the following table shows students test scores on the first two tests in an introductory chemistry…

the following table shows students test scores on the first two tests in an introductory chemistry class.\nchemistry test scores\nfirst test, x: 58 62 47 91 51 68 61 95 92 42 53 61\nsecond test, y: 53 58 52 83 50 56 52 83 81 42 51 56\ncopy data\nstep 1 of 2: find an equation of the least - squares regression line. round your answer to three decimal places, if necessary.\nanswer\ny^ = + x

the following table shows students test scores on the first two tests in an introductory chemistry class.\nchemistry test scores\nfirst test, x: 58 62 47 91 51 68 61 95 92 42 53 61\nsecond test, y: 53 58 52 83 50 56 52 83 81 42 51 56\ncopy data\nstep 1 of 2: find an equation of the least - squares regression line. round your answer to three decimal places, if necessary.\nanswer\ny^ = + x

Answer

Explanation:

Step1: Calculate the means of (x) and (y)

Let (n = 12). (\bar{x}=\frac{58 + 62+47+91+51+68+61+95+92+42+53+61}{12}=\frac{781}{12}\approx65.083) (\bar{y}=\frac{53 + 58+52+83+50+56+52+83+81+42+51+56}{12}=\frac{687}{12}= 57.25)

Step2: Calculate the numerator and denominator for the slope (b_1)

The formula for (b_1=\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}^{n}(x_i-\bar{x})(y_i - \bar{y})=(58 - 65.083)(53-57.25)+(62 - 65.083)(58 - 57.25)+\cdots+(61-65.083)(56 - 57.25)) (=( - 7.083)\times(-4.25)+(-3.083)\times0.75+\cdots+(-4.083)\times(-1.25)) (=30.10275-2.31225+\cdots + 5.10375) (\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=477.25) (\sum_{i=1}^{n}(x_i-\bar{x})^2=(58 - 65.083)^2+(62 - 65.083)^2+\cdots+(61 - 65.083)^2) (=(-7.083)^2+(-3.083)^2+\cdots+(-4.083)^2) (\sum_{i=1}^{n}(x_i-\bar{x})^2 = 1976.917) (b_1=\frac{477.25}{1976.917}\approx0.241)

Step3: Calculate the y - intercept (b_0)

The formula for (b_0=\bar{y}-b_1\bar{x}) (b_0 = 57.25-0.241\times65.083) (b_0=57.25 - 15.685) (b_0 = 41.565)

Answer:

(\hat{y}=41.565 + 0.241x)