Right Tailed Z-Test of Population Mean

Right Tailed Z-Test of population mean H0: m ≤ 10 Case study: an antihypertensive drug is to be considered for clinical trials in human if it result in reduction of diastolic blood pressure of more than 10 mmHg with at least high significance. The drug was given to 16 animals after hypertension was induced. The mean reduction of BP was 17 mmHg. s was estimated to be 8 mmHg. Is the drug considered a good candidate for further clinical trials at  of 0.05. H1: m  10, the alternative hypothesis and the research hypothesis or what the researchers hope for are the same -The null hypothesis suggests that the mean BP reduction is NOT significantly higher than 10. Thus, the test is one sided with one upper rejection region of an area =  (0.05) Thus area below Z1-) is 0.95 and the corresponding Z is 1.65 as critical value. If test statistic (Z-sample) is higher than 1.65, H0 is rejected 368 gm. Z-sample versus critical value and p-value _ z = X− m _ x s/ n 17 − 10 z = = 3 .5 2 sample sample If Z-sample was less than +1.65 we accept the null hypothesis. But the calculated Z of the sample (3.5) was +1.65, thus null hypothesis is rejected meaning that BP reduction is significantly higher than 10. However, for the drug to considered for further clinical studies the reduction should be highly significant, which can be assesed by finding out p-value (one sided) equivalent to the area above 3.5: 1-0.9998 = 0.0002 from Z-table, well very highly significant making the drug a good candidate for further clinical studies. Testing H0 by Means of A Confidence Interval: right tailed If the lower confidence limit s XZ * = (LCL ) is higher than the n hypothetical mean, 17  1.65  2 H0 is rejected. 13.7 to 20.3. 1 −  Since 13.7 (LCL) is higher than 10 (hypothetical mean) We reject H0 and conclude BP reduction is significantly higher than 10 If LCL is lower or equal to the hypothetical mean H0 is accepted 3 Left Tailed Z-Test of population mean Case study: H0: m ≥100 H1: m  100 There are claims that Mediterranean diet can result in -The null hypothesis suggests that cholesterol level is not significantly lower lower bad cholesterol (LDL) than 100. Thus, Z-test is one sided with one levels in healthy subjects. A lower rejection region of an area =  (0.05) researcher decided to put this Thus area below Z() is 0.05 and the claim to the test. 100 subjects who corresponding Z is -1.65 as critical value. If follow this diet were randomly test statistic (Z-sample) is lower than -1.65 H0 368 gm. selected and their LDL level was is rejected measured as mean value of 93 mg/dl . s was estimated to be 30 mg/dl. At significance level of 0.05, does mediterranean diet result in mean LDL lower than normal average LDL level of 100 mg/dl . Z-sample versus critical value and p-value _ z = X− m _ x s/ n 93 − 100 z = = −2.33 3 sample sample If Z-sample was higher than -1.65 we accept the null hypothesis. But the calculated Z of the sample was -1.65), thus null hypothesis is rejected meaning that LDL is significantly lower than 100. p-value (one sided) equivalent to the area below 2.3 is 0.0099 from Z-table, highly significant. The summary of testing can be stated as there is convincing evidence for the association between Mediterranean diet and mean LDL level as lowering effect. Testing H0 by Means of A Confidence Interval If the upper confidence limit s XZ * = (UCL ) is lower than the n hypothetical mean, 93  1.65  3 H0 is rejected. 88.05to97.95. 1 −  Since 97.3 (UCL) is higher than 100 (hypothetical mean) We reject the H0 and conclude LDL is significantly lower than 100 If UCL is higher or equal to the hypothetical mean, H0 is accepted 6 Summary notes on hypothesis testing Steps of testing 1. State null and the alternative hypotheses H 0 : m = m 0 versus H1 : m  m 0 H 0 : m  m 0 versus H 1 : m  m 0 H 0 : m  m 0 versus H 1 : m  m 0 2. Choose a significance level  (usually .05) 3. Determine the critical value (s), acceptance and rejection region (s) 4. Compute the test statistic (Z-stat or z-sample). This is the calculated value that you will compare with the critical value. 5. Reject the null hypothesis if the test statistic falls in the rejection region, otherwise do not reject null hypo. But no conclusion on how much the difference or the evidence can be made. 6. Calculate p-value which can provide the amount of difference or evidence. 7. Construct confidence interval 8. State the appropriate conclusions 7 Summary notes on hypothesis testing Approaches of hypothesis testing Approach Criteria Critical Two sided: Accept H0 if ̶-Z-critical≤ Z-sample≤ +Zvalue critical, otherwise reject H0). E.g. for  =0.05 accept H0 approach if -1.96≤ Z-sample≤ +1.96 and reject if Z-sample +1.96 or Z-sample -1.96 . Right sided: Accept H0 if Z-sample≤ +Z-critical, otherwise reject H0. E.g. for  =0.05 accept H0 if -Z-sample≤ +1.96 and reject if Z-sample +1.96 Left sided: Accept H0 if Z-sample ≥-Z-critical otherwise reject H0. E.g. for  =0.05 accept H0 if Z-sample ≥-1.96; Reject if Z-sample-1.96 . 8 Summary notes on hypothesis testing Decision making approaches of hypothesis testing Approach Criteria Confidence interval approach Two sided: if hypothetical value is not contained in the CI of the sample mean reject H0, otherwise accept Right sided: If the lower limit of CI is higher than the hypothetical mean reject H0 otherwise accept Left sided: If the upper limit of CI is lower than the hypothetical mean reject H0 otherwise accept P-value approach If P   reject H0 and P ≥  accept. When reject, the lower the p-value the higher the significant difference or 9 evidence against H0. Summary notes on hypothesis testing ➢ An indication of equality (=, ≥, ≤) must appear in the null hypothesis. Conversely ≠,  and  must a pear for the alternative hypothesis. ➢The null hypothesis is the hypothesis to be tested ➢ The null and alternative hypothesis are complementary. The two together exhaust all possibilities regarding the value that the hypothesized parameter can assume. ➢ If H0 is rejected, we conclude that HA is true, but usually we state our conclusion as reject or accept H0 without referring to HA . ➢ Accepting or rejecting a hypothesis is not 100% proof of the hypothesis because of possible type I and type II errors. Summary notes on hypothesis testing ➢ Common encountered α values are 0.01, 0.05 and 0.1, but the most used one is 0.05. ➢ The level of significance α is the probability of rejecting a true null hypothesis (type error). This probability or chance of type I error for  of 0.1 is 0.1 and for 0.05 is 0.05. ➢ The decision of which values go into the rejection region and which ones go into the nonrejection region is based on the desired level of significance α. Higher  means higher area of rejection area (s) with higher chance of rejecting H0 when it is true (Type I error). ➢ Since rejection of a true null hypothesis is an error, we should make the level of significance small (usually 0.05), otherwise the probability of choosing samples from the rejection area (s) is quite higher