University:
Kentucky State University
Course:
MAT 321 | Probability/Statistics I
Academic year:

2020
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459

Pages:

19
Author:

plenceplubparun2

Hypothesis Testing: Difference Between Two Populations: Independent samples (z and t tests) • Assumptions 1. Two populations are normally distributed, If not normal, requires large samples 2. Standard deviations of populations A. known, z-stat can be calculated and ztest is performed B. Unknown standard deviations B1) Pooled t-test: assuming equal standard deviations B2) Unpooled t-test: unequal standard deviations Continued. 1 3. The samples are independent: the scores of one sample do not affect those of the other as the following scenarios A. Random samples from different populations: Testing the difference of the means between two fixed populations we test the differences between samples from each population. When both samples are randomly selected, the samples are independent because they came from two different populations randomly. B. Random assignment to different treatments or groups: When working with subjects, if we select a random sample and then randomly assign half of the subjects to one group and half to another we can make inferences about the populations by z or t test. The following illustrate dependent samples for which z or t-test can not be used: If we need to compare STAT scores for your section between first and second exam ( same students and thus dependent samples). Similarly, If the same sample was subjected to two different treatments (first the sample was given drug A and after washout period the same sample was given drug B). When the samples are dependent paired t-test rather than t-test is performed. Z-test of Independent Samples (known ) 1. State the null and alternative hypotheses. Two tauked: H0: μ1 – μ2 = 0 or μ1 = μ2 , HA: μ1 – μ2 ≠ 0 or μ1 ≠ μ2 Right tailed: H0: μ1 – μ2 ≤ V, HA: μ1 – μ2  V Left tailed: H0: μ1 – μ2 ≥ V, HA: μ1 – μ2  V V can be zero or any other value 2. Calculate the standard error SE =  2 1 n 1 +  n 2 2 2 3. Choose α and set the criterion (critical values) for rejecting the null Hypothesis as follows: a 0.1 0.05 0.01 Two tailed ±1.65 ±1.96 ±2.58 Z-Critical Right tailed 1.28 1.65 2.33 Left tailed -1.28 -1.65 -2.33 4. Compute the test statistic: ( x − x ) − ( −  ) z= SE 1 2 1 2 0 The subscript 0 indicates that the difference is a hypothesized parameter. 5. Compare z with the critical value (s) 6. Make a decision: reject or fail to reject the null hypothesis. E.g. Researchers wish to know if the data they collected provide sufficient evidence to indicate a difference (two sided) in the mean uric acid levels between normal individuals and individuals with Down’s syndrome. The data consist of uric acid readings on 12 individuals with Down’s syndrome and 15 normal individuals. The means are: _ _ x1 = 4.5 mg/100 mL and x2 = 3.4 mg/100 mL The data constitute two independent simple random samples each drawn from normally distributed population with variance equal to 1 for the Down’s syndrome population and 1.5 for the normal population. H0: μ1 – μ2 = 0, HA: μ1 – μ2 ≠ 0 An alternative way of stating the hypothesis is as follows H0: μ1 = μ2, HA: μ1 ≠ μ2 • Let α = 0.05, the critical values of z are 1.96 and -1.96. Reject H0 unless -1.96 ≥ z ≥ 1.96   1 1.5 SE = + = + = 0.4282 n n 12 15 2 2 1 2 1 2 ( x − x ) − ( −  ) z= SE 1 2 1 z= 2 0 (4.5 − 3.4) − 0 1.1 = = 2.57 0.4282 0.4282 Statistical decision to reject H0 since 2.57  1.96 Conclude that on the basis of the data, there is an indication that the two population means are not equal. p value: 0.0102 (total area to the right of 2.57 and the left of -2.57) • Confidence Interval x x z 1 − 2 (1−a / 2 )  SE = 1.1  1.96  0.4282 From 0.26 to 1.94 Since the confidence interval does not capture the hypothetical mean difference (0), It is suggested that the two populations (normal and downs syndrome) means (uric acid) are significantly different at a = 0.05 -Suppose in the previous example the question of interest was is mean uric acid is higher in the population of downs syndrome than in that of normal subjects: -The z-test becomes right sided with single critical value of 1.65. H0 and HA are to be restated as follows H0: μ1 – μ2 ≤ 0, HA: μ1 – μ2  0 Or H0: μ1 ≤ μ2, HA: μ1  μ2 -Because the hypothesized difference is still zero Z-stat is not changed and still calculated as 2.57, well higher than the critical value and thus H0 is rejected. • Confidence Interval x x z 1 − 2 (1−a )  SE = 1.1  1.65  0.4282 From 0.39 to 1.81 -LCL is higher than the hypothesized difference of 0, Thus consistent with above conclusion that H0 is rejected -Suppose in the previous example the question of