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Dahasolnce [82]
3 years ago
9

Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minim

al cardiovascular and respiratory changes, and a quick, smooth recovery with minimal aftereffects so that horses can be left unattended. An article reports that for a sample of n = 69 horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was 18.94 min and the standard deviation was 8.3 min. Does this data suggest that true average lateral recumbency time under these conditions is less than 20 min? Test the appropriate hypotheses at level of significance 0.10.State the appropriate null and alternative hypotheses.State the rejection region(s) for an ? = 0.10 test. If the critical region is one-sided, enter NONE for the unused region.z ?___z ?___Compute the test statistic value.z=___
Mathematics
1 answer:
iren [92.7K]3 years ago
5 0

Answer:

We accept  H₀

Step-by-step explanation:

Normal Distribution

size sample   n  = 69

sample mean  18.94

standard deviation  8.3

Is a one tailed-test to the left we are traying of find out is we have enough evidence to say that the mean is less than 20 min.

1.-Test hypothesis                              H₀         ⇒   μ₀  =  20

   Alternative hypothesis                  Hₐ         ⇒    μ₀  < 20

2.- Critical value

for α = 0.1  we find from z Table

z(c) = - 1.28

3.-We compute z(s)

z(s) = [ (  μ  -  μ₀ ) / (σ/√n)                ⇒    z(s) = [( 18.94  -  20  )*√69)/8.3]

z(s) = ( -1.06)*8.31/8.3                    

z(s) =  - 1.061

4.- We compare

z(c)   and z(s)           -1.28 > -1.061

Then z(c)    >  z(s)

z(s) in inside acceptance region  so we accept H₀

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x => 3 can be anything greater than 3 and 3.

Now for the dots.

The open dots in signs are (< and >).

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The closed dots in signs are (<= and =>).

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Arrows.

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Arrows pointing to the right mean that the values are getting smaller.

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