Comparing Two Populations or Treatments

Comparing Two Populations or Treatments
Chapter 11 Comparing Two Populations or Treatments

Section 11.1 Inferences Concerning the Difference between Two Populations or Treatment Means Using Independent Samples

On the next slide we will investigate this distribution.
Suppose we have a population of adult men with a mean height of 71 inches and standard deviation of 2.5 inches. We also have a population of adult women with a mean height of 65 inches and standard deviation of 2.3 inches. Assume heights are normally distributed. Suppose we take a random sample of 30 men and a random sample of 25 women from their respective populations and calculate the difference in their heights (man’s height – woman’s height). If we did this many times, what would the distribution of differences be like? On the next slide we will investigate this distribution.

71 Male Heights 65 Female Heights Randomly take one of the sample means for the males and one of the sample means for the females and find the difference in mean heights. sM = 2.5 sF = 2.3 Suppose we took repeated samples of size n = 25 from the population of female heights and calculated the sample means. We would have the sampling distribution of xF Suppose we took repeated samples of size n = 30 from the population of male heights and calculated the sample means. We would have the sampling distribution of xM. 71 65 6 Doing this repeatedly, we will create the sampling distribution of (xM – xF) xM - xF

The sampling distribution is normally distributed with
Heights Continued . . . Describe the sampling distribution of the difference in mean heights between men and women. What is the probability that the difference in mean heights of a random sample of 30 men and a random sample of 25 women is less than 5 inches? The sampling distribution is normally distributed with 6

Notation - Comparing Two Means

More Notation - Comparing Two Means

The variance of the differences is the sum of the variances.
Properties of the Sampling Distribution of x1 – x2 If the random samples on which x1 and x2 are based are selected independently of one another, then 1. and The sampling distribution of x1 – x2 is always centered at the value of m1 – m2, so x1 – x2 is an unbiased statistic for estimating m1 – m2. Mean value of x1 – x2 3. In n1 and n2 are both large or the population distributions are (at least approximately) normal, x1 and x2 each have (at least approximately) normal distributions. This implies that the sampling distribution of x1 – x2 is also (approximately) normal. The variance of the differences is the sum of the variances.

We must know s1 and s2 in order to use this procedure.
The properties for the sampling distribution of x1 – x2 implies that x1 – x2 can be standardized to obtain a variable with a sampling distribution that is approximately the standard normal (z) distribution. When two random samples are independently selected and n1 and n2 are both large or the population distributions are (at least approximately) normal, the distribution of We must know s1 and s2 in order to use this procedure. If s1 and s2 is unknown we must use t distributions. is described (at least approximately) by the standard normal (z) distribution.

Two-Sample t Test for Comparing Two Populations
Null Hypothesis: H0: m1 – m2 = hypothesized value Test Statistic: The appropriate df for the two-sample t test is The computed number of df should be truncated to an integer. A conservative estimate of the P-value can be found by using the t-curve with the number of degrees of freedom equal to the smaller of (n1 – 1) or (n2 – 1). The hypothesized value is often 0, but there are times when we are interested in testing for a difference that is not 0. where and

Two-Sample t Test for Comparing Two Populations Continued . . .
Null Hypothesis: H0: m1 – m2 = hypothesized value Alternative Hypothesis: P-value: Ha: m1 – m2 > hypothesized value Area under the appropriate t curve to the right of the computed t Area under the appropriate t curve to the left of the computed t Ha: m1 – m2 < hypothesized value 2(area to right of computed t) if +t or 2(area to left of computed t) if -t Ha: m1 – m2 ≠ hypothesized value

Another Way to Write Hypothesis Statements:
H0: m1 = m2 H0: m1 - m2 = 0 Ha: m1 - m2 < 0 Ha: m1 - m2 > 0 Ha: m1 - m2 ≠ 0 When the hypothesized value is 0, we can rewrite these hypothesis statements: Be sure to define BOTH m1 and m2! Ha: m1 < m2 Ha: m1 > m2 Ha: m1 ≠ m2

Two-Sample t Test for Comparing Two Populations Continued . . .
Assumptions: The two samples are independently selected random samples from the populations of interest The sample sizes are large (generally 30 or larger) or the population distributions are (at least approximately) normal. When comparing two treatment groups, use the following assumptions: Individuals or objects are randomly assigned to treatments (or vice versa) The sample sizes are large (generally 30 or larger) or the treatment response distributions are approximately normal.

