Section 6 Comparing Two Samples William Christensen, Ph.D.

Comparing Two Samples In the last section we learned how to test hypotheses, or in other words, use statistics to test a claim about a population parameter. We build on that knowledge in this section and learn how to compare two sets of sample data. That is, we learn how to test a claim that two samples come from the same (or different) populations.

Large Independent Samples (n>30)
Comparing two means Large Independent Samples (n>30)

Assumptions The two samples are independent
The samples are not related or paired with each other If the samples are related (dependent) they are often referred to as matched pairs or paired samples – we’ll learn to deal with them later The two sample sizes are large. That is, n1  30 and n2  30. Both samples are random samples

Test Statistic for Comparing Two Population Means
The “test statistic” for comparing the two population means is calculated as follows (there is no Excel shortcut): The formula requires: The means of both samples (x-bar1 and x-bar2) The population (or sample) standard deviations from each sample Most of the time we don’t have σ, so we use s The sample size (n) for each sample Note: we ALWAYS assume that the population means (µ1 and µ2) net out to 0. In other words, always assume that (µ1 - µ2) = 0

Critical Value(s) for Comparing Two Population Means
The “critical value(s)” for comparing two population means are found exactly the same as in Section 5 (hypothesis testing of means with large samples) – see the next slide for a review of how to find those critical values using Excel We again use HYPOTHESIS TESTING when we compare two population means. The 3 possible sets of hypotheses are: H0: µ1 = µ2 and H1: µ1 ≠ µ2 (two-tailed test) H0: µ1  µ2 and H1: µ1  µ2 (right-tail test) H0: µ1  µ2 and H1: µ1  µ2 (left-tail test)

µ Critical Value Example α = 0.05
CV = -1.96 Null Hypothesis Alternative Hypothesis Two-tail test = ≠ Right-tail test   Left-tail test   CV = 1.96 Area = α/2 = 0.025 Area = α/2 = 0.025 CV = CV = 1.645 Area = α = 0.05 Area = α = 0.05 Notice how critical values for the left-tail are ALWAYS negative and critical values for the right-tail are ALWAYS positive

Comparing Two Means: EXAMPLE
The Coke vs. Pepsi data set on the class website includes the weights (in pounds) of samples of regular Coke and regular Pepsi. Sample statistics are shown. Use the 0.01 significance level to test the claim that the mean weight of regular Coke is the same as the mean weight of regular Pepsi (claim is H0: µcoke = µpepsi and H1: µcoke ≠ µpepsi )

Coke Versus Pepsi Example
The following info (you can download data off website or the CD-ROM that comes with the Triola text) can easily be calculated from the data sets using Excel mean/average and standard deviation functions. Regular Coke Regular Pepsi n x-bar s

Claim: µ1 = µ2 with α = 0.01 H0 : µ1 = µ2 and H1 : µ1  µ2 Accept H1 Accept H0 Accept H1 Z = Z = 2.575 z = 0 Change left-side neg. value to pos. or use =NORMSINV(0.995)

Next, calculate the test statistic using the formula: Z = Z = 2.575 z = 0 Conclusion: Accept H1: µ1 ≠ µ2 because the test statistic is in the critical region (outside the critical value)

Finally, we conclude that the mean weights of Coke and Pepsi are different (not equal to each other). In fact, you can see from the sample means that Pepsi weighs more than Coke, and we found that difference to be statistically significant. Z = Z = 2.575 z = 0

Small Independent Samples (n30)
Comparing two means Small Independent Samples (n30)

Assumptions The two samples are independent
At least one of the sample sizes is small. That is, n1  30 OR n2  30. Both samples are random samples from normally distributed populations

Test Statistic for Comparing Two Population Means
The “test statistic” for comparing two population means where at least one sample size  30 is calculated as follows: The formula requires: The means of both samples (x-bar1 and x-bar2) The sample standard deviations from each sample The sample size (n) for each sample, where df = smaller of n1-1 or n2-1 Note: we ALWAYS assume that the population means (µ1 and µ2) net out to 0. In other words, always assume that (µ1 - µ2) = 0

Critical Value(s) for Comparing Two Population Means
The “critical value(s)” for comparing two population means are found exactly the same as in Section 5 (hypothesis testing of means with small samples) – we use Excel function TINV(probability,degrees_freedom) - see the next slide for a review and example. Important note: we use the full alpha value (not alpha/2) for a 2-tailed test and must use 2 x alpha for a 1-tailed t-test. We again use HYPOTHESIS TESTING when we compare two population means. The 3 possible sets of hypotheses are: H0: µ1 = µ2 and H1: µ1 ≠ µ2 (two-tailed test) H0: µ1  µ2 and H1: µ1  µ2 (right-tail test) H0: µ1  µ2 and H1: µ1  µ2 (left-tail test)