interest was is mean uric acid is1 mg% higher in the population of downs syndrome than in that of normal subjects: -The z-test is right sided with single critical value of 1.65. H0 and HA are to be restated as follows H0: μ1 – μ2 ≤ 1, HA: μ1 – μ2  1 -Because the hypothesized difference is 1. z= (4.5 − 3.4) − 1 0.1 = = 0.234 0.4282 0.4282 -z-stat is lower than the critical value, H0 is not rejected. -Confidence Interval is the same as in the previous slide: 0.39 to 1.81. LCL is lower than the hypothesized difference of 1,thus consistent with above conclusion that H0 is not rejected t-test of Independent Samples (unknown and equal  of populations), (Pooled ttest) • Critical values: Total degrees of freedom (tdf = (n1-1)+(n2-1) = n1+ n2 -2 Two tailed: t(1-a/2, n1+ n2 -2) Right tailed: t(1-a, n1+ n2 -2) Left tailed: t (a, n1+ n2 -2) or -t(1-a, n1+ n2 2) t-Test (Pooled t-test) of Independent Samples (unknown and equal  of populations) • SE: When the population standard deviations are unknown but assumed to be equal we may “pool” the standard deviations (S1 and S2) using the following equation to give pooled S (Sp): Sp is weighted averages of sample standard deviations accounting for the contribution of their sizes and thus the test is termed pooled t-test (n − 1)s + (n − 1)s s = n +n −2 2 1 1 2 2 2 p 1 SE = s 2 p n 1 + s 2 p n 2 2 SE = s p (df 1)s12 + (df 2)s 22 sp = Tdf 1 1 + n1 n 2 E.g. A new formulation of fast disintegrating tablets was designed to perform faster on the average than an old formulation. To test that hypothesis, the times in seconds it took each tablet formulation to disintegrate were measured are recorded. The results are shown in the following table. New formulation Old formulation 42.1 41.3 42.4 43.2 41.8 42.7 43.8 42.5 43.1 44.0 41.0 41.8 42.8 42.3 42.7 43.6 43.3 43.5 41.7 44.1 Mean = 42.14, s1 = 0.683 Mean = 43.23, s2 = 0.750mg Do the data provide sufficient evidence to conclude that, on the average, the new formulation disintegrate faster? Meaning does it have shorter disintegration time. Perform the required hypothesis test at the 5% level of significance assuming equal variability of disintegration time between the two formulations Critical value: Left-tailed test total degrees of freedom =10+10−2 =18 Critical value = tα,18=−t0.05,18 = −1.734 or = -t1-α,18=−t0.95,18 = −1.734 Rejection region t* <−1.734 H0: μ1 – μ2 ≥0, HA: μ1 – μ2  0 (n − 1)s + (n − 1)s s = n +n −2 2 1 1 2 2 2 p 1 2 9  (0.683) + 9  (0.75) s = = 0.717 10 + 10 − 2 2 2 p ( x − x ) − ( −  ) tstat = = SE (42.14 − 43.23) − 0 tstat = = −3.4 0.447 1 2 1 2 0 Since t-stat is less than t-critical (− 3.4<−1.734). H0 is rejected at a = 0.05. Thus the mean disintegration time for the new formulation is significantly shorter (faster disintegration) than the old one SE = s p 1 1 + n n 1 SE = 0.717 2 1 1 + = 0.447 10 10 • Confidence Interval x1 − x 2  t (a ,n1+ n2 − 2)  SE = − 1.09  1.734  0.447 From - 1.87 to - 0.32 Since UCL is less than the hypothetical mean difference (0), there is enough evidence to reject the null hypothesis, and thus the new formulation significantly disintegrates faster than the old formulation at a = 0.05 t-Test (Unpooled t-test) of Independent Samples (unknown and unequal  of populations) • SE is not calculated using the pooled SD, but rather the SDs or variances of the samples, and thus the test is termed as unpooled t-test. • 2 s 2 s 1 SE = + n n 2 1 2 • SE is used in the calculation of t-statics as follows ( x − X ) − ( −  ) ( x − X ) − ( −  ) t − stat = = s s SE + n n 1 2 1 2 2 1 2 1 2 2 0 1 2 1 2 0 Critical value(s) for Unpooled t-test We can not use the t-table to find the critical values for total df= n1+n2-2. Instead of finding the critical values from the tables for each sample associated with sampled degrees of freedom, we need to compute it taking into consideration the critical values for each sampling distributions and the weight of each sample as follows. Example Researchers wish to know if two populations differ with respect to the mean value of the total serum complement activity (CH50). The data consist of CH50 determinations on apparently normal subjects (n2=20 ) and subjects with disease (n1=10 ). The sample means and standard deviations are: x1 = 62.6 s1 = 33.8 __ x 2 = 47.2 s 2 =10.1 The research team want to know if they can conclude that the two population means are different. Assuming that both populations are approximately normally distributed, construct the proper hypothesis and test it?. The populations are normally distributed with unknown variances that are unequal. With this in mind we will use t’-stat.

Difference Between Two Populations Independent Samples (Z and T Tests)

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