Lesson 1: Comparison of Population Means Part c: Comparison of Two Means


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1 Lesson : Comparison of Population Means Part c: Comparison of Two Means Welcome to lesson c. This third lesson of lesson will discuss hypothesis testing for two independent means.
2 Steps in Hypothesis Testing. State the null hypothesis H 0 and the alternative hypothesis H a.. Calculate the value of the test statistic on which the test will be based. 3. Find the pvalue for the observed data. 4. State a conclusion. Recall the steps in hypothesis testing. First, we state the null and alternative hypothesis to address our research question. Next, calculate the test statistic. We then compare the test statistic to a density curve to find the pvalue. Finally, we compare this pvalue to the type I error probability to determine our conclusion.
3 Hypothesis Testing: Comparing Two Means Identify two independent populations. Draw a simple random sample of size n from population and a simple random sample of size n from population. Compute the mean for each sample. Formulate hypothesis test based on the difference of the means 3 Let s discuss the setting for hypothesis testing of two means. First, we take random samples from two independent populations. We wish to compare the average value mu from these two independent populations to see if they are similar or different. From each sample we compute an average value, xbar. We will then use these sample means to determine if the populations are centered in the same location or different locations. We look at a linear combination (xbarxbar) to help us make this decision. If xbarxbar is close to zero, the populations probably have the same center. If xbarxbar is not close to zero the populations may have different centers. 3
4 Hypothesis Testing: Comparing Two Means Step : State your hypotheses H 0 : μ  μ = 0 H a : μ  μ 0 (twosided) Or H a : μ  μ < 0 (onesided) Or H a : μ  μ > 0 (onesided) 4 First step is to represent our scientific question is the null and alternative hypothesis. The null: Ho: mm = 0 represents the condition that the populations are centered in the same spot. As with onesample and matched pairs hypothesis testing we can have a onesided or twosided alternative. The two sided alternative is that the means differ. That is mm does not equal zero. We could also look at the alternative that m is greater than m. This would be if we subtract m from m we would obtain a negative number. Or We could look at the alternative that m is greater than m. This would be if we subtract m from m we would obtain a positive number. 4
5 Two sample Problem with σ and σ known. z = ( x x) σ σ + n n 5 If the population standard deviation for both populations is known then the statistic we use is the z statistic. You can see the standard error of the difference of xbar and xbar is in the denominator. You will recall that when we make a linear combination of two means that the variances are additive. This is why the standard error in the denominator has a plus sign instead of a minus sign. As before, knowing the variance of the population is not typical. So we usually substitute the value of the sample variance in where the population variance is in this equation. 5
6 Use this t when s larger /s smaller > Two sample Problem with σ and σ unknown (do not assume σ =σ ). df= smaller of n  or n  t = ( x x) s n + s n 6 When we do this substitution we switch to using the student s t statistic. Notice in the title it also says do not assume s=s. There are two ways to calculate the standard error of the difference of samples means. The first is what you see in the denominator here. This is when we cannot assume that the population standard deviations are the same from the two independent populations. We do not know the populations standard deviations so we look at our sample standard deviations. If the larger sample standard deviation divided by the smaller sample standard deviation is greater than two, we do not assume sigma=sigma. This is a rule of thumb. It says our sample standard deviations are different enough we cannot assume sigma=sigma. Our degrees of freedom for this ttest are the smaller of n or n. 6
7 Two sample Problem with σ and σ unknown and assumption σ = σ. Use this t when s larger /s smaller < ( x x) μo t = s + p n n withs p with n + n ( n ) s + ( n ) s = n + n degreesof freedom 7 So what happens when s larger over s smaller is less than two? Well, we can assume sigma=sigma. If we make this assumption we use the t test given in this slide. Notice we pool the variances to create a common variance. This is called sp. We then use sp in the denominator to calculate the standard error of the difference of sample means. So why worry about whether or not to assume sigma=sigma? Notice the degrees of freedom when we make the assumption are n+n. This is more degrees of freedom than in the previous t test. When we have more degrees of freedom in the ttest we have more power to detect a difference should there be one. We want to make the assumption that the population variances are equal when we can. 7
