Question

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 aftereffects so that horses can be left unattended. An article reports that for a sample of n = 75 horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was 18.81 min and the standard deviation was 8.4 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.

Answer 1

Answer:

$p_v =P(t_{74}<-1.227)=0.112$    

If we compare the p value and a significance level for example $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.

Step-by-step explanation:

Data given and notation    

$\bar X=18.81$ represent the average lateral recumbency for the sample    

$s=8.4$ represent the sample standard deviation    

$n=75$ sample size    

$\mu_o =20$ represent the value that we want to test    

$\alpha$ represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)    

$p_v$ represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to apply a left tailed  test.  

What are H0 and Ha for this study?    

Null hypothesis:  $\mu \geq 20$  

Alternative hypothesis :$\mu < 20$  

Compute the test statistic  

The statistic for this case is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

$t=\frac{18.81-20}{\frac{8.4}{\sqrt{75}}}=-1.227$

The degrees of freedom are given by:

$df=n-1=75-1=74$    

Give the appropriate conclusion for the test  

Since is a one side left tailed test the p value would be:    

$p_v =P(t_{74}<-1.227)=0.112$    

Conclusion    

If we compare the p value and a significance level for example $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.