µ Critical Value Example α = 0.05, smaller sample size =15
CV = -2.14 Null Hypothesis Alternative Hypothesis Two-tail test = ≠ Right-tail test   Left-tail test   CV = 2.14 Area = α/2 = 0.025 Area = α/2 = 0.025 Note: you must make left-side t-values negative CV = -1.76 CV = 1.76 Area = α = 0.05 Area = α = 0.05 Notice how critical values for the left-tail are ALWAYS negative and critical values for the right-tail are ALWAYS positive

Comparing Two Means: EXAMPLE
People spend huge sums of money for the purchase of magnets to treat pain. Researchers conducted a study to determine whether magnets are effective in treating back pain. Pain was measured using the visual analog scale, with the results given below (larger numbers mean more effective pain reduction). Use α=0.05 to test the whether those treated with magnets had greater pain reduction than those given a fake/sham treatment (similar to a placebo). Does it appear that magnets are effective in treating back pain? How might larger samples effect our results? Reduction in pain level after magnet treatment: n=20, mean=0.49, s=0.96 Reduction in pain level after sham treatment: n=20, mean=0.44, s=1.4 Note: our hypotheses (related to the amount of pain reduction) are: H0: µmagnet  µsham H1: µmagnet  µsham )

Magnets & Pain Reduction Example
α=0.05 (use 2x α for 1-tail test), df=smaller of n1-1 or n2-1 =20-1=19 H0 : µmagnet  µsham and H1 : µmagnet  µsham (right-tail test) Accept H0 Accept H1 Z = 1.729 t = 0

Next, calculate the test statistic using the formula: Z = 1.729 t = 0 Conclusion: Accept H0: µmagnet  µsham

Conclusion: Accept H0: µmagnet  µsham The results do not support the claim that magnets are effective in reducing back pain. Even though the mean pain reduction with magnets (0.49) was greater than with the sham treatment (0.44), the standard deviations were very high (0.96 and 1.4), making it difficult to statistically find any meaningful difference between the means. Even with larger samples, unless the standard deviations came down, it would be difficult to show a difference between the means.

Comparing two means Matched pair Samples

Assumptions The sample data consist of matched pairs.
The samples are simple random samples. If the number of pairs of sample data is small (n  30), then the population of differences in the paired values must be approximately normally distributed. Note: taking “before-and-after” measurements to create and then compare two sample sets is a typical example of matched pairs

Notation for Matched Pairs
µd = mean value of the differences d between the population of paired data d-bar = mean value of the differences d between the paired sample data (e.g., the average difference between “before” and “after” measurements) sd = standard deviation of the differences d for the paired sample data n = number of pairs of data

Test Statistic for Comparing Matched Pairs
The “test statistic” for comparing matched pairs is calculated as follows (there is no Excel shortcut): The formula requires: The mean difference between the matched pairs from the two samples (often a “before” and an “after” sample), represented by d-bar Usually, the hypotheses assume there is no difference between the populations. In other words, the most common null hypothesis for matched pairs is H0: µd = 0. Whether µd = 0 or some other value, the value stated in the hypotheses is what goes into the test statistic formula. Of course, with µd =0, µd simply drops out of the formula. The standard deviation of the differences between the matched pairs of the two samples (see Example) n represents the number of matched pairs we have

Critical Value(s) for Testing Matched Pairs
The “critical value(s)” for comparing matched pairs are found as follows: Use Excel function =NORMSINV(probability) when n>30 Use Excel function =TINV(probability,df) when n  30, where df (degrees of freedom) = n – 1 (note: for TINV ALWAYS use α for 2-tailed tests and αx2 for 1-tailed tests) We use HYPOTHESIS TESTING to compare matched pairs. The 3 most common hypotheses are: H0: µd = 0 and H1: µd ≠ 0 (two-tailed test) H0: µd  0 and H1: µd  0 (right-tail test) H0: µd  0 and H1: µd  0 (left-tail test)

Do Male Students Exaggerate Their Heights? EXAMPLE
Use the following data to test the claim that male students exaggerate their heights (i.e., the difference between what they report and their actual heights is greater than 0). Use a 95% confidence level. 7.9 This value is so large that it seems there must be some kind of mistake. How could anyone reasonably say the are almost 8 inches taller than they really are. Because of this fact, and the fact that such an abnormally large value would drastically affect our results, we toss out this “outlier” and are left with 11 useable differences d-bar and std.dev of d do not include the outlier value

