A. Describe the relationship from the graph and assess the validity of the infer

Question

A. Describe the relationship from the graph and assess the validity of the inference for these data.

B. Give the value of R^2 and the equation of the least-squares line.

C. Based on the information given, test the hypothesis that there is no straight-line relationship between parent and offspring nest size. Give a test-statistic, its approximate p-value and your conclusion. State the hypotheses also for null and alternative. (How can we test this hypothesis on a graphing calculator without knowing all of the data points?)

D. Give a 95% confidence interval for the true population heritability (slope B) of nest size on a logarithmic scale.

Please answer entirely A,B, C, and D.

The figure below shows the relationship between parent nest size and offspring nest size for 100 parent offspring observations of barn swallows, Hirundo rustica. The line is the least-squares-regression line. Note that the graph actually plots the logarithm of nest size, because nest sizes, like many measures of volume, are otherwise strongly skewed to the right. Below is a partial Excel regression output for these data. 2.8 m 2.6 2.4 2.2 1.8 2.3 1.8 2.8 Logarithm of parent nest size (cm3)

Explanation / Answer

Part-A :From the scatter plot we observe that points are equally distributed around the regression line and not too far so it seems that the regression analysis between logarithm of dependent and independent variables is appropriate, however, the original variables has not good regression validity.

Part-B: R2=0.5226

Equation of fitted line is

Offspring nest size= 0.5764+00.6762 Parents Nest Size

Part-c:

Null hypothesis H0: 1=0 versus the alternative hypothesis H1: 10

   Test statistic t= coefficient/error= 0.6762/0.0653=10.35528331

Degree of freedom =n-2=100-2=98

p-value=2.05294E-17 using excel function =TDIST(10.35528331,98,2)

As p-value is less than 0.05, we conclude that there is significant straight line relationship.

Part-D: 95% confidence interval =slope±t(0.05,n-2)*SE(slope)

=0.6762±1.9845*0.0653

=(0.54661215     0.80578785)

In log scale 95% confidence interval =log(0.54661215)      log(0.80578785)

=(-0.604015777 -0.215934785)

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