Question

Bob Michels owns a business in an emerging agricultural market, medicinal cannabis. He is interested in running a regression analysis with fertilizer potency as the dependent variable and manufacturing process temperature as the predictor variable. After taking a sample of 25 batches and recording these two variables, Bob developed the following results:

Potency = -8.78 + .48897 TEMP

(.0792)

S_e = 3.473 SS(total)= 737.73

Potency is the percent of maximum fertilizer potency

Temp is the temperature in degrees in Fahrenheit.

Interpret the regression coefficient.

Test whether ther is a positive linear relationship between potency and Temp. Use alpha = .05

Develop the Mission ANOVA table

Answer 1

Interpretation of slope coefficient -

For a 1 degree Fahrenheit increase in temperature, we estimate the percent of maximum fertilizer potency to increase by 0.48897%

Interpretation of intercept coefficient -

The estimated maximum fertilizer potency at 0 degree Fahrenheit is -8.78

H0:

H1;

Test statistic, t = 0.48897 / 0.0792 = 6.17

Degree of freedom = n-2 = 25-2 = 23

P-value = P(t > 6.17) = 0.000001

Since p-value is less than 0.05 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that there is a positive linear relationship between potency and Temp.

S_e = 3.473 SS(total)= 737.73

df Total = n-1 = 25 -1 = 24

df Error = n - 2 = 25 - 2 = 23

df Group = k = 1 (k is number of predictors in the model)

S_e = 3.473

=> MSE = 3.473² = 12.06173

SSE = DF error * MSE = 23 * 12.06173 = 277.4198

SS Group = SS Total - SSE = 737.73 - 277.4198 = 460.3102

MS Group = SS Group / df Group = 460.3102 / 1 = 460.3102

F = MS Group / MS error = 460.3102 / 12.06173 = 38.16287

P-value = P(F > 38.16287, df = 1,23) =

Source	DF	SS	MS	F	Pr(>F)
Group	1	30.08	460.3102	38.16287	2.66 x 10-6
Error	23	277.4198	12.06173
Total	24	737.73