Are women still paid less than men for comparable work
Are women still paid less than men for comparable work? A study was carried out in which salary data was collected from a random sample of men and from a random sample of women who worked as purchasing managers and who were subscribers to Purchasing magazine. Annual salaries (in thousands of dollars) appear below (the actual sample sizes were much larger). Use a = .05 to determine if there is convincing evidence that the mean annual salary for male purchasing managers is greater than the mean annual salary for female purchasing managers. H0: m1 – m2 = 0 Ha: m1 – m2 > 0 If we had defined m1 as the mean salary for female purchasing managers and m2 as the mean salary for male purchasing managers, then the correct alternative hypothesis would be the difference in the means is less than 0. Men 81 69 76 74 79 65 Women 78 60 67 61 62 73 71 58 68 48 Where m1 = mean annual salary for male purchasing managers and m2 = mean annual salary for female purchasing managers State the hypotheses:

Verify the assumptions
Salary War Continued . . . H0: m1 – m2 = 0 Ha: m1 – m2 > 0 Assumptions: Given two independently selected random samples of male and female purchasing managers. Men 81 69 76 74 79 65 Women 78 60 67 61 62 73 71 58 68 48 Where m1 = mean annual salary for male purchasing managers and m2 = mean annual salary for female purchasing managers Even though these are samples from subscribers of Purchasing magazine, the authors of the study believed it was reasonable to view the samples as representative of the populations of interest. Men Women 60 80 2) Since the sample sizes are small, we must determine if it is plausible that the sampling distributions for each of the two populations are approximately normal. Since the boxplots are reasonably symmetrical with no outliers, it is plausible that the sampling distributions are approximately normal. Verify the assumptions

What potential type of error could we have made with this conclusion?
Salary War Continued . . . H0: m1 – m2 = 0 Ha: m1 – m2 > 0 Test Statistic: P-value = a = .05 Since the P-value < a, we reject H0. There is convincing evidence that the mean salary for male purchasing managers is higher than the mean salary for female purchasing managers. Men 81 69 76 74 79 65 Women 78 60 67 61 62 73 71 58 68 48 Where m1 = mean annual salary for male purchasing managers and m2 = mean annual salary for female purchasing managers What potential type of error could we have made with this conclusion? Truncate (round down) this value. Type I Now find the area to the right of t = 3.11 in the t-curve with df = 15. Compute the test statistic and P-value To find the P-value, first find the appropriate df.

Mean Fill Example We would like to compare the mean fill of 32 ounce cans of beer from two adjacent filling machines. Past experience has shown that the population standard deviations of fills for the two machines are known to be s1 = and s2 = respectively. A sample of 35 cans from machine 1 gave a mean of and a sample of 31 cans from machine 2 gave a mean of State, perform and interpret an appropriate hypothesis test using the 0.05 level of significance.

Mean Fill Example Continued
m1 = mean fill from machine 1 m2 = mean fill from machine 2 H0: m1 - m2 = 0 Ha: m1  m2 Significance level: a = 0.05

Since n1 and n2 are both large (> 30) we do not have to make any assumptions about the nature of the distributions of the fills. This example is a bit of a stretch, since knowing the population standard deviations (without knowing the population means) is very unusual. Accept this example for what it is, just a sample of the calculation. Generally this statistic is used when dealing with “what if” type of scenarios and we will move on to another technique that is somewhat more commonly used when s1 and s2 are not known.

Calculation: P-value: P-value = 2 P(z > 1.86) = 2 P(z < -1.86) = 2(0.0314) =

Since the P-value > a, we fail to reject H0. There is not convincing evidence that the two machines produce bottles with different mean fills.

Cold Medicine Example Brand A Brand B
In an attempt to determine if two competing brands of cold medicine contain, on the average, the same amount of acetaminophen, twelve different tablets from each of the two competing brands were randomly selected and tested for the amount of acetaminophen each contains. The results (in milligrams) follow. Use a significance level of 0.01. Brand A Brand B 517, 495, 503, , 508, 513, 521 503, 493, 505, , 533, 500, 515 498, 481, 499, , 498, 515, 515 State and perform an appropriate hypothesis test.

