The data provided is from houses for sale in Salinas. Prices are in the $1,000.South SalinasPrice6055707021025705665580707539608950855429700489652629499639562805655615720Square Feet155511222336345820881564170816021536160528452661895195197120431848866213014082650155621052003Bedrooms324533ww33ww344WNA234432435343North SalinasPrice585480489565499321642565405408605808555509608689568705515544545870699409Square Feet247313451025125512057201569188012099551855258414451215129913441398187713081701142726021823936Bedrooms43234) Use the regression equation to predict the price of a 2000 square foot house in North Salinas.5) Test for a linear correlation between price and bedrooms in South Salinas.6) Test for a linear correlation between price and bedrooms in North Salinas.N/mM|N|3|M|N|N|34|mm|m22334343542

4) The objective is to use the regression equation to predict the price of a 2000 square foot house in North Salins. The regression equation can be obtained using Excel. Enter the data of North Salins in Excel sheet. Go to DATA > Data analysis Select Regression, click ok Enter Y range i.e., price Enter X range i.e., square feet Select the output range Click ok Excel output is given below: Thus, the estimated regression equation is: Price=249.5762+0.2085 Square feet Substitute 2000 for square feet in the equation to obtain the predicted Price of a 2000 square foot house in North Salins: Price=249.5762+0.2085×2000=666.5762≈666 5) State the hypotheses as follows: Null Hypothesis H0:ρ=0 i.e., there is no linear correlation between price and bedrooms in south Salins. Alternative Hypothesis H1:ρ≠0 i.e., there is a linear correlation between price and bedrooms in south Salins. Enter the data of South Salins in Excel sheet. Go to DATA > Data analysis Select Regression, click ok Enter Y range i.e., price Enter X range i.e., bedrooms Select the output range Click ok Excel output is given below: From the output, the multiple R indicates the correlation coefficient r. That is: 0.68519989 The test statistic is given by the formula: t=rn-21-r2=0.6851998924-21-0.685199892=3.2138723630.7283550719=4.4125077≈4.4125 Degrees of freedom: df=n-2=24-2=22 Use Excel function "=T.DIST.2T(4.4125,22)", to obtain the p-value. The result is given below: Since the p-value is too small i.e., less than the significance level (0.05), reject the null hypothesis. Therefore, it can be concluded that there is a significant linear correlation between Price and bedrooms in South Salins. 6) State the hypotheses as follows: Null Hypothesis H0:ρ=0 i.e., there is no linear correlation between price and bedrooms in north Salins. Alternative Hypothesis H1:ρ≠0 i.e., there is a linear correlation between price and bedrooms in north Salins. Enter the data of North Salins in Excel sheet. Go to DATA > Data analysis Select Regression, click ok Enter Y range i.e., price Enter X range i.e., bedrooms Select the output range Click ok Excel output is given below: From the output, the multiple R indicates the correlation coefficient r. That is: 0.84398803 The test statistic is given by the formula: t=rn-21-r2=0.8439880324-21-0.843988032=3.9586547570.5363620095=7.380565≈7.381 Degrees of freedom: df=n-2=24-2=22 Use Excel function "=T.DIST.2T(7.381,22)", to obtain the p-value. The result is given below: The p-value is 0.00000022 Since the p-value is too small i.e., less than the significance level (0.05), reject the null hypothesis. Therefore, it can be concluded that there is a significant linear correlation between Price and bedrooms in North Salins.