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Question

Creating a Model from a Dataset Quick Check
5 of 55 of 5 Items

Question
Use the image and table to answer the question.

An illustration shows a scatterplot titled Number of Laps the Swim Team Completes Per Week. The x-axis shows weeks ranging from 1 through 10 in increments of one. The y-axis shows number of laps and ranges from 100 to 1,000 in 100 unit increments.

Week Number of Laps
1 50
2 150
3 200
4 300
5 500
6 600
7 700
8 800
Which of the following is an appropriate regression model, rounded to the nearest tenth, given the bivariate dataset?

(1 point)
Responses

f(x)∼−111.9x−91.1
f left parenthesis x right parenthesis is similar to negative 111.9 x minus 91.1

f(x)∼111.9x+50
f left parenthesis x right parenthesis is similar to 111.9 x plus 50

f(x)∼111.9x
f left parenthesis x right parenthesis is similar to 111.9 x

f(x)∼111.9x−91.1
f left parenthesis x right parenthesis is similar to 111.9 x minus 91.1
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Answer 1

To determine the appropriate regression model from the provided scatterplot data, we need to analyze the relationship between the number of laps completed by the swim team per week.

From the data in the table:

• Week 1: 50 laps
• Week 2: 150 laps
• Week 3: 200 laps
• Week 4: 300 laps
• Week 5: 500 laps
• Week 6: 600 laps
• Week 7: 700 laps
• Week 8: 800 laps

To fit a linear regression model, we generally express it in the form:

\[ f(x) = mx + b \]

where \( m \) is the slope, and \( b \) is the y-intercept.

Observing the data, there appears to be a positive trend, where an increase in weeks (x) corresponds to an increase in the number of laps (y). By inferring from the table, it seems like the relationship is roughly linear but we should approximate the slope and intercept.

From the data:

• The increase from week 1 to week 2 is 100 laps.
• The increase continues with somewhat regular increments, suggesting a slope close to 100.

Given that the maximum is 800 laps at week 8, we can estimate a linear increase.

Now looking at the answers provided:

1. \( f(x) \sim -111.9x - 91.1 \) (not appropriate due to the negative slope)
2. \( f(x) \sim 111.9x + 50 \) (looks plausible with positive slope and positive intercept)
3. \( f(x) \sim 111.9x \) (no intercept, which may not fit well if starting at 0)
4. \( f(x) \sim 111.9x - 91.1 \) (not appropriate as it has a negative intercept)

Focusing on the option that would best fit the data where a positive relationship exists with a slope around 100—although not exact—we can select:

The appropriate regression model appears to be \( f(x) \sim 111.9x + 50 \).

Final answer: \( f(x) \sim 111.9x + 50 \)