = 115.39

Sales = $115.39 million

CI =

1 yo ta/2,n-21/ MSE (0r)

critical value = 3.355

99% CI for sales = ( 111.55, 119.23)

e) minitab results for correlation

Correlation: Advertisement (X), Sales (Y)

Correlations

Pearson correlation	0.900
P-value	0.000

correlation is significant since p-value is '0' which is less than level of significance alpha 0.05.