In: Statistics and Probability
Run a regression analysis on the following bivariate set of data with y as the response variable.
x | y |
---|---|
81 | 81.3 |
92.6 | 90.8 |
80.1 | 94.9 |
77.8 | 53.4 |
89.4 | 102.9 |
70.3 | 38.2 |
90.2 | 98 |
81.4 | 94.6 |
94.9 | 122.4 |
77.2 | 42.1 |
70.6 | 47.8 |
71 | 50.6 |
Find the correlation coefficient and report it accurate to three
decimal places.
r =
What proportion of the variation in y can be explained by
the variation in the values of x? Report answer as a
percentage accurate to one decimal place. (If the answer is
0.84471, then it would be 84.5%...you would enter 84.5 without the
percent symbol.)
r² = %
Based on the data, calculate the regression line (each value to
three decimal places)
y = x +
Predict what value (on average) for the response variable will be
obtained from a value of 95.7 as the explanatory variable. Use a
significance level of α=0.05α=0.05 to assess the strength of the
linear correlation.
What is the predicted response value? (Report answer accurate to
one decimal place.)
y =
correlation coefficient r= | Sxy/(√Sxx*Syy) = | 0.881 |
r² = 0.881^2*100 = 77.7 %
3)
sample size n= | 12 | |
y̅ = | 76.4167 | |
x̅ = | 81.3750 | |
Sxx= | Σ(Xi-X̅)2= | 829.583 |
Sxy = | Σ(Xi-X̅)(Yi-Y̅)= | 2386.435 |
slope= β̂1 = | Sxy/Sxx= | 2.8767 |
intercept= β̂0 = | y̅-β1x̅= | -157.6723 |
Least square line equation: ŷ =2.877*x +(-157.672) |
4)
predicted val=-157.672+95.7*2.877= | 117.7 (please try 117.6 if this comes wrong) |