Cold Medicine Example Continued
m1 = the mean amount of acetaminophen in cold tablet brand A m2 = the mean amount of acetaminophen in cold tablet brand B H0: m1 = m2 (m1 - m2 = 0) Ha: m1  m2 (m1 - m2  0) Significance level: a = 0.01 Test Statistic

Assumptions: The samples were selected independently and randomly. Since the samples are not large, we need to be able to assume that the populations (of amounts of acetaminophen) are both normally distributed.

Assumptions (continued): As we can see from the normality plots and the boxplots, the assumption that the underlying distributions are normally distributed appears to be quite reasonable.

Calculation:

Calculation: We truncate the degrees of freedom to give df = 17.

P-value: From the table of tail areas for t curve (Table IV) we look up a t value of 3.5 with df = 17 to get Since this is a two-tailed alternate hypothesis, P-value = 2(0.001) = Conclusion: Since P-value = < 0.01 = a, H0 is rejected. The data provides strong evidence that the mean amount of acetaminophen is not the same for both brands. Specifically, there is strong evidence that the average amount per tablet for brand A is less than that for brand B.

The Two-Sample t Confidence Interval for the Difference Between Two Population or Treatment Means
The general formula for a confidence interval for m1 – m2 when The two samples are independently selected random samples from the populations of interest The sample sizes are large (generally 30 or larger) or the population distributions are (at least approximately) normal. is The t critical value is based on df should be truncated to an integer. For a comparison of two treatments, use the following assumptions: 1) Individuals or objects are randomly assigned to treatments (or vice versa) 2) The sample sizes are large (generally 30 or larger) or the treatment response distributions are approximately normal. where and

Find the mean and standard deviation for each treatment.
In a study on food intake after sleep deprivation, men were randomly assigned to one of two treatment groups. The experimental group was required to sleep only 4 hours on each of two nights, while the control group was required to sleep 8 hours on each of two nights. The amount of food intake (Kcal) on the day following the two nights of sleep was measured. Compute a 95% confidence interval for the true difference in the mean food intake for the two sleeping conditions. 4-hour sleep 3585 4470 3068 5338 2221 4791 4435 3099 3187 3901 3868 3869 4878 3632 4518 8-hour sleep 4965 3918 1987 4993 5220 3653 3510 3338 4100 5792 4547 3319 3336 4304 4057 Find the mean and standard deviation for each treatment. x4 = s4 = x8 = s8 =

Verify the assumptions.
Food Intake Study Continued . . . 4-hour sleep 3585 4470 3068 5338 2221 4791 4435 3099 3187 3901 3868 3869 4878 3632 4518 8-hour sleep 4965 3918 1987 4993 5220 3653 3510 3338 4100 5792 4547 3319 3336 4304 4057 x4 = s4 = x8 = s8 = Assumptions: Men were randomly assigned to two treatment groups Verify the assumptions. 2) The assumption of normal response distributions is plausible because both boxplots are approximately symmetrical with no outliers. 4000 4-hour 8-hour

Food Intake Study Continued . . .
4-hour sleep 3585 4470 3068 5338 2221 4791 4435 3099 3187 3901 3868 3869 4878 3632 4518 8-hour sleep 4965 3918 1987 4993 5220 3653 3510 3338 4100 5792 4547 3319 3336 4304 4057 x4 = s4 = x8 = s8 = Based upon this interval, is there a significant difference in the mean food intake for the two sleeping conditions? No, since 0 is in the confidence interval, there is not convincing evidence that the mean food intake for the two sleep conditions are different. Calculate the interval. We are 95% confident that the true difference in the mean food intake for the two sleeping conditions is between Kcal and Kcal. Interpret the interval in context.

Thread Example Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. The sample data is given in the following table. Construct a 98% confidence interval for the difference of the population means.

Thread Example Continued
Truncating, we have df = 96.

Thread Example Continued
Looking on the table of t critical values (table III) under 98% confidence level for df = 96, (we take the closest value for df , specifically df = 120) and have the t critical value = 2.36. The 98% confidence interval estimate for the difference of the means of the tensile strengths is

Octane Example A student recorded the mileage he obtained while commuting to school in his car. He kept track of the mileage for twelve different tanksful of fuel, involving gasoline of two different octane ratings. Compute the 95% confidence interval for the difference of mean mileages. His data follow: 87 Octane 90 Octane 26.4, 27.6, , 30.9, 29.2 28.9, 29.3, , 32.8, 29.3

Octane Example Continued
Let the 87 octane fuel be the first group and the 90 octane fuel the second group, so we have n1 = n2 = 6 and Truncating, we have df = 9.