8 SAS Example We have 0 students, 5 are randomly assigned to control and 5 are randomly assigned to treatment. Response times to a stimulus is measured for all 0 participants. Research question: Do the treatment scores come from a population whose mean is different from which the control scores were drawn? Control mean = 88.6 millisec Treatment mean = 0.6 millisec 8 Let s try an example. We will also introduce a bit of SAS to understand our example. Recall: SAS is a computer language that helps us analyze data. At this point you may have tried the SAS tutorial for the first assignment. If not, you may want to do this before moving on. Consider the example above. You have 0 students, 5 are randomly assigned to control and 5 are randomly assigned to treatment. The outcome of interest is response times to a stimulus. We are wondering of the treatment group and the control group come from populations whose mean values are different. The control sample mean was 88.6 milliseconds. The treatment sample mean was 0.6 milliseconds. 8
9 Data Response; Input Group $ Time; Datalines; C 80 C 93 C 83 C 89 C 98 T 00 T 03 T 04 T 99 T 0 ; Proc ttest data = response; Title "Ttest example"; class group; var time; run; SAS: Proc ttest This is a SAS program that reads in data and performs a two sample ttest. The first line Data response; tells SAS that we want to create a temporary data set called response. The next line Input Group $ Time; tells SAS that we have two variables. One is called Group and it is categorical. This is indicated by the dollar sign following the word Group. The second variable is called time. There is no designation of type of variable after time. SAS will assume it is quantitative if there is not a designation. Next we have Datalines; this tells SAS: Here comes the actual data. Following datalines is the data C and T are control and treatment. The data is separated by a single space with a new line for each person s data. Notice the semicolon is a line below the final piece of data. The data step is now complete. We can then do a procedure on this data. 9
10 Data Response; Input Group $ Time; Datalines; C 80 C 93 C 83 C 89 C 98 T 00 T 03 T 04 T 99 T 0 ; Proc ttest data = response; Title "Ttest example"; class group; var time; run; SAS: Proc ttest The procedure is Proc ttest SAS will analyze the data using a ttest. We tell SAS which data set by data=response; We can insert a title with the title command followed by the title in quotes. Next class group; tells SAS between which two groups we would like to perform the ttest. In our case it is the variable called Groups with control or treatment as group. var time; tells SAS that we want to analyze the outcome time; Finally, run; tells SAS go ahead and analyze using the procedure. Again, for more on SAS programming see the class SAS tutorials. 0
11 The TTEST Procedure Statistics Lower CL Upper CL Lower CL Upper Variable Group N Mean Mean Mean Std Dev Std Dev Std Dev Std Err Time C Time T Time Diff () Note: SAS Tests >Ho: μ=μ vs. Ha: μ not equal μ If everything ran without error you will see output. This is the first piece of output you see. We have variable time divided by our two groups, Control and treatment. We also have a row indicating the difference of control treatment outcomes. Let s start with row. We see this is our control group with n=5. We have the sample mean of 88.6 with a 95% confidence interval from 79.5 to We have a sample standard deviation of 7.3. SAS also gives us the 95% confidence interval for the standard deviation. Finally, SAS gives us the standard error of 7.3 divided by the square root of 5 with value 3.7. We have the same information in row for our treatment group. The third row has the confidence interval for mm from 0.8 to 5.. Notice this confidence interval does not contain zero. Later we will see why this is important to us. Finally, notice SAS is always testing the two sided hypothesis test: Ho: u=u vs. Ha: u not equal u.
12 TTests Variable Method Variances DF t Value Pr > t Time Pooled Equal Time Satterthwaite Unequal Note: SAS tests both with assumption σ =σ and σ not equal σ This is the output for the ttest statistic. Notice we have two results Pooled and Satterthwaite. These correspond to the two choices that we have for the t statistic. We need to decide which is appropriate for our analysis based on our data. If we go back to the previous slide we see the standard deviation for the control group over the standard deviation for the treatment group is greater than. This means we cannot assume the population standard deviations are the same. We need to choose the variances unequal option (Satterthwaite). Our degrees of freedom when we do this calculation by hand are the smalelr of n or n. SAS uses a formula to calculate more exact degrees of freedom, so our number will not match. The tvalue is with a two sided pvalue of At a = 0.05 we would reject the null and conclude response times are different.
