The equation the least squares regression line as: y = 0.34x + 22.32 Is the corr
ID: 3181747 • Letter: T
Question
The equation the least squares regression line as: y = 0.34x + 22.32 Is the correlation coefficient significant at the 5% level of significance? A. Yes, it is significant. B. No, it is not significant. C. The results are inconclusive. D. More data points are needed to test significance. Ariel's parents asked you to predict Ariel's height when she becomes 6 years old (72 months), based on the line of regression. How would you respond? A. 47 inches B. 48 inches C. 46 to 49 inches D. Ariel's height cannot be predicted from the line of regression. The interpretation of the slope is: A. On average, for every additional will increase by about 0.34 inches. B. At birth, Ariel's height was about 22.34 inches, for every additional inch increase in height, Ariel's age will increase by about 0.34 months. D. On average, Ariel's height increases 0.34 inches per additional month of age. The following table shows a sample of scores on a Math 10 quiz from three different classes. We are interested in knowing if the mean quiz scores are the same for each class. Assuming the assumptions for doing one-way ANOVA are met, what is the distribution for this test? A. F_3, 5 B. F_2, 4 C. F_2, 12 D. F_3, 15 If a test of a hypothesis has a Type I Error probability of 0.01, this means that A. if the null hypothesis is true, we reject it 1% of the time. B. if the null hypothesis is true, we don't reject it 1% of the time. C if the null hypothesis is false, we reject it 1% of the time. D. if the null hypothesis is false, we don't reject it 1% of the time.Explanation / Answer
Step 1
Null Hypothesis Ho : µ1 =µ2 =µ3
Alternative Hypothesis : µ1 µ2 µ3
Step 2
Degrees of freedom between = k - 1 = 3 - 1 = 2
Degrees of freedom Within = n - k = 15 - 3 = 12
Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 3.885
Step 3
Grand Mean = G / N = 16.8+18+7.8 / 3 = 14.2
SST = ( Xi - GrandMean)^2 = (16-14.2)^2 + (18-14.2)^2 + (16-14.2)^2 + ……..& so on = 516.04
SS Within = (Xi - Mean of Xi ) ^2 =,(16-16.8)^2 + (18-16.8)^2 + (16-16.8)^2 + ……..& so on = 285.84
SS Between = SST - SS Within = 516.04 - 285.84 = 230.2
Step 4
Mean Square Between = SS Between / df Between = 230.2/2 = 115.1
Mean Square Within = SS Within / df Within = 285.84/12 = 23.82
Step 5
F Cal = MS Between / Ms Within = 115.1/23.82 = 4.832
We got |F cal| = 4.832 & |F Crit| =3.885
MAKE DECISION
Hence Value of |F cal| > |F Crit|and Here We Reject Ho
[ANSWERS]
Q16.
F 2,12
ONE WAY ANOVA Treatments Mean = X /n class1 16 18 16 6 9 16.8 class2 16 19 0 18 18 18 class3 6 4 3 0 13 7.8Related Questions
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