Minor surgery on horses under field conditions requires a reliable short-term an
ID: 3226995 • Letter: M
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 = 66 horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was 18.73 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. H_0:mu = 20 H_a: mu notequalto 20 H_0: mu notequalto 20 H_a: mu = 20 H_0: mu notequalto 20 H_a: mu > 20 H_0:mu = 20 H_a:muExplanation / Answer
Solution:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 20
Alternative hypothesis: < 20
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = 8.4 / sqrt(66) = 20 / sqrt(50) = 1.0339685
DF = n - 1 = 66 - 1 = 65
t = (x - ) / SE = (18.73 - 20)/1.0339685 = -1.2283
z = (x - ) / s = (18.73 - 20) / 8.4 = -0.15119
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Calculating p-value from z-value, for two-tailed test,
The P-Value is 0.879976.
Interpret results. Since the P-value is greater than the significance level (0.10), we cannot reject the null hypothesis.
Do not reject the null hypothesis. There is not sufficient evidence that the average lying-down time is less than 20 minutes.