Question

A study was performed among patients with glaucoma, an important eye disease usually manifested by high intraocular pressure (OP); left untreated, glaucoma can lead to blindness. The patients were currently on two medications (A and B) to be taken together for this condition. The purpose of this study was to determine whether the patients could drop medica- tions A and B and be switched to a third medication (medi- cation C) without much change in their IOP. Ten patients were enrolled in the study. They received medications A+B for 60 days and had their IOP measured at the end of the 60-day period (referred to as IOPA then taken off medications A and B and switched to medication C, which they took for an additional 60 days. IOP was mea- sured a second time at the end of the 60-day period while the patient was on medication C (referred to as IOP). The results were as shown in Table 7.12. TABLE 7.12 Effect of medication regimen on IO P among glaucoma patients IOPA B (mm Hg) 18.0 16.0 17.0 18.0 20.0 19.0 19.0 12.0 17.0 21.5 Patient number IOPc (mm Hg) 14.5 11.5 21.0 14.0 10 19.0 7.101 What procedure can be used to test the hypothesis that there has been no mean difference in IOP after 60 days between the two drug regimens? 7.102 Perform the procedure mentioned in Problem 7.101, and report a two-tailed p-value. A goal of the study was to establish whether switching to medication C is "equivalent" to the original regimen of medication A and B. "Equivalence" here means that the underlying mean IOP after switching to medication C has not changed by more than 2 mm Hg in either direction. 7.103 What procedure can be used to establish equiv- alence? Is equivalence the same as accepting the null hypothesis in Problem 7.101? Why or why not? 7.104 Implement the procedure in Problem 7.103 to address the question of whether the regimens are equivalent.

Answer 1

IOP(A+B)		IOP(c`)
Mean	17.75	Mean	17.8
Standard Error	0.81394103	Standard Error	1.22972445
Median	18	Median	18
Mode	18	Mode	18
Standard Deviation	2.573907535	Standard Deviation	3.888730155
Sample Variance	6.625	Sample Variance	15.12222222
Kurtosis	2.281408398	Kurtosis	-0.704318489
Skewness	-1.044590456	Skewness	0.000566833
Range	9.5	Range	12.5
Minimum	12	Minimum	11.5
Maximum	21.5	Maximum	24
Sum	177.5	Sum	178
Count	10	Count	10

We are going to use 2 sample t-test. As we are checking if there is a significant difference between the two samples.

The provided sample means are shown below: X 17.75 X2 17.8 Also, the provided sample standard deviations are: 81-2.573907535 82 3.888730155 and the sample sizes aren 10 and n2 10. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is α-005, and the degrees of freedom are df 18. In fact, the degrees of freedom are computed as follows, assuming that the population

variances are equal: 2.101, for α = 0.05 and df- Hence, it is found that the critical value for this two-tailed test is tc 18. The rejection region for this two-tailed test is Rt: t>2.101 Since it is assumed that the population variances are equal, the t-statistic is computed as follows: n1+n2-2 2 17.75 17.8 -0.034 (10-1)2.5739075352+10-1)3.8887301552 (1 1010 Since it is observed that t = 0.034 < tcー2101, ît is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p 0.9733, and since p 0.973320.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ| is different than μ2, at the 0.05 significance level