Question

Twenty-one mature flowers of a particular species were dissected, and the number of stamens and carpels present in each flower were counted.

x, Stamens	52	68	70	38	61	51	56	65	43	37	36	74	38	35	45	72	59	60	73	76	68
y, Carpels	21	32	29	19	20	30	31	29	20	24	21	30	27	24	28	22	34	26	32	34	35

(a) Is there sufficient evidence to claim a linear relationship between these two variables at α = .05?
(i) Find r. (Give your answer correct to three decimal places.)

(iii) State the appropriate conclusion.

Reject the null hypothesis, there is not significant evidence to claim a linear relationship.Reject the null hypothesis, there is significant evidence to claim a linear relationship.    Fail to reject the null hypothesis, there is significant evidence to claim a linear relationship.Fail to reject the null hypothesis, there is not significant evidence to claim a linear relationship.

(b) What is the relationship between the number of stamens and the number of carpels in this variety of flower?. (Give your answers correct to two decimal places.)

=

+  x

(c) Is the slope of the regression line significant at α = .05?(i) Find t. (Give your answer correct to two decimal places.)


(ii) Find the P-value. (Give your answer bounds exactly.)
< p <
(iii) State the appropriate conclusion.

Reject the null hypothesis, there is not evidence of a significant slope.Reject the null hypothesis, there is evidence of a significant slope.    Fail to reject the null hypothesis, there is evidence of a significant slope.Fail to reject the null hypothesis, there is not evidence of a significant slope.

(d) Find the 98% prediction interval for the number of carpels that one would expect to find in a mature flower of this variety if the number of stamens were 61. (Give your answers correct to one decimal place.)

Lower Limit
Upper Limit

Answer 1

a)

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
52	21	16.3832	36.5737	24.4785
68	32	142.8594	24.5261	59.1927
70	29	194.6689	3.8118	27.2404
38	19	325.7166	64.7642	145.2404
61	20	24.5261	49.6689	-34.9025
51	30	25.4785	8.7166	-14.9025
56	31	0.0023	15.6213	-0.1882
65	29	80.1451	3.8118	17.4785
43	20	170.2404	49.6689	91.9546
37	24	362.8118	9.2880	58.0499
36	21	401.91	36.57	121.24
74	30	322.29	8.72	53.00
38	27	325.72	0.00	0.86
35	24	443.00	9.29	64.15
45	28	122.05	0.91	-10.52
72	22	254.48	25.48	-80.52
59	34	8.72	48.34	20.53
60	26	15.62	1.10	-4.14
73	32	287.38	24.53	83.95
76	34	398.10	48.34	138.72
68	35	142.86	63.24	95.05

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	1177.00	568.00	4064.95	532.95	855.95
mean	56.05	27.05	SSxx	SSyy	SSxy

i) correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.5815

ii) Ho:   ρ = 0  
Ha:   ρ ╪ 0  
n=   21  
alpha,α =    0.05  
correlation , r=   0.5815  
t-test statistic = r*√(n-2)/√(1-r²) =        3.116
DF=n-2 =   19  
p-value =    0.0057  
Decison:   p value < α , So, Reject Ho  
critical t-value =    2.0930  

iii) Reject the null hypothesis, there is significant evidence to claim a linear relationship

b)

sample size ,   n =   21          
here, x̅ = Σx / n=   56.048   ,     ȳ = Σy/n =   27.048  
                  
SSxx =    Σ(x-x̅)² =    4064.9524          
SSxy=   Σ(x-x̅)(y-ȳ) =   856.0          
                  
estimated slope , ß1 = SSxy/SSxx =   856.0   /   4064.952   =   0.21057
                  
intercept,   ß0 = y̅-ß1* x̄ =   15.24574          
                  
so, regression line is   Ŷ =   15.25 +   0.21 *x

c)

i)

Ho:   ß1=   0          
H1:   ß1╪   0          
n=   21              
alpha =   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    4.309   /√   4064.95   =   0.0676
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.2106   /   0.0676   =   3.12

ii) 0.001< p value <0.01

iii) Reject the null hypothesis, there is evidence of a significant slope.

d)

X Value=   61              
Confidence Level=   98%              
                  
                  
Sample Size , n=   21              
Degrees of Freedom,df=n-2 =   19              
critical t Value=tα/2 =   2.539   [excel function: =t.inv.2t(α/2,df) ]          
                  
X̅ =    56.05              
Σ(x-x̅)² =Sxx   4064.95              
Standard Error of the Estimate,Se=   4.3086              
                  
Predicted Y at X=   61   is          
Ŷ =   15.2457   +   0.2106   *61=   28.090
                  

                  
For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   4.423              
margin of error,E=t*std error=t*S(ŷ)=    2.539   *   4.423   =   11.2313
                  
Prediction Interval Lower Limit=Ŷ -E =   28.090   -   11.231   =   16.9
Prediction Interval Upper Limit=Ŷ +E =   28.090   +   11.231   =   39.3