Looking on the table under 95% with 9 degrees of freedom, the critical value of t is 2.26. The 95% confidence interval for the true difference of the mean mileages is (-3.99, -0.57).

Comments: We had to assume that the samples were independent and random and that the underlying populations were normally distributed since the sample sizes were small. If we randomized the order of the tankfuls of the two different types of gasoline we can reasonably assume that the samples were random and independent. By using all of the observations from one car we are simply controlling the effects of other variables such as year, model, weight, etc.

By looking at the following normality plots, we see that the assumption of normality for each of the two populations of mileages appears reasonable. Given the small sample sizes, the assumption of normality is very important, so one would be a bit careful utilizing this result.

Pooled t Test Used when the variances of the two populations are equal (s1 = s2) Combines information from both samples to create a “pooled” estimate of the common variance which is used in place of the two sample standard deviations Is not widely used due to its sensitivity to any departure from the equal variance assumption P-values computed using the pooled t procedure can be far from the actual P-value if the population variances are not equal. When the population variances are equal, the pooled t procedure is better at detecting deviations from H0 than the two-sample t test.

Difference in Means with Paired Samples
Section 11.2 Difference in Means with Paired Samples

Suppose that an investigator wants to determine if regular aerobic exercise improves blood pressure. A random sample of people who jog regularly and a second random sample of people who do not exercise regularly are selected independently of one another. Can we conclude that the difference in mean blood pressure is attributed to jogging? What about other factors like weight? One way to avoid these difficulties would be to pair subjects by weight then assign one of the pairs to jogging and the other to no exercise.

Where md is the mean of the differences in the paired observations
Summary of the Paired t test for Comparing Two Population or Treatment Means Null Hypothesis: H0: md = hypothesized value Test Statistic: Where n is the number of sample differences and xd and sd are the mean and standard deviation of the sample differences. This test is based on df = n – 1. Alternative Hypothesis: P-value: Ha: md > hypothesized value Area to the right of calculated t Ha: md < hypothesized value Area to the left of calculated t Ha: md ≠ hypothesized value 2(area to the right of t) if +t or 2(area to the left of t) if -t Where md is the mean of the differences in the paired observations The hypothesized value is usually 0 – meaning that there is no difference.

Summary of the Paired t test for Comparing Two Population or Treatment Means Continued
Assumptions: The samples are paired. The n sample differences can be viewed as a random sample from a population of differences. The number of sample differences is large (generally 30 or more) or the population distribution of differences is (at least approximately) normal.

Is this an example of paired samples?
An engineering association wants to see if there is a difference in the mean annual salary for electrical engineers and chemical engineers. A random sample of electrical engineers is surveyed about their annual income. Another random sample of chemical engineers is surveyed about their annual income. No, there is no pairing of individuals, you have two independent samples

Is this an example of paired samples?
A pharmaceutical company wants to test its new weight-loss drug. Before giving the drug to volunteers, company researchers weigh each person. After a month of using the drug, each person’s weight is measured again. Yes, you have two observations on each individual, resulting in paired data.

First, find the differences: pre-test minus post-test.
Can playing chess improve your memory? In a study, students who had not previously played chess participated in a program in which they took chess lessons and played chess daily for 9 months. Each student took a memory test before starting the chess program and again at the end of the 9-month period. Test the claim at the 0.05 level of significance If we had subtracted Post-test minus Pre-test, then the alternative hypothesis would be the mean difference is greater than 0. Student 1 2 3 4 5 6 7 8 9 10 11 12 Pre-test 510 610 640 675 600 550 625 450 720 575 Post-test 850 790 775 700 690 540 680 Difference -340 -180 -210 -100 -225 -90 -240 -55 35 -5 H0: md = 0 Ha: md < 0 Where md is the mean memory score difference between students with no chess training and students who have completed chess training First, find the differences: pre-test minus post-test. State the hypotheses.

