Refer to the Minitab output below in which x NO3 wet deposi

Refer to the Minitab output below in which x = NO3? wet deposition (g N/m2) and y = lichen N (% dry weight).

(a) Carry out the model utility test at level 0.01, using the rejection region approach.
State the appropriate null and alternative hypotheses.

H0: B1 not equal to0


Ha: B1 = 0

H0: B1 = 0


Ha: B1 not equal to 0    

H0: B1 = 0


Ha: B1 > 0

H0: B1 = 0


Ha: B1 < 0
State the rejection region(s). If the critical region is one-sided, enter NONE for the unused region. Round your answers to three decimal places.


Compute the test statistic value. Round your answer to two decimal places.
t =  

State the conclusion in the problem context.

Reject H0. The model is useful.Reject H0. The model is not useful.    Fail to reject H0. The model is useful.Fail to reject H0. The model is not useful.


(b) Repeat part (a) using the P-value approach. (Round your answer to three decimal places.)
P-value =  

State the conclusion in the problem context.

Reject H0. The model is useful.Reject H0. The model is not useful.    Fail to reject H0. The model is useful.Fail to reject H0. The model is not useful.

The regression equation is
lichen N = 0.389 + 0.959 no3 depo
Predictor Coef Stdev t-ratio p
Constant 0.38932 0.09869 3.94 0.002
no3 depo 0.9593 0.1823 5.26 0.000

s = 0.1925   R-sq = 71.6%   R-sq (adj) = 69.0%

Solution

State the appropriate null and alternative hypotheses.

H0: B1 = 0

Ha: B1 not equal to 0    

State the rejection region(s). If the critical region is one-sided, enter NONE for the unused region. Round your answers to three decimal places.

Degrees of freedom for t=11

t ?

-3.106

       t ?

3.106

If calculated t < -3.106 or t>3.106 reject the null hypothesis


Compute the test statistic value. Round your answer to two decimal places.
t =  5.26

State the conclusion in the problem context.( Calculated t=5.26 >3.106 table value)

Reject H0. The model is useful.

Reject H0. The model is not useful.    

Fail to reject H0. The model is useful.

Fail to reject H0. The model is not useful.


(b) Repeat part (a) using the P-value approach. (Round your answer to three decimal places.)
P-value =  0.000

State the conclusion in the problem context.   ( P =0.000 < 0.01 level)

Reject H0. The model is useful.

Reject H0. The model is not useful.    

Fail to reject H0. The model is useful.

Fail to reject H0. The model is not useful.

t ?

-3.106

       t ?

3.106

Refer to the Minitab output below in which x = NO3? wet deposition (g N/m2) and y = lichen N (% dry weight). (a) Carry out the model utility test at level 0.01,
Refer to the Minitab output below in which x = NO3? wet deposition (g N/m2) and y = lichen N (% dry weight). (a) Carry out the model utility test at level 0.01,

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