Question

Please help!!!!

This is worth 10 points if done correctly. Be sure to state the null and alternative hypothesis as well as the test statistic, critical value, and conclusion in plain terms. Test for a difference in the Central venous oxygen saturation on admission and 6 hours after admission for these ten patients. Use alpha = .05 Patient On Admission 6 hrs after admission 59.1 56.7 58.2 56.0 60.7 59.5 59.8 65.3 56.1 37.8 50.0 71.4 51.3

Answer 1

Let us denote the difference

d = oxygen saturation on admission - oxygen saturation 6 hours after admission

Let ud denotes the population mean of difference. Hypotheses : To test null hypothesis Ho Hd =0 against alternative hypothesis H1 pd 70 This is a two tailed test as alternative hypothesis H1 contains only #sign. Let d denotes the sample mean of difference, sd denotes the sample standard deviation of difference and n denotes the sample size. Here, d =-6.7800, số =55.882, S4=7.4754, n = 10 The test statistic can be written as: t- Vn(d – 0.0) - sd which under Ho follows at distribution with n - 1 df. Decision rule / Rejection region : We reject Hat 0.05 level of significance if P-value < 0.05 or if tobs > to.025,n–1 = 2.262157 density IT Th 0.0 - 02 Now, Value of test statistic: The value of the test statistic tobs = V10(-6.7800 - 0) 2 = -2.868102 2 -2.868 7.4754 7 associated degrees of freedom = 9 The critical value tcritical = +2.262157 =+ 2.262 P-value = P(tn-11 > tobs) = 2 * P(tg < -2.868102) = 2 * 0.009267 = 0.018534 0.0185 Conclusion : Since p-value < 0.05 and tobs > tcritical = 2.262157, so we reject H, at 0.05 level of significance Hence, we can conclude that there is significant difference between two population means.