the table shows the population of center city in various years. use the data from 1990 and 2005 to create a linear model that predicts the population of the city (y) in a given year (x). in which year was the actual population of center city most different from the value predicted by this model?

\begin{tabular}{|c|c|}
\hline
year & city population \\
\hline
1985 & 194,957 \\
\hline
1990 & 197,800 \\
\hline
1992 & 199,532 \\
\hline
2000 & 203,750 \\
\hline
2005 & 206,561 \\
\hline
2012 & 210,600 \\
\hline
\end{tabular}

- 1985
- 1992
- 2000
- 2012

Question

the table shows the population of center city in various years. use the data from 1990 and 2005 to create a linear model that predicts the population of the city (y) in a given year (x). in which year was the actual population of center city most different from the value predicted by this model?

\begin{tabular}{|c|c|}
\hline
year & city population \\
\hline
1985 & 194,957 \\
\hline
1990 & 197,800 \\
\hline
1992 & 199,532 \\
\hline
2000 & 203,750 \\
\hline
2005 & 206,561 \\
\hline
2012 & 210,600 \\
\hline
\end{tabular}

- 1985
- 1992
- 2000
- 2012

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [ "Linear Modeling", "Residual Analysis" ], "current_concepts": [ "Linear Modeling", "Residual Analysis" ] } ## Find the linear model using 1990 and 2005 data $$ \begin{aligned} &(x_1, y_1) = (1990, 197800) \ &(x_2, y_2) = (2005, 206561) \ &m = \frac{206561 - 197800}{2005 - 1990} = \frac{8761}{15} \approx 584.067 \ &y - 197800 = 584.067(x - 1990) \ &y = 584.067x - 964493.33 \end{aligned} $$ ## Calculate predicted values and residuals for the options $$ \begin{aligned} &\text{For } 1985: \ &\hat{y} = 197800 + 584.067(1985 - 1990) = 197800 - 2920.33 = 194879.67 \ &|\text{Actual} - \text{Predicted}| = |194957 - 194879.67| = 77.33 \ \ &\text{For } 1992: \ &\hat{y} = 197800 + 584.067(1992 - 1990) = 197800 + 1168.13 = 198968.13 \ &|\text{Actual} - \text{Predicted}| = |199532 - 198968.13| = 563.87 \ \ &\text{For } 2000: \ &\hat{y} = 197800 + 584.067(2000 - 1990) = 197800 + 5840.67 = 203640.67 \ &|\text{Actual} - \text{Predicted}| = |203750 - 203640.67| = 109.33 \ \ &\text{For } 2012: \ &\hat{y} = 197800 + 584.067(2012 - 1990) = 197800 + 12849.47 = 210649.47 \ &|\text{Actual} - \text{Predicted}| = |210600 - 210649.47| = 49.47 \end{aligned} $$ ## Identify the year with the largest absolute residual $$ \begin{aligned} &\max(77.33, 563.87, 109.33, 49.47) = 563.87 \quad (\text{at } x = 1992) \end{aligned} $$

1985

1992

2000

2012

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Statistics", "Residual Analysis" ] }

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QUESTION IMAGE

Question

Explanation:

Find the linear model using 1990 and 2005 data

Calculate predicted values and residuals for the options

Identify the year with the largest absolute residual

Answer: