Two-Sided AlternativeOne-Sample TestTwo-Sided AlternativeExample: In a cardiology study, one variable of interest is Body Mass Index (BMI) which is measured as weight (kg)/height2 (m2). A total of 14 adult males were studied. The data are given in the table below. Can we conclude that the mean BMI of the population from which the sample was drawn is 35?Subject BMI Subject BMI Subject BMI1 23 6 21 11 232 25 7 23 12 263 21 8 24 13 314 37 9 32 14 455 39 10 57n = 14, o = 35, X = 30.5, s = 10.6392Ho:  = 35H1:   35Let  = 0.05Def: A two-tailed or two-sided test is one in which the parameter under study in the alternative hypothesis can be less than or greater than the value of the parameter in the null hypothesis. Reject Ho if t > c2 or t < c1Accept Ho if c1  t  c2To determine c1, c2 we need to specify . Pr (reject Ho | Ho true) = Pr (t < c1 | Ho true) + Pr (t > c2 | Ho true)= /2 + /2 = t ~ tn-1 under HoTherefore, c1 = tn-1, /2 and c2 = tn-1, 1-/2 are the upper and lower critical values.Therefore, reject Ho if t > = tn-1, 1-/2 or if t < tn-1, /2Accept Ho if tn-1, /2  t  tn-1, 1-/2Example:  = 0.05tn-1, 1-/2 = t13, 0.975 = 2.160tn-1, /2 = t13, 0.025 = -2.160Reject Ho if t > 2.160 or if t < -2.16058184342541463921035530.../..ns/μXtoDecision: Do not reject Ho.Conclusion: We have insufficient evidence to state that the mean BMI of the population from which the sample came is anything other than 35.p-value: Probability of getting a value of the test statistic this extreme or more extreme if Ho is true.a. If t  0, p = 2 * Pr (tn-1  t)b. If t > 0, p = 2 * [1 – Pr (tn-1  t)]For the example, this is 2 * the area to the left of the test statistic, t, with 13 df.t13, 0.05 = -1.771t = -1.58t13, 0.10 = -1.350Therefore, 0.05 * 2 < p < 0.10 * 2or 0.10 < p < 0.20One-sample z-testIf n is very large (>200) or if  is known, we can substitute a z-test for the t-test.),~N(nσ/μXzo10For a two-tailed test, reject Ho ifz > z1-/2orz < z/2Example: Suppose we are interested in the mean age of a certain population. Suppose, we know that 2 = 20 and that the population is approximately normally distributed. Can we conclude that the mean age is different from 30?Ho:  = 30H1:   30Let  = 0.05Reject Ho if z > 1.96 or if z < -1.96.Suppose we sample 10 individuals and get X = 27.12241421310203027../nσ/μXz~oDecision: Reject HoConclusion: We have evidence that the mean age is not equal to 30 years.p-value: a. If z  0, p = 2(z)b. If z > 0, p = 2[1 - (z)]Therefore,p = 2(-2.12)= 2 * 0.017= 0.034Note: There are one-tailed z-tests also. The critical value is determined such that allof  is at the upper end or at the lower end (but not both). p = (z)Confidence Intervals:Let’s go back to the first example about BMI.o = 35, X = 30.5, s = 10.6392We could construct a 95% Confidence Interval).,.(...*../.*..n*s/tXα,n64363624146530843421602530146392101602530211If we reject Ho at the 5% level of significance, then o = 35 must lie outside the 95% Confidence Interval. If we accept Ho, then o = 35 must lie within the 95% Confidence Interval. Let’s calculate a 95% Confidence Interval for the second example which used the z-test. ).,.(..*./*.n*σzXα7729232477227414219612710209612721/Note: o = 30 is not contained in the Confidence Interval. This is consistent with thefact that we rejected Ho. 3 Approaches1. Critical value2. p-value3. Confidence IntervalSteps to consider for a Statistical Test of a Hypothesis1. State the problem2. Formulate hypotheses—one- or two-tailed.3. Choose . This determines critical value.4. Determine the test statistic.5. Calculate the test statistic.6. Decision and