Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

annual high temperatures in a certain location have been tracked for se…

Question

annual high temperatures in a certain location have been tracked for several years. let x represent the year and y the high temperature. at the 0.05 significance level, does the data below show significant (linear) correlation between x and y?

question help: video post to forum

Explanation:

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

The mean of \(x\) values: \(\bar{x}=\frac{2 + 3+4+5+6+7+8+9+10}{9}=\frac{54}{9} = 6\)
The mean of \(y\) values: \(\bar{y}=\frac{20.16+23.69+22.42+16.35+19.78+23.61+30.54+27.27+29.6}{9}=\frac{213.42}{9}=23.713\) (approx)

Step2: Calculate the numerator and denominator for the correlation coefficient formula

The formula for the correlation coefficient \(r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}\)
\(\sum_{i = 1}^{9}(x_{i}-\bar{x})(y_{i}-\bar{y})=(2 - 6)(20.16-23.713)+(3 - 6)(23.69 - 23.713)+(4 - 6)(22.42-23.713)+(5 - 6)(16.35 - 23.713)+(6 - 6)(19.78-23.713)+(7 - 6)(23.61-23.713)+(8 - 6)(30.54-23.713)+(9 - 6)(27.27-23.713)+(10 - 6)(29.6-23.713)\)
\(=(- 4)(-3.553)+(-3)(-0.023)+(-2)(-1.293)+(-1)(-7.363)+0\times(-3.933)+1\times(-0.103)+2\times6.827+3\times3.557+4\times5.887\)
\(=14.212+0.069 + 2.586+7.363+0-0.103+13.654+10.671+23.548\)
\(=71.9\) (approx)

\(\sum_{i = 1}^{9}(x_{i}-\bar{x})^{2}=(2 - 6)^{2}+(3 - 6)^{2}+(4 - 6)^{2}+(5 - 6)^{2}+(6 - 6)^{2}+(7 - 6)^{2}+(8 - 6)^{2}+(9 - 6)^{2}+(10 - 6)^{2}\)
\(=16 + 9+4+1+0+1+4+9+16=60\)

\(\sum_{i = 1}^{9}(y_{i}-\bar{y})^{2}=(20.16-23.713)^{2}+(23.69 - 23.713)^{2}+(22.42-23.713)^{2}+(16.35 - 23.713)^{2}+(19.78-23.713)^{2}+(23.61-23.713)^{2}+(30.54-23.713)^{2}+(27.27-23.713)^{2}+(29.6-23.713)^{2}\)
\(=(-3.553)^{2}+(-0.023)^{2}+(-1.293)^{2}+(-7.363)^{2}+(-3.933)^{2}+(-0.103)^{2}+(6.827)^{2}+(3.557)^{2}+(5.887)^{2}\)
\(=12.624+0.001+1.672+54.214+15.468+0.011+46.598+12.652+34.667\)
\(=177.9\) (approx)

\(r=\frac{71.9}{\sqrt{60\times177.9}}=\frac{71.9}{\sqrt{10674}}\approx\frac{71.9}{103.3}\approx0.696\)

The degrees of freedom \(df=n - 2=9-2 = 7\). For a significance level of \(\alpha = 0.05\) and \(df = 7\), the critical value \(r_{c}\) (from the correlation coefficient critical value table) is \(r_{c}=0.666\)
Since \(|r|=0.696>0.666\)

Answer:

Yes, significant correlation