A straight line is fitted to some data using least squares. Summary statistics a

Question

A straight line is fitted to some data using least squares. Summary statistics are below n« 10, = 5, y =12, ss,,-140, ss,,-123, ss,,-155 The least squares intercept and slope are 7.60 and 0.88, respectively, and the ANOVA table is below. Source DF Ss Regression 1 108.06 108.06 ub 6.94 5.87 otal 9 155 What is the estimated mean value for y when x-9? [2 pt(s)) Submit Anowe Tries 0/3 If the fitted value for y is 18, what is the corresponding value for x? 12 pt(s)1 Submit Anwer Tries 0/3 What proportion of the variation in y is explained by the regression in x? (2 pt(s) Submt Answe Tries 0/3 What is the standard error of the slope estimate? 12 pt(s) Submt Answer Tries 0/3 To test whether there is any association between X and Y, what are the null and alternative hypotheses? [1 pt(s)] Submit Answer Tries 0/1 Test whether there is a significant linear association between X Y. First, compute the value of the test statistic (based on 1). [3 pt(5) Submit Answer Tries 0/3

Explanation / Answer

Solution:

From the given information n = 10, x = 5, y-bar = 12, SSxx = 140, SSxy = 123, SSyy = 155
Given that intercept b0 = 7.60 and slope b1 = 0.88
Then the required regression equation is,
y-hat = 7.60 + 0.88x

The estimated mean value of Y when X = 9 is,
y-hat = 7.60 + 0.88 (9)
= 15.52
The corresponding value of X if, the fitted value for Y is 18 is,
y-hat = 7.60 + 0.88x
18 = 7.60 + 0.88x
10.40 = 0.88 x
x = 11.8181

The portion of the variation in Y is explained by the regression in X is calculated in X is calculated as follows:
From the given information
The sum of square of regression(SSR) is 108.06
The sum of squares of total (SST) is 155
Then the coefficient of determination is calculated as follows:
R^2 = SSR/SST = 108.06/155 = 0.6972
-------------------------------------------------
The standard error of the slope estimate is calculated as follows:
SE(y-hat) = sqrt(1/n + (x-x)^2/SSxx)
= sqrt(1/10 + (9-5)^2/140)
= 0.4629
----------------------------------------------------
Setup the null and alternative hypothesis:

H0: 1 = 0, HA: 1 0
-----------------------------------------------------
The test statistic used to test the hypothesis is,
t = b1- 1/SE = 0.88 - 0/0.4629 = 1.9011
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Degrees of freedom Df = n-2
= 10- 2 = 8
P-value: tdist(1.9011, 8, 2) = 0.0938
Therfore, 95% confidence interval is calculated as follows:
CI = y-hat ± t(/2, n-p) [SE(y-hat)]
= 15.52 ± 2.75(0.4629)
= (14.2470, 16.7929)

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