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A study was conducted to monitor the emissions of a noxious substance from a che

ID: 3054075 • Letter: A

Question

A study was conducted to monitor the emissions of a noxious substance from a chemical plant and the concentration of the chemical at a location in close proximity to the plant at various times throughout the year. A total of 14 measurements were made. Computer output for the simple linear regression least-squares fit is provided. (Some entries have been omitted and replaced with **.)

Linear Fit

Concentration = 1.5429211 + 1.8247687 emissions

Summary of Fit

RSquare                                                0.793919

RSquare Adj                                         0.776745

Root Mean Square Error                       1.513979

Mean of Response                                8.810714

Observations (or Sum Wgts)                               14

Analysis of Variance

Source               DF          Sum of Squares            Mean Square         FRatio

Model                1             105.96390                      105.9639                 46.2294

Error 12 27.50559 2.29 Prob >F

C. Total              13           133.46949                                                      <.0001

Parameter Estimates

Term                      Estimate                 Std Error           t Ratio            Prob>|t|

Intercept                 1.5429211               1.142937               ****            0.2019

Emissions                1.8247687               0.268379               ****            <.0001

A) Calculate the 95% confidence interval for slope coefficient B1

B) Do the hypothesis test H0: B1 =0 against Ha: B1 ? 0. (State the null and alternative hypothesis, test statistics, p-value and conclusion, please also provide the meaning of this conclusion.)

Explanation / Answer

Answer below:

A.

95% CI is :

Coeff +/- t*SE

= 1.8247687 +/- t* 0.268379

t value for 95% CI at df = 13 is 2.160

So, = 1.8247687 +/- 2.160* 0.268379

=1.2451 to 2.4045

B.

Its already done in the regression table.

Null and alternate hypo:

Ho: B1=0

Ha: B1!=0

Test stat = (Estimate-0)/SE = 1.8247687/.268379 = 6.7992

Since our test statistic is more than the critical t of 2.160 we have the

fact the we reject Ho and conclude that Emissions is statistically significant

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