Question

Tobramycin is a powerful antibiotic. To minimize its toxic side effects, the dose can be individ- ualized for each patient. Thirty patients participated in a study of the accuracy of this individualized dosing. For each patient, the actual peak concentration (ug/ml) of Tobramycin in the blood serum was measured, and the predicted peak concentration of Tobramycin was calculated, based on the patient's age, sex, weight, and other characteristics. The results were reported as in the table.

Summary	Predicted	actual
mean	4.52	4.40
SD	0.90	.85
n	30	30

Dose the reported summary give enough information for you to judge whether the individualized dosing is, on the whole, accurate in its predicted of peak concentration? In other words, dose the reported summary give enough information for you to complete paired-sample t test? If so, describe how you would make this judgment. If not, describe what additional information you would need and why.

Answer 1

here we use two independent sample t-test not paired t-test. since both type of observations are not dependent on the patients.

The paired sample t-test, sometimes called the dependent sample t-test, is a statistical procedure used to determine whether the mean difference between two sets of observations is zero. In a paired sample t-test, each subject or entity is measured twice, resulting in pairs of observations. Common applications of the paired sample t-test include case-control studies or repeated-measures designs.

• The sampling method for each sample is simple random sampling.
• The test is conducted on paired data. (As a result, the data sets are not independent.)
• The sampling distribution is approximately normal,

here we use t-test with

null hypothesis H0:mean1=mean2 and alternate hypothesis H1:mean1≠mean2

statistic t=|(mean1-mean2)|/((s_p*(1/n1 +1/n2)^1/2) =0.5309 with df is n=n1+n2-2 =58

and s_p²=((n1-1)s1²+(n2-1)s2²)/n=0.7663

since the two tailed p-value is more than typical value of alpha=0.05, so we fail to reject H0 (or accept H0) and conclude that there is sufficient evidence that there is no difference between predicted and actual value.

	sample	mean	s	s2	n	(n-1)s2
	predicted	4.5200	0.9000	0.8100	30	23.4900
	actual	4.4000	0.8500	0.7225	30	20.9525
	difference=	0.1200	sum=	1.5325	60	44.4425
	sp2=	0.7663
	sp=	0.8754
	SE=	0.2260
	t=	0.5309
one tailed	p-value=	0.2987
two tailed	p-value=	0.5975
critical	t(0.05)	2.0017