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
f left parenthesis x right parenthesis is similar to 111.9 x
f(x)∼111.9x+50
f left parenthesis x right parenthesis is similar to 111.9 x plus 50
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−91.1
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
f left parenthesis x right parenthesis is similar to 111.9 x
f(x)∼111.9x+50
f left parenthesis x right parenthesis is similar to 111.9 x plus 50
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−91.1
Answers
GPT-4o mini
To determine the appropriate regression model for the given data, we should first examine the trend in the provided values for the number of laps completed by the swim team over the weeks.
The dataset shows:
- 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
The number of laps appears to increase with each week, indicating a positive linear correlation.
To find a linear regression model, we can estimate the slope (m) and intercept (b) of the linear equation of the form \( f(x) = mx + b \).
1. **Calculate the Slope (m):**
From the data, we can see that for every increase of 1 in x (weeks), the number of laps increases in a somewhat linear manner. For example, between weeks 1 and 2, it increases by 100 laps.
However, to compute the exact slope:
If we look at the change from week 1 (50) to week 8 (800), the increase is:
\( 800 - 50 = 750 \) laps over \( 8 - 1 = 7 \) weeks
⇒ Average increase per week = \( 750 / 7 \approx 107.1 \)
However, the dataset suggests that as you go further in weeks, the increase rate itself has been kind of jumpy, which might suggest a slightly more complex model (but we are examining linear regression here).
2. **Estimate Intercept (b):**
To find the intercept, we can take one of the points (x, y) and solve for b:
Using Week 1: \( f(1) = 111.9(1) + b \)
\( 50 = 111.9 + b \) ⇒ \( b = 50 - 111.9 = -61.9\)
3. **Construct the regression line:**
Using slope of about 111.9 from your options and the intercept found:
\( f(x) \approx 111.9x - 61.9 \)
Given the response options provided, the closest representation for a linear model that captures this trend would be:
- **f(x) = 111.9x - 91.1** (while our value was -61.9, this is the closest option to suggest a positive linear relationship)
Thus, the appropriate regression model given the options is:
**f(x)∼111.9x−91.1**.
The dataset shows:
- 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
The number of laps appears to increase with each week, indicating a positive linear correlation.
To find a linear regression model, we can estimate the slope (m) and intercept (b) of the linear equation of the form \( f(x) = mx + b \).
1. **Calculate the Slope (m):**
From the data, we can see that for every increase of 1 in x (weeks), the number of laps increases in a somewhat linear manner. For example, between weeks 1 and 2, it increases by 100 laps.
However, to compute the exact slope:
If we look at the change from week 1 (50) to week 8 (800), the increase is:
\( 800 - 50 = 750 \) laps over \( 8 - 1 = 7 \) weeks
⇒ Average increase per week = \( 750 / 7 \approx 107.1 \)
However, the dataset suggests that as you go further in weeks, the increase rate itself has been kind of jumpy, which might suggest a slightly more complex model (but we are examining linear regression here).
2. **Estimate Intercept (b):**
To find the intercept, we can take one of the points (x, y) and solve for b:
Using Week 1: \( f(1) = 111.9(1) + b \)
\( 50 = 111.9 + b \) ⇒ \( b = 50 - 111.9 = -61.9\)
3. **Construct the regression line:**
Using slope of about 111.9 from your options and the intercept found:
\( f(x) \approx 111.9x - 61.9 \)
Given the response options provided, the closest representation for a linear model that captures this trend would be:
- **f(x) = 111.9x - 91.1** (while our value was -61.9, this is the closest option to suggest a positive linear relationship)
Thus, the appropriate regression model given the options is:
**f(x)∼111.9x−91.1**.