Question

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 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

Answer 1

To determine an appropriate regression model for the given dataset, let's first analyze the data:

Week Number (x): 1, 2, 3, 4, 5, 6, 7, 8
Number of Laps (y): 50, 150, 200, 300, 500, 600, 700, 800

We can observe the relationship between the week number and the number of laps. The number of laps increases as the week number increases.

To find the regression model, we can examine how the number of laps grows in relation to the week number:

1. The data indicates a linear relationship, since the increases in laps appear to be consistent.
2. Plotting the points would give us a clearer picture, but we can also infer from the values that they increase by approximately 100 laps for every additional week after the third week.

To calculate the slope (m) of the regression line, we can take the differences in y divided by the differences in x between the known points. Looking at the differences:

• From week 1 to week 2: (150 - 50) / (2 - 1) = 100
• From week 2 to week 3: (200 - 150) / (3 - 2) = 50
• From week 3 to week 4: (300 - 200) / (4 - 3) = 100
• From week 4 to week 5: (500 - 300) / (5 - 4) = 200
• From week 5 to week 6: (600 - 500) / (6 - 5) = 100
• From week 6 to week 7: (700 - 600) / (7 - 6) = 100
• From week 7 to week 8: (800 - 700) / (8 - 7) = 100

The total change in the number of laps is approximately linear, which suggests a generally consistent increase around 100.

Using regression calculation methods (like least squares), we would typically arrive at an equation of the form \( f(x) = mx + b \).

From the responses given:

• \( f(x) \sim 111.9x - 91.1 \) (seems plausible)
• \( f(x) \sim 111.9x + 50 \) (seems plausible)
• \( f(x) \sim 111.9x \) (does not include an intercept, but slope is similar)
• \( f(x) \sim -111.9x - 91.1\) (negative slope, not consistent with data trend)

Given the observed linear growth of about 100, the first two options seem reasonable due to the slope value close to 111.9. The intercept (whether it’s -91.1 or +50) could affect how closely the line fits the actual data.

Given the overall trend is positive and given slope suggests a significant positive relationship, the best choice amongst the options appears to be:

\( f(x) \sim 111.9x - 91.1 \)

This is rounded to the nearest tenth and reflects a supportive analysis of the dataset.