Question

kathleen is a math teacher at a large school. she wonders if her test problems are too wordy. kathleen is curious whether the wordiness is affecting student performance.
for the last several tests, kathleen computes the average number of words in each question, x, as well as the average percentage scores on the tests, y.
round your answers to the nearest thousandth.
y = x +

Explanation:

Step1: Calculate the slope ($m$)

Use the formula \(m=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^{2})-(\sum x)^{2}}\)
First, calculate \(\sum x = 17.73 + 19.28+21.96 + 38.52+41.75+45.86+48.94=233.04\)
\(\sum y=80.80 + 69.47+86.57 + 70.46+66.08+65.31+64.77 = 503.46\)
\(\sum xy=(17.73\times80.80)+(19.28\times69.47)+(21.96\times86.57)+(38.52\times70.46)+(41.75\times66.08)+(45.86\times65.31)+(48.94\times64.77)\)
\(=1432.584+1339.382+1891.073+2714.019+2750.84+2995.017+3177.574 = 16300.49\)
\(\sum x^{2}=17.73^{2}+19.28^{2}+21.96^{2}+38.52^{2}+41.75^{2}+45.86^{2}+48.94^{2}\)
\(=314.353+371.718+482.242+1483.790+1743.063+2103.139+2395.124 = 9293.429\)
\(n = 7\)
\(m=\frac{7\times16300.49-233.04\times503.46}{7\times9293.429-(233.04)^{2}}\)
\(=\frac{114103.43-117200.16}{65054.003 - 54207.682}\)
\(=\frac{-3096.73}{10846.321}\approx - 0.285\)

Step2: Calculate the y - intercept ($b$)

Use the formula \(b=\frac{\sum y - m\sum x}{n}\)
\(b=\frac{503.46-(-0.285)\times233.04}{7}\)
\(=\frac{503.46 + 66.416}{7}\)
\(=\frac{569.876}{7}\approx81.411\)

Answer:

\(y=-0.285x + 81.411\)