Male Height Example Claim: µd  0 with α = 0.05
H0 : µd  0 and H1 : µd  0 (right-tail test) Accept H0 Accept H1 t = 1.81 t = 0 Step 1: write out the hypotheses, graph the problem, and find the critical value

Male Height Example Conclusion: Accept H1
Step 1: calculate the test statistic and form conclusion Accept H0 Accept H1 t = 1.81 t = 0 Conclusion: Accept H1 The test statistic is outside the critical value (in the critical region)

Male Height Example Conclusion:
Accept H1: µd  0. In other words, we find there is a statistically significant difference between what male students say their height is compared to their actual height. Accept H0 Accept H1 t = 1.81 t = 0

Comparing two proportions

Assumptions We have proportions from two independent random samples
For both samples, the conditions n*p  5 and n*q  5 are satisfied See Section 5, Hypothesis Testing for Population Proportions for a review

Notation for Proportions
p = population proportion (always between 0 and 1) p-hat = = sample proportion q-hat = n = size of the sample x = number of successes in the sample. Sometimes we are given p-hat directly and sometimes we must calculate p-hat by using the simple formula p-hat = x / n. (e.g., if 12 out of 24 cars are volkswagens, then the proportion of volkswagens in the sample is x/n or 12/24 = 0.50) With two samples we mark these variables as 1 or 2, which designates which sample they came from

Notation for Proportions
When comparing two population proportions, our test statistic requires us to calculate a “pooled estimate” of p1 and p2, which we call p-bar The formula for p-bar is as follows. You should note this formula (along with all test statistic formulas) since you must know it for the exam.

Test Statistic for Comparing Population Proportions
The “test statistic” for comparing two population proportions is calculated as follows (there is no Excel shortcut): The formula requires: The sample proportion (p-hat) from each sample ALWAYS assume there is no difference between the population proportions. In other words, always assume p1 - p2 = 0 in the formula The pooled p estimate (p-bar) and q-bar calculated using the formulas in the previous slide The size of each of the two sample (n1 and n2)

Critical Value(s) for Comparing Two Population Proportions
The “critical value(s)” for comparing two population proportions are found as exactly the same as any other z-values, using Excel function NORMSINV(probability): We use HYPOTHESIS TESTING to compare two population proportions. The 3 possible hypotheses are: H0: p1 = p2 and H1: p1 ≠ p2 (two-tailed test) H0: p1  p2 and H1: p1  p2 (right-tail test) H0: p1  p2 and H1: p1  p2 (left-tail test)

Desire for Marriage EXAMPLE
In a Time/CNN survey, 24% of 205 single women said that they “definitely want to get married.” In the same survey, 27% of 260 single men gave that same response. Using α = 0.05 test the claim that there is no difference between single men and single women regarding their desire to get married. H0: pmen = pwomen (who want to get married) H1: pmen ≠ pwomen (who want to get married)

α 0.05, H0 : pm = pw and H1 : pm  pw (2-tail test) Accept H1 Accept H0 Accept H1 Step 1: write out the hypotheses, graph the problem, and find the critical value Z = Z = 1.96 z = 0 Change left-side neg. value to pos. or use =NORMSINV(0.975)

Next, calculate the test statistic using the formula: BUT, before we can use this test statistic formula we must first calculate pooled p (p-bar) and q-bar using the formulas we learned. We find pooled p = (x1+x2)/(n1+n2), where x1=0.24*205=49 single women and x2=0.27*260=70 single men, so pooled p = (49+70)/( )=0.256, so pooled q=1-(pooled p)= =0.744

Now we can calculate the test statistic using the formula: Z = Z = 1.96 z = 0 Conclusion: Accept H0: pm=pw

Conclusion: Accept the null hypothesis (H0) that the population proportions are equal. In other words, we find no difference between the proportion of single men and the proportion of single women who “definitely want to get married”. Z = Z = 1.96 z = 0

Comparing population variances or standard deviations
Using two samples to compare population variances

Assumptions We have variances from two independent random samples
The two populations are each normally distributed See Section 5, Hypothesis Testing for Variances and Standard Deviation for a general review. However, note that in Section 5 we used the 2 distribution, whereas in this Section we use a similar, but new distribution called the “F” distribution

Notation for Variance and Standard Deviation Testing
s = standard deviation of sample σ = standard deviation of population s2 = variance of sample σ2 = variance of population Since we have two samples we are using to compare two populations, we also label these variables as “1” or “2”, with the larger sample variance/standard deviation ALWAYS labeled s1 and the smaller ALWAYS labeled s2.

F - distribution Not symmetric
nonnegative values only Not symmetric  Use Excel function FINV to find the F critical value. All one-tailed F tests are right-tailed tests and there are no negative F-distribution values since the origin is at 0.

