Hypothesis Testing for Population Proportions and Means: Independent Samples, Study notes of Business Statistics

Download Hypothesis Testing for Population Proportions and Means: Independent Samples and more Study notes Business Statistics in PDF only on Docsity! Chapter 9 Key Ideas Hypothesis Test (Two Populations) Section 9-1: Overview In Chapter 8, discussion centered around hypothesis tests for the proportion, mean, and standard deviation/variance of a single population. However, often researchers want to compare two different populations. For example, surgeons may want to try out a new surgical technique for a certain ailment, but they are not sure if it will be better. To test this, they could take two samples of people. In one sample, they could use the existing technique, and in the other, they could use the new technique. By comparing survival proportions of the two groups, they could then determine whether the new sample is any better. In this case, the first population is all people who would receive the existing technique. The second population is all people who would receive the new technique. By using the methods discussed in this chapter, such inference can be done. Section 9-2: Inferences About Two Proportions To use the methods described in this section, we first need to rely on a few conditions that must be met for everything to work properly. Conditions 1. The proportions are taken from two simple random samples which are independent. Here, independent means that observations from the first population are not related to, or paired with, observations from the second population. 2. For each of the two samples, there are at least 5 successes and 5 failures. Notation p1 = Population Proportion (Population 1) p2 = Population Proportion (Population 2) n1 = Sample Size (Population 1) n2 = Sample Size (Population 2) x1 = Number of Successes (Population 1) x2 = Number of Successes (Population 2) 1 1 1ˆ n x p  = Sample Proportion (Population 1) 2 2 2ˆ n x p  = Sample Proportion (Population 2) 11 ˆ1ˆ pq  22 ˆ1ˆ pq  21 21 nn xx p    = Pooled Sample Proportion pq 1 The Test The goal is to test the hypotheses given by: H0: p1 = p2 H1: p1 ≠ p2 (or p1 > p2 or p1 < p2) In other words, we test whether the proportions for each population are equal vs. whether they are different in some way. Test Statistic: 21 21 21 2121 )ˆˆ(or, )()ˆˆ( n qp n qp pp Z n qp n qp pppp Z       The critical values and P-Values come from the standard normal distribution. Decisions are made in exactly the same way as in Chapter 8. (For the test statistic, we assume H0 is true, which means p1 – p2 = 0) Example Close to an election, ballot issues #3 and #4 are very controversial. Researchers want to see whether there is a difference in the proportions of people who support issue #3 and those who support issue #4. They randomly sample 200 total people. They ask 100 of these people whether they support issue #3, to which 56 say “yes”. They ask the other 100 people whether they support issue #4, to which 45 say “yes”. Is there a difference in the proportions of supporters for each issue? Test this with α = 0.05. Solution From the information given, we see that: n1 = 100, x1 = 56, 56.0ˆ1 p , n2 = 100, x2 = 45, 45.0ˆ 2 p 495.0,505.0 200 101 100100 4556 21 21        q nn xx p H0: p1 = p2 H1: p1 ≠ p2 Test Statistic: 56.1 0707.0 11.0 100 )495.0(505.0 100 )495.0(505.0 45.056.0)ˆˆ( 21 21        n qp n qp pp Z Traditional Method  Since H1 has a “≠” sign, we want to find the critical value that has an area of α/2 above it and α/2 below it on the standard normal distribution (i.e. we want the value 2 Z ).  From the table, this cut-off value with an area of 0.025 above is 1.96.  Now we compare the test statistic to 1.96 and find that Z = 1.56 < 1.96.  This means that Z is not in the critical region (shaded area).  Therefore, we do not reject H0. In other words, we conclude that there is not enough evidence to claim that there is a difference in proportions of supporters for the two issues. P-Value Method  Since H1 has a “≠” sign, we want to find area above |Z| = 1.56 and below -|Z| = -1.56 for the standard normal distribution.  From the Z-Table, this area is 2(0.0594) = 0.1188.  Now we compare this area to α and see that 0.1188 > 0.05.  Therefore, we do not reject H0. In other words, we conclude that there is not enough evidence to claim that there is a difference in proportions of supporters for the two issues. Section 9-3: Inferences About Two Means: Independent Samples In similar fashion to testing for differences in proportions, one may also wish to test for a difference in the means of two populations. In the interests of time, we will consider the most general case, where the population standard deviations are unknown, and no assumptions are made about them. Better hypothesis tests exist when these value are both known, or are unknown but assumed to be equal. Consult the textbook for more information on these tests. Again, certain requirements must be met for the techniques discussed in this section to be theoretically sound. Conditions 1. σ1 and σ2 are unknown and no assumption is made about the equality of σ1 and σ2. 