Question

# Consider the following data for two variables, x and y. x 22 24 26 30 35...

Consider the following data for two variables, x and y.

 x y 22 24 26 30 35 40 13 22 34 34 41 35

(a) Develop an estimated regression equation for the data of the form ŷ = b0 + b1x.  (Round b0 to one decimal place and b1 to three decimal places.)

ŷ =

(b) Use the results from part (a) to test for a significant relationship between x and y. Use α = 0.05

Find the value of the test statistic. (Round your answer to two decimal places.)

F =

p-value =

Is the relationship between x and y significant?

Yes, the relationship is significant.

No, the relationship is not significant.

(c)Develop a scatter diagram for the data.

Does the scatter diagram suggest an estimated regression equation of the form ŷ = b0 + b1x + b2x2? Explain.

Yes, the scatter diagram suggests that a linear relationship may be appropriate.

Yes, the scatter diagram suggests that a curvilinear relationship may be appropriate.

No, the scatter diagram suggests that a linear relationship may be appropriate.

No, the scatter diagram suggests that a curvilinear relationship may be appropriate.

(d) Develop an estimated regression equation for the data of the form ŷ = b0 + b1x + b2x2.  (Round b0 to one decimal place and b1 to two decimal places and b2 to four decimal places.)

ŷ =

(e) Use the results from part (d) to test for a significant relationship between x, x2,  and y. Use α = 0.05. Is the relationship between x, x2, and y significant?

Find the value of the test statistic. (Round your answer to two decimal places.)

p-value =

Is the relationship between x, x2, and y significant?

Yes, the relationship is significant.

No, the relationship is not significant.

(f)Use the model from part (d) to predict the value of y when x = 25. (Round your answer to three decimal places.)

a
)Applying regression from excel: data -data analysis:

y^ =-3.7+1.138x

b)F =5.62

p value =0.077

No, the relationship is not significant.

Yes, the scatter diagram suggests that a curvilinear relationship may be appropriate.

d)

y^ =-165.5+12.07x-0.1766x2

e)value of the test statistic =17.51

p-value =0.022

.

Yes, the relationship is significant.

f)

predicted value =-165.5+12.07*25-0.1766*25^2=25.875 (please try 25.887 if this comes wrong)

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