Question

In: Statistics and Probability

2. Data was collected where a weightlifter was asked to do as many repetitions as possible...

2. Data was collected where a weightlifter was asked to do as many repetitions as possible using different amounts of weight. Below is a table that shows how much weight was on the bar, and how many repetitions the weightlifter could do: Weight 200 300 400 500 Reps 42 27 12 3

a. Calculate the correlation for this data. What does this value tell you about the relationship between these two variables?

b. Determine the least squares regression line for this data. Interpret the values for the y-intercept and the slope within this scenario.

c. Calculate r2 for this data and describe what it represents.

d. Using the regression line from part (b), calculate the predicted number of repetitions for this weight lifter if the weight is 400 pounds, and then calculate and interpret the residual for that weight using the data.

Solutions

Expert Solution

Data was collected where a weightlifter was asked to do as many repetitions as possible using different amounts of weights.

a) Correlation for the given data is calculated using R code. The value of correlation is  -0.9938587. This value tells us that there is a strong negative linear association between weight and number of repetitions. That means if the weight increases the number of repetitions will decrease.

b) The least squares regression line for this data is given by

   y = 67.2 - 0.13200 * x

where y denotes number of repetitions and x denotes amount of weight.

The y intercept is the value of the response when the predictor value is zero. The slope of the line denotes change in the value of response for unit change in the value of the predictor.

c) R squared for this data is 0.9878. This value denotes proportion of the total variation of the response explained by the linear regression line of y on x.

d) the predicted number of repetitions for this weight lifter if the weight is 400 pounds is 14.4 so approximately 14. The residual for that weight is ( y - ypred) = 12-14.4 = -2.4 . Here ypred is the predicted value of y. The residual value is small and hence the fit is good.

The R code is attached.


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