2. The two samples are independent. 3. Both samples are simple random samples. 4. Either both populations are normally distributed or n1 and n2 are both greater than 30. Notation μ1 = Population Mean (Population 1) μ2 = Population Mean (Population 2) n1 = Sample Size (Population 1) n2 = Sample Size (Population 2) 1x = Sample Mean (Population 1) 2x = Sample Mean (Population 2) s1 = Sample Standard Deviation (Population 1) s2 = Sample Standard Deviation (Population 2) Degrees of Freedom = df = min(n1, n2) – 1 (so df is still n – 1, but here n is the smaller of the two sample sizes) The Test The goal is to test the hypotheses given by: H0: μ1 = μ2 H1: μ1 ≠ μ2 (or μ1 > μ2 or μ1 < μ2) In other words, we test whether the means for each population are equal vs. whether they are different in some way. Test Statistic: 2 2 2 1 2 1 21 2 2 2 1 2 1 2121 )(or, )()( n s n s xx t n s n s xx t        The critical values and P-Values come from the Student t distribution with Degrees of Freedom = min(n1, n2) – 1. Decisions are made in exactly the same way as in Chapter 8. (For the test statistic, we assume H0 is true, which means μ1 – μ2 = 0) Example Researchers want to see if the average money spent per week by tourists in Chicago is less than the average money spent per week by tourists in New York City. They take a sample of 60 tourists from Chicago and 56 tourists from NYC. Of the Chicago tourists, average spending was $635, with a standard deviation of $50. Of the NYC tourists, the average spending was $650, with a standard deviation of $30. Run a hypothesis test with α = 0.05. Solution From the information given, we see that: n1 = 60, 6351 x , s1 = 50, n2 = 56, 6502 x , s2 = 30, df = min(n1, n2) – 1 = 56 – 1 = 55 H0: μ1 = μ2 H1: μ1 < μ2 Test Statistic: 974.1 599.7 15 56 30 60 50 650635)( 22 2 2 2 1 2 1 21          n s n s xx t Traditional Method  Since H1 has a “<” sign, we want to find the critical value that has an area of α below it on the t distribution (i.e. we want the value t ).  From the table, this cut-off value with an area of 0.05 below is -1.673.  Now we compare the test statistic to -1.673 and find that t = -1.974 < -1.673.  This means that t is in the critical region (shaded area).  Therefore, we reject H0. The evidence suggests that Chicago tourists pay less per week than those in New York City. P-Value Method  Since H1 has a “<” sign, we want to find area below t = -1.974 for the t distribution.  Now, notice that the t-Table does not allow one to directly find this area. However, for df = 55 we see that -2.004 has an area below of 0.025 and -1.673 has an area below of 0.05.  Since -2.004 < t < -1.673, the p-value will fall between 0.025 and 0.05. (see picture)  Now we compare this area to α and see that 0.025 < p < 0.05 = α.  Since p < 0.05, we reject H0. The evidence suggests that Chicago tourists pay less per week than those in New York City. Example One question on everyone’s mind is whether there is a difference in the average number of pets owned by Columbus families and the average number of pets owned by Cleveland families. Researchers set out to answer this important question. They sampled 35 Columbus families and 48 Cleveland families. Of the Columbus families, the average number of pets was 2.4, with a standard deviation of 1.4. Of the Cleveland families, the average number of pets was 1.9, with a standard deviation of 0.9. Run a hypothesis test (α = 0.05) to see if there is a difference in the average number of pets owned by families in the two cities. Solution From the information given, we see that: n1 = 35, 4.21 x , s1 = 1.4, n2 = 48, 9.12 x , s2 = 0.9, df = min(n1, n2) – 1 = 35 – 1 = 34 H0: μ1 = μ2 H1: μ1 ≠ μ2 Test Statistic: 852.1 2700.0 5.0 48 9.0 35 4.1 9.14.2)( 22 2 2 2 1 2 1 21        n s n s xx t Traditional Method  Since H1 has a “≠” sign, we want to find the critical value that has an area of α/2 above it and α/2 below it on the t distribution (i.e. we want the value 2 t ).  From the table, this cut-off value with a one-tailed area of 0.025 is 2.032.  Now we compare the test statistic to 2.032 and find that t = 1.852 < 2.032.  This means that t is not in the critical region (shaded area).  Therefore, we do not reject H0. There is not sufficient evidence to conclude that the average number of pets is different in the two cities. P-Value Method  Since H1 has a “≠” sign, the p-value will be the area above |t| = 1.852 and below -|t| = -1.852 for the t distribution.  Again, notice that the t-Table does not allow one to directly find this area. However, for df = 34 we see that 1.691 has a two-tailed area of 0.10 and 2.032 has a two-tailed area of 0.05.  Since 1.691 < t < 2.032, the p-value will fall between 0.05 and 0.10. (see picture)  Now we compare this area to α and see that p > 0.05 = α.  Since p > 0.05, we do not reject H0. There is not sufficient evidence to conclude that the average number of pets is different in the two cities.