the number of newly reported crime cases in a county in new york state is shown in the accompanying table…

the number of newly reported crime cases in a county in new york state is shown in the accompanying table, where x represents the number of years since 1995, and y represents number of new cases. write the linear regression equation that represents this set of data, rounding all coefficients to the nearest hundredth. using this equation, estimate the calendar year in which the number of new cases would reach 1282.\n\n|years since 1995 (x)|new cases (y)|\n|----|----|\n|0|1053|\n|1|1059|\n|2|1122|\n|3|1156|\n|4|1145|\n|5|1194|

the number of newly reported crime cases in a county in new york state is shown in the accompanying table, where x represents the number of years since 1995, and y represents number of new cases. write the linear regression equation that represents this set of data, rounding all coefficients to the nearest hundredth. using this equation, estimate the calendar year in which the number of new cases would reach 1282.\n\n|years since 1995 (x)|new cases (y)|\n|----|----|\n|0|1053|\n|1|1059|\n|2|1122|\n|3|1156|\n|4|1145|\n|5|1194|

Answer

Explanation:

Step1: Calculate the means of x and y

Let (n = 6). (\bar{x}=\frac{0 + 1+2+3+4+5}{6}=\frac{15}{6}=2.5) (\bar{y}=\frac{1053 + 1059+1122+1156+1145+1194}{6}=\frac{6729}{6}=1121.5)

Step2: Calculate the slope (m)

[ \begin{align*} m&=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}\ \sum_{i = 1}^{6}(x_i-\bar{x})(y_i - \bar{y})&=(0 - 2.5)(1053-1121.5)+(1 - 2.5)(1059 - 1121.5)+(2 - 2.5)(1122-1121.5)+(3 - 2.5)(1156-1121.5)+(4 - 2.5)(1145-1121.5)+(5 - 2.5)(1194-1121.5)\ &=(- 2.5)(-68.5)+(-1.5)(-62.5)+(-0.5)(0.5)+(0.5)(34.5)+(1.5)(23.5)+(2.5)(72.5)\ &=171.25 + 93.75-0.25 + 17.25+35.25+181.25\ &=498.5\ \sum_{i=1}^{6}(x_i-\bar{x})^2&=(0 - 2.5)^2+(1 - 2.5)^2+(2 - 2.5)^2+(3 - 2.5)^2+(4 - 2.5)^2+(5 - 2.5)^2\ &=6.25+2.25 + 0.25+0.25+2.25+6.25\ &=17.5\ m&=\frac{498.5}{17.5}\approx28.49 \end{align*} ]

Step3: Calculate the y - intercept (b)

We know that (y=mx + b), substituting (\bar{x}) and (\bar{y}) and (m) into the equation: (1121.5=28.49\times2.5 + b) (b=1121.5-28.49\times2.5=1121.5 - 71.225=1050.275\approx1050.28)

The linear - regression equation is (y = 28.49x+1050.28)

Step4: Estimate the value of x when (y = 1282)

Set (y = 1282) in the equation (y = 28.49x+1050.28) (1282=28.49x+1050.28) (28.49x=1282 - 1050.28=231.72) (x=\frac{231.72}{28.49}\approx8.13)

Since (x) represents the number of years since 1995, the year is (1995 + 8.13\approx2003)

Answer:

The linear - regression equation is (y = 28.49x+1050.28) and the year is approximately 2003.