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Chapter 3: Q 22. (page 174)

A student wonders if tall women tend to date taller men than do short

women. She measures herself, her dormitory roommate, and the women in the adjoining dorm rooms. Then she measures the next man each woman dates. Here are the data (heights in inches):

a. Make a scatterplot of these data. Describe what you see.

b. Find the correlation r step by step, using the formula on page 166. Explain how your value for r matches your description in part (a).

Short Answer

Expert verified

Part (a) It appears that there is a positive correlation between women and men.

Part (b) The value of r is 0.5653

Step by step solution

01

Part (a) Step 1: Given information

02

Part (a) Step 2: Explanation

There are the following steps into excel.

Put the data into excel sheet.

Click on

Insert.

Select the data set.

Click on

Scatter

It appears that there is a positive correlation between women and men.

03

Part (b) Step 1: Calculation

The sample size (n)is 6

The linear correlation coefficient (r)can be calculated as:

r=n(Σxy)−(Σx)(Σy)nΣx2(Σx)2(nΣy2(Σy)2)=6(27339)−(396)(414)6(26158)-(396)26(28598)−4142=0.5653

Thus, the value of ris 0.5653

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Most popular questions from this chapter

Each year, students in an elementary school take a standardized math test at the end of the school year. For a class of fourth-graders, the average score was 55.1 with a standard deviation of 12.3. In the third grade, these same students had an average score of 61.7 with a standard deviation of 14.0. The correlation between the two sets of scores is r = 0.95. Calculate the equation of the least-squares regression line for predicting a fourth-grade score from a third-grade score.

a. y^=3.58+0.835x

b. y^=15.69+0.835x

c. y^=2.19+1.08x

d. y^=−11.54+1.08x

What’s my grade? In Professor Friedman’s economics course, the correlation

between the students’ total scores prior to the final examination and their final exam scores is r = 0.6. The pre-exam totals for all students in the course have a mean of 280 and a standard deviation of 30. The final exam scores have a mean of 75 and a standard deviation of 8. Professor Friedman has lost Julie’s final exam but knows that her total before the exam was 300. He decides to predict her final exam score from her pre-exam total.

a. Find the equation for the least-squares regression line Professor Friedman should use to make this prediction.

b. Use the least-squares regression line to predict Julie’s final exam score.

c. Explain the meaning of the phrase “least squares” in the context of this question.

d. Julie doesn’t think this method accurately predicts how well she did on the final exam. Determine r2. Use this result to argue that her actual score could have been much higher (or much lower) than the predicted value.

Husbands and wives The mean height of American women in their early twenties is 64.5inches and the standard deviation is 2.5inches. The mean height of men the same age is 68.5inches, with standard deviation2.7inches. The correlation between the heights of husbands and wives is about r=0.5

(a) Find r2and interpret this value in context.

(b) For these data, s=1.2. Explain what this value mean

Teenagers and corn yield Identify the explanatory variable and the response variable for the following relationships, if possible. Explain your reasoning.

a. The height and arm span of a sample of 50 teenagers.

b. The yield of corn in bushels per acre and the amount of rain in the growing season.

Scientists examined the activity level of 7 fish at different temperatures. Fish activity was rated on a scale of 0 (no activity) to 100 (maximal activity). The temperature was measured in degrees Celsius. A computer regression printout and a residual plot are provided. Notice that the horizontal axis on the residual plot is labeled “Fitted value,” which means the same thing as “predicted value.”

Which of the following gives a correct interpretation of s in this setting?

a. For every 1°C increase in temperature, fish activity is predicted to increase by 4.785 units.

b. The typical distance of the temperature readings from their mean is about 4.785°C.

c. The typical distance of the activity level ratings from the least-squares line is about 4.785 units.

d. The typical distance of the activity level readings from their mean is about 4.785 units.

e. At a temperature of 0°C, this model predicts an activity level of 4.785 units.
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