13 Equality of Variances Variable Method Num DF Den DF F Value Pr > F Time Folded F Note: If Pr > F is < 0.05 do NOT assume σ =σ do not pool sample variances. 3 So why this final piece of output when we are able to draw our conclusion from the previous output? Well, SAS performs a test of equality for population variances instead of using our ratio as a rule of thumb. Generally, the results will agree. Here SAS has the Ho: sigma=sigma vs. Ha: sigma not equal sigma. The test statistic is an f with a pvalue of At a = 0.05 we would reject the null and conclude the population variance cannot be assumed to be equal. This would lead us to use the Satterthwaite test above as we had decided before. Recall: we said that the confidence interval for the difference of means does not contain zero. This corresponds to the two sided hypothesis test of the difference of means. If the confidence interval does not contain zero we would reject the null hypothesis of equal means. Note however; SAS calculates this confidence interval is based on the assumption that sigma=sigma. Your results may not always match if you cannot make this assumption. 3
14 Hypothesis Testing Comparing Two Means: An Example The effect of environmental exposure to lead on intellectual development is investigated using two randomly selected samples of 7 year old children from similar backgrounds but with different lead exposures. 4 Here is an example not using SAS to do our analysis. Lead has detrimental effect on intellectual development, especially when young children are exposed. The effect of environmental exposure to lead on intellectual development is investigated using two randomly selected samples of 7 year old children from similar backgrounds but with different lead exposures. 4
15 Hypothesis Testing Comparing Two Means: An Example Serum lead levels in group > 30 ug/dl Serum lead levels in group < 30 ug/dl 5 The two groups of children have different lead levels. One group had lead levels above 30 micrograms per deci liter. The other group had lead levels below 30 micrograms per deci liter. Researchers are wondering if children with lead levels above 30 micrograms per deci liter will score differently on intelligence tests than the children with lead levels that are lower. 5
16 Hypothesis Testing Comparing Two Means: An Example Does a significant difference exist between the mean intelligence test score in these two groups? The data for intelligence test score is summarized below: n =6 n =4 x =94 x =0 s =7 s =8 6 A random sample was drawn from each populations and the children were given an establish intelligence test. The results are as follows. Of the 6 kids in the higher lead level group the average score was 94 with a standard deviation of 7 points. Of the 4 kids with the lower lead level the average score was 0 with a standard deviation of 8 points. Is there a difference between mean intelligence test scores for the different populations? The sample averages are different, but this could have happened by chance. We can do a hypothesis test of two means to see if the means are significantly different. 6
17 Hypothesis Testing Comparing Two Means: An Example Step : State your hypotheses (set α=.0) H 0 : μ  μ = 0 H a : μ  μ 0 (twosided) Step : Calculate your test statistic t = ( x x ) ( μo) (94 0) 0 = = s n 6 4 s n The first step is to write our null and alternative hypothesis. Remember this is a twosided hypothesis. We did not specify that either group would be lower. Next we need to decide with test statistic to use. We do not know sigma for either populations so we will use a t statistic, but which one? If we look at the ratio of the sample standard deviations, the larger over the smaller, we see that this value is greater than two. We do not assume the population standard deviations are the same and we do not pool the variance. The unpooled t value is
18 Hypothesis Testing Comparing Two Means: An Example Step 3: Calculate the pvalue * p( t.789) = *.005 =.005 Degrees of freedom = 40 Step 4 : Make a conclusion pvalue < α, then reject H o The data suggests a significant mean difference exists in intelligence scores for the two groups. 8 Step 3 is calculate our pvalue. We look at times the probability that a t with 40 degrees of freedom is less than Our conclusion is to reject Ho. This means the scores are significantly differenct at the alpha 0.05 level. Our sample means were not likely different by chance. It is likely the average score for the populations would have different locations. In other words, the data suggests a significant mean difference exists in intelligence scores for the two groups. 8
19 Hypothesis Testing: A Pooled Ttest Example Independent random samples selected from two normal populations produced the sample means and standard deviations shown in the table: Sample size Mean Sample Standard deviation Sample Test the null hypothesis that the population means are equal vs. the alternative that they are not equal. Let α=0.05, this means we will reject the null hypothesis when it is true 5% of the time. 7 Sample 9 Here is an example where we would choose to use the pooled t test. Independent random samples selected from two normal populations produced the following results. There were 7 subjects in sample with a mean value of 5.4 and a sample standard deviation of 3.4. There were subjects in sample with a mean value of 7.9 and a standard deviation of 4.8. I want to test the null hypothesis that the population means are equal vs. the alternative that they are not equal. Let =0.05, this means we will reject the null hypothesis when it is true 5% of the time. Notice the ratio of the larger sample standard deviation divided by the smaller sample standard deviation is less than two. 9
20 Hypothesis Testing: A Pooled Ttest Example. Ho: μ  μ =0 Ha: μ  μ 0. Calculate the test statistic: s p and ( n = ) s + ( n ) s n + n ( x x) μo t = = s + p n n ( ) 6.4 (7 )(3.4 ) + ( )(4.8 = =.645 ) = We choose to use the pooled t test. First, we need to figure out what the pooled estimate of the variance would be. We call this sp. It s value is 6.4. We use this value in the denominator for the t statistic. Our statistic yields a value of
21 Hypothesis Testing: A Pooled Ttest Example. This test statistic follows the tdistribution with 7 degrees of freedom. 3. Pvalue=*P(T>.645) =0. (answer from calculator). 4. Therefore we fail to reject the null hypothesis based on an αlevel of 0.05 and conclude that the two population means are not likely different. This test statistic follows the tdistribution with 7 degrees of freedom. This is n + n. The Pvalue=*P(T>.645) =0.. I obtained this answer from a calculator of a computer program. You cannot get an exact value using the table in your book. Our pvalue leads us to fail to reject the null hypothesis based on an?level of 0.05 and conclude that the two population means are not likely different. This assumes we had the power to detect a difference should there be one. This ends lesson c. Please go to self assessment c.
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