In: Statistics and Probability
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 | 30 | 29 | 19 | 20 | 30 | 29 | 31 | 18 | 24 | 23 | 28 | 27 | 26 | 28 | 20 | 36 | 28 | 34 | 36 | 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
a) i) r = 0.511
iii) option B is right.
Reject the null hypothesis, there is significant evidence to claim a linear relationship.
b) Y =16.02 + 0.20 X
c) t = 2.59 ( from the regression output )
ii) 0.01<p value <0.02
d) Reject the null hypothesis, there is evidence of a significant slope.
Since the P value is less than the significance level 0.05