F–distribution Finding Critical Value
Probability = α for one-tailed tests and α/2 for two-tailed tests. FOR ALL F-TESTS (one or two-tailed), the right critical value is the ONLY VALUE we need to check and the only value returned by the Excel function FINV deg_freedom1 = the degrees of freedom for sample1, which is ALWAYS the sample with the larger variance/standard deviation. Degrees of freedom is simply sample size minus 1 (df=n-1) for sample 1. deg_freedom2 = the degrees of freedom for sample2, which is ALWAYS the sample with the smaller variance/standard deviation. Degrees of freedom is simply sample size minus 1 (df=n-1) for sample 2.

Critical Value(s) for Comparing Two Population Variances or Std
Critical Value(s) for Comparing Two Population Variances or Std. Deviations The “critical value” for comparing two population variances or standard deviations are found using the F-distribution just described and the Excel function FINV(probability, df1, df2): We use HYPOTHESIS TESTING to compare two population variances. The 3 possible hypotheses are: H0: σ1 = σ2 and H1: σ1 ≠ σ2 (right-tail test with probability=α/2) H0: σ1  σ2 and H1: σ1  σ2 (right-tail test with probability=α) H0: σ1  σ2 and H1: σ1  σ2 (right-tail test with probability=α) Note: whether we compare variances or standard deviations, the results will be the same. Thus, these hypotheses can be written either for variances or standard deviation (shown here).

F–distribution Finding Critical Value Two Examples
F-critical value for α=0.05, one-tail test (left or right), where the sample with the larger variance has a sample size of 30 (df=n-1=30-1=29) and the sample with the smaller variance has a sample size of 50 (df=n-1=50-1=49) F-critical value for α=0.05, two-tail test, where the sample with the larger variance has a sample size of 45 (df=n-1=45-1=44) and the sample with the smaller variance has a sample size of 40 (df=n-1=40-1=39)

Test Statistic for Comparing Population Variances or Standard Deviations
The “test statistic” for comparing two population variances or standard deviations is as follows: The formula requires: The standard deviation or variance from each sample Note: if you are given sample variance (s2), do not square it again since variance is already the square of standard deviation. On the other hand, if given standard deviation (s) for each sample, make sure you square it as shown in the formula

Comparing Two Variances
Accept H0 (no difference between population variances) if the F-test statistic is less than the F-critical value Accept H1 (there is a difference between population variances) if the F-test statistic is greater than the F-critical value Accept H0 Accept H1 F-critical value (FINV) Probability (area in the tail) equals alpha for ALL one-tailed tests (left or right) and alpha/2 for all two-tailed tests. This area or probability is the first entry in the Excel FINV function

Discussion: Comparing Population Variances or Standard Deviations
If two samples have exactly equal variances, then the F test statistic would be 1 (see formula – if numerator and denominator are equal then F=1) Likewise, if the samples come from populations with dramatically different variances then we would expect the F test statistic to be a large number (the larger variance is always in the numerator, making big variance differences result in relatively large numbers, much greater than 1)

Coke vs. Pepsi (weight) Variance EXAMPLE
The Coke/Pepsi data set included in the text and listed on the class website includes sample weights of regular Coke and Pepsi. Each sample contains 36 observations (n1=36 and n2=36). The standard deviation for the weight of Coke is lbs., and the standard deviation for the weight of Pepsi is lbs. Use an alpha of 0.05 to test whether the variance in the weight of Coke and Pepsi is significantly/statistically different. Note: since the variance in the weight of Coke is greater than that of Pepsi, we must list it first and put it in the numerator (top) of the F test statistic calculation. H0: σ2coke = σ2pepsi (weight of cans) H1: σ2coke ≠ σ2pepsi (weight of cans) Note: this is “technically a two-tailed test, but we only have the ability to find, and are only interested in, the right-side critical value

Note: both samples are of the same size (36), so df1=36-1=35, and df2=36-1=35 Since this is “technically” a two-tailed test, the probability (area) equals alpha/2 = 0.05/2 = 0.025 Accept H0 Accept H1 F-critical value (FINV) F critical value = 1.96

The F test statistic is calculated using the formula: Conclusion: Since the F test statistic is less than the F critical value, we accept the null hypothesis (H0) that there is no statistically significant difference between the VARIANCES in the weight of Coke and Pepsi Accept H0 Accept H1 F-critical value (FINV) F critical value = 1.96

Section 6 Comparing Two Samples E N D William Christensen, Ph.D.

Section 6 Comparing Two Samples William Christensen, Ph.D.

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