Minor surgery on horses under field conditions requires a re

Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal after effects so that horses can be left unattended. An article reports that for a sample of n = 64 horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was 18.92 min and the standard deviation was 7.2 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 a = 0.10 test. If the critical region is one-sided, enter NONE for the unused region. Round your answers to two decimal places. z Compute the test statistic value. Round your answer to two decimal places. State the conclusion in the problem context. Reject the n ll hypothesis. There is not sufficient evidence that the average lying-down time is less than 20 minutes. Do not reject the null hypothesis. There is sufficient evidence that the average lying-down time is less than 20 minutes. Do not reject the null hypothesis. There is not sufficient evidence that the average lying-down time is less than 20 minutes. Reject the null hypothesis. There is sufficient evidence that the average lying-down time is less than 20 minutes.

Solution

NOTE:
When Population S.D is unknown we use t-test statistic , but in the problem it used Z symbols. So I\'m calculating both. Please use whichever is feasible

Z-TEST

Test Used: Z-Test For Single Mean
Set Up Hypothesis
Null Hypothesis H0: U=20
Alternate Hypothesis H1: U<20
Test Statistic
Population Mean(U)=20
Given That X(Mean)=18.92
Standard Deviation(S.D)=7.2
Number (n)=64
we use Test Statistic (Z) = x-U/(s.d/Sqrt(n))
Zo=18.92-20/(7.2/Sqrt(64)
Zo =-1.2
P-Value : Left Tail - Ha : ( P < -1.2 ) = 0.1151
Critical Value
The Value of |Z | at LOS 0.1% is -1.28
We got |Zo| =1.2 & | Z | =1.28
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho

ANS:
H0: U=20 , H1: U<20
Z<=-1.28
Z>=NONE
Z = -1.2
Do not reject the null hypothesis. There is not suffìcient evidence that the average lying-down time is less than 20 minutes.

T-TEST
Set Up Hypothesis
Test Statistic
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =18.92-20/(7.2/Sqrt(63))
to =-1.2
| to | =1.2
Critical Value
The Value of |t | with n-1 = 63 d.f is 1.295
We got |to| =1.2 & | t | =1.295
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Left Tail -Ha : ( P < -1.2 ) = 0.11732
Hence Value of P0.1 < 0.11732,Here We Do not Reject Ho

ANS:
H0: U=20 , H1: U<20
t<=-1.295
t>=NONE
to = -1.2
Do not reject the null hypothesis. There is not suffìcient evidence that the average lying-down time is less than 20 minutes.