Name: ______
Quiz #2 (Modules 10, 11, 12 and 13)
1. The following chart reports a summary waist girths, measured in centimeters (cm), from 247 men and 260 women who exercise regularly. Waist girth is the length around the waist. Use this information to answer the questions below.
Gender / n / Min / Mean / Max / Standard Dev.Women / 260 / 57.9 / 69.8 / 101.5 / 7.6
Men / 247 / 67.1 / 84.6 / 113.2 / 8.8
a) (4 pts.) Are there any outliers among the females when looking at the pelvic diameter? Are they high or low outliers (or both)? Show all relevant work and explain how you can tell.
To find outliers, we find the upper and lower fences, where the upper fence is the mean plus 2 standard deviations and the lower fences is the mean minus 2 standard deviations. For the females, we get:
Upper Fence: 69.8 + 2(7.6) = 85 cm Lower Fence: 69.8 – 2(7.6) = 54.6 cm
Since the maximum value is larger than the upper fence, there is at least 1 high outlier (could be more).
Since the minimum is larger than the lower fence, there are NO low outliers.
b) (6 pts.) Compare the typical ranges of the men and women to see if one group generally has larger waist girth than the other. Show all relevant work and explain how you can tell.
To find the “typical range” for the men and women, we take the mean and add and subtract 1 standard deviation.
Men’s typical range: 84.6 – 8.8 to 84.6 + 8.8, after calculating we get (75.8 cm , 93.4 cm)
Women’s typical range: 69.8 +/- 7.6 would make 62.2 cm to 77.4 cm.
Comparing these two intervals, we can see that there is very little overlap between men’s and women’s typical waist girth measurements, which indicates that one group is larger than another. We can see that the men’s waist girths typically measuring larger than women’s if we compare the lower bound of the male’s typical range and the upper bound of the female’s typical range. The lower bound for men’s typical waist girth is 75.8 cm, which is just barely larger than the upper bound of the female’s typical range – 77.4 cm. These two ideas together indicate that the men have generally larger waist girth measurements than women.
Thigh Girth / SD
Thigh Girth / Mean
Weight / SD Weight / Correlation
(Thigh girth vs. Weight)
Female / 260 / 57.20 cm / 4.64 cm / 60.60 kg / 9.62 kg / 0.857
Male / 247 / 56.50 cm / 4.25 cm / 78.14 kg / 10.51 kg / 0.772
2 a) (5 pts.) Use the table above to find the least squares regression line for WOMEN, where Thigh Girth is the explanatory variable and Weight is the predicted variable. (You should get something like: ) Since I am giving you the answer, you must show work in order to receive any credit.
Slope (m) =
y-intercept (b) =
So the equation of the line should be given by y = mx + b or y = 1.8x – 41 OR we could write it:
Weight = 1.8(Thigh Girth) - 41
b) (5 pts.) Identify the slope of the regression line and explain in the context of this problem what the slope means. (Remember, you should get as your regression line)
Based on this data, the slope is about 1.8 kg/cm. This means that for an increase of 1 cm in thigh girth, the line predicts an increase of 1.8 kg in weight for women.
c) (5 pts.) If a woman had a thigh girth of 64.1 cm, what is her predicted weight? Write your answer in a sentence with appropriate units.
Weight = 1.8 ( 64.1) – 41 = 74.38 kg.
If a woman had a thigh girth of 64.1 cm, the line predicts her weight to be about 74 kg.
3. Here we have measurements from 252 men. Body Fat is the percentage of a man’s weight that comes from fat, as opposed to muscle. Weight is measured in pounds. Abdomen is a measurement (cm) around the body at the stomach.
a) (5 pts.) Using the correlation, explain which variable is a better predictor of Percent of Body Fat. Be sure to explain why (what about the correlation indicates to you that the variable you chose is a better predictor?) In your description, to get full credit, you need to make it clear to me that you know what r measures.
The abdomen is a better predictor of Percent Body Fat, since its correlation (0.81) is larger than the correlation for weight/percent body fat (0.61). A correlation closer to 1 indicates a stronger linear relationship.
b) (5 pts.) Calculate for each scatterplot. Use this to explain which variable is a better predictor and explain what it means about the relationship between the explanatory variable and response variable (just for the better predictor). In your description, to get full credit, you need to make it clear to me that you know what measures.
The value of r^2 is also larger for the variables abdomen/percent body fat (66%), than it is for weight/percent body fat (37.5%). This indicates that abdomen is a better predictor of percent body fat than weight. About 66% of the variation in percent body fat is explained by abdomen, leaving about 34% of the variation due to other variables that we are not considering. Weight, however, accounts for only 37.5% of the variation in percent body fat.
c) (5 pts.) Using the standard error, explain which variable is a better predictor of Percent of Body Fat. Be sure to explain why (what about the standard error indicates to you that the variable you chose is a better predictor?) In your description, to get full credit, you need to make it clear to me that you know what the standard error measures.
The standard error measures the approximate distance of the points to the regression line and it approximates the average error in using the regression line to estimate the percent body fat. The smaller the value of Se, the closer the points are to the line, which indicates that the line is a better estimate. The abdomen/percent body fat relationship has a smaller Se (4.86%) than the weight/percent body fat relationship (6.63%). This again shows that the abdomen is a better predictor of % body fat than weight. When predicting percent body fat using the regression line, we add and subtract a Se with our estimate to yield a range of values to approximate the percent body fat (which is why our Se values are percentages – it estimates the error in approximating percent body fat).