150 is 15% of what number
2 answers:
You might be interested in
Answer:
- f(x) = 34.59 +3.74x
- 0.96
- very strong
- 8.93
Step-by-step explanation:
1. The selection of answers offers a choice of positive or negative slope, a slope of about 4 or about 35, and y-intercepts from the same set of values.
If you consider a couple of points (at the ends of the domain is usually a good choice), you see the slope is about 4 and the y-intercept is around 30.
For example, the points (5, 50) and (15, 91) have a slope of (91-50)/(15-5) = 4.1 so will give rise to the equation ...
y -50 = 4.1(x -5)
Filling in x=0 gives y = 50 -20.5 = 29.5
So, the nearest of the offered choices is ...
f(x) = 34.59 +3.74x
Please note that no calculator is involved. We simply estimated the parameters of the regression line based on a couple of the data points. This estimate is sufficient to allow us to find the correct answer choice.
__
2. If we look at a few other data points, we see that they have a generally upward trend that matches the above equation pretty well. That suggests a very high value of correlation, so the appropriate choices would be 0.93 or 0.96. Of course, the correlation is positive, matching the sign of the slope of the regression line. (Since the data points do not fall exactly on a line, the correlation is not 1.00. My assessment is they fall close enough to the line to rate a correlation of .96 instead of .93.)
At this point, the output of your calculator can confirm that the correlation coefficient is closer to 0.96. Mine says r = 0.9624.
r = 0.96
__
3. We have already determined by examining the data points that the correlation is "very strong." The correlation coefficient of 0.96 confirms this.
__
4. Using the equation of part 1, we can find the number of study hours:
68 = 34.59 +3.74x
(68 -34.59)/3.74 = x ≈ 8.933
We expect a score of 68 is associated with about 8.93 study hours.
_____
Please note that we have used very little calculation to determine the answers to this question. The only requirement is a general understanding of what linear regression and correlation coefficients mean.
