Question

1.) Consider the data in the following table. In this study, the authors were interested in the use of erythrocyte sodium (ES) concentration as a potential biomarker for the response to lithium treatment in patients with bipolar illness. The ES levels were measured in 10 bipolar patients before and after treatment with lithium. The ES levels before and after treatment with lithium, as well as the after - before differences are given in Table 1. The authors wished to determine if there was a significant increase in ES level following lithium treatment.

            Table 1. Erythrocyte Sodium Concentrations (mmol/l) in Bipolar Patients

Phase	Erythrocyte Sodium Concentration (mmol/l)
Ill ("Before")	5.67, 5.89, 5.46, 7.30, 6.74, 5.80, 8.30, 6.01, 5.37, 5.45
On Lithium ("After")	6.02, 5.87, 7.42, 7.50,   6.24, 6.55, 8.65, 6.27, 5.64, 6.38
Difference	0.35, -0.02, 1.96, 0.20, -0.50, 0.75, 0.35, 0.26, 0.27, 0.93

Test statistic = __2.19____   d.f. = __9__   p-value = ___ 0.0560___

One sample t test was most appropriate to perform

??? State your conclusion in terms that a layperson would understand. ???

     

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d = 0

Alternative hypothesis: u_d > 0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 0.65614

SE = s / sqrt(n)

S.E = 0.20749

DF = n - 1 = 10 -1

D.F = 9

t = [ (x₁ - x₂) - D ] / SE

t = 2.19

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a one-tailed test, the P-value is the probability that a t statistic having 9 degrees of freedom is greater than 2.19.

Thus, the P-value = 0.028.

Interpret results. Since the P-value (0.028) is less than the significance level (0.10), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there was a significant increase in ES level following lithium treatment.