Playing Chess Continued . . .
Student 1 2 3 4 5 6 7 8 9 10 11 12 Pre-test 510 610 640 675 600 550 625 450 720 575 Post-test 850 790 775 700 690 540 680 Difference -340 -180 -210 -100 -225 -90 -240 -55 35 -5 H0: md = 0 Ha: md < 0 Assumptions: 1) Although the sample of students is not a random sample, the Where md is the mean memory score difference between students with no chess training and students who have completed chess training Verify assumptions investigator believed that it was reasonable to view the 12 sample differences as representative of all such differences. 2) A boxplot of the differences is approximately symmetrical with no outliers so the assumption of normality is plausible.

Playing Chess Continued . . .
Student 1 2 3 4 5 6 7 8 9 10 11 12 Pre-test 510 610 640 675 600 550 625 450 720 575 Post-test 850 790 775 700 690 540 680 Difference -340 -180 -210 -100 -225 -90 -240 -55 35 -5 H0: md = 0 Ha: md < 0 Test Statistic: Where md is the mean memory score difference between students with no chess training and students who have completed chess training State the conclusion in context. Compute the test statistic and P-value. P-value ≈ 0 df = 11 a = .05 Since the P-value < a, we reject H0. There is convincing evidence to suggest that the mean memory score after chess training is higher than the mean memory score before training.

Paired t Confidence Interval for md
When The samples are paired. The n sample differences can be viewed as a random sample from a population of differences. The number of sample differences is large (generally 30 or more) or the population distribution of differences is (at least approximately) normal. the paired t interval for md is Where df = n - 1

Playing Chess Revisited . . .
Student 1 2 3 4 5 6 7 8 9 10 11 12 Pre-test 510 610 640 675 600 550 625 450 720 575 Post-test 850 790 775 700 690 540 680 Difference -340 -180 -210 -100 -225 -90 -240 -55 35 -5 Compute a 90% confidence interval for the mean difference in memory scores before chess training and the memory scores after chess training. We are 90% confident that the true mean difference in memory scores before chess training and the memory scores after chess training is between and

Weight Loss Example A weight reduction center advertises that participants in its program lose an average of at least 5 pounds during the first week of the participation. Because of numerous complaints, the state’s consumer protection agency doubts this claim. To test the claim at the 0.05 level of significance, 12 participants were randomly selected. Their initial weights and their weights after 1 week in the program appear on the next slide. Set up and perform an appropriate hypothesis test.

Weight Loss Example Continued

md = mean of the individual weight changes (initial weight–weight after one week) This is equivalent to the difference of means: md = m1 – m2 = minitial weight - m1 week weight H0: md = 5 Ha: md < 5 Significance level: a = 0.05 Test statistic:

Assumptions: According to the statement of the example, we can assume that the sampling is random. The sample size (12) is small, so from the boxplot we see that there is one outlier but never the less, the distribution is reasonably symmetric and the normal plot confirms that it is reasonable to assume that the population of differences (weight losses) is normally distributed.

Calculations: According to the statement of the example, we can assume that the sampling is random. The sample size (10) is small, so P-value: This is a lower tail test, so looking up the t value of 3.0 under df = 11 in the table of tail areas for t curves (table IV) we find that the P-value =

Conclusions: Since P-value = < 0.05 = a, we reject H0. We draw the following conclusion. There is strong evidence that the mean weight loss for those who took the program for one week is less than 5 pounds.

Minitab returns the following when asked to perform this test. This is substantially the same result. Paired T-Test and CI: Initial, One week Paired T for Initial - One week N Mean StDev SE Mean Initial One week Difference 95% upper bound for mean difference: 3.720 T-Test of mean difference = 5 (vs < 5): T-Value = P-Value = 0.003

Section 11.3 Large-Sample Inferences Concerning the Difference Between Two Population or Treatment Proportions

Some people seem to think that duct tape can fix anything
Some people seem to think that duct tape can fix anything even remove warts! Investigators at Madigan Army Medical Center tested using duct tape to remove warts versus the more traditional freezing treatment. Suppose that the duct tape treatment will successfully remove 50% of warts and that the traditional freezing treatment will successfully remove 60% of warts. Let’s investigate the sampling distribution of pfreeze - ptape

pfreeze = the true proportion of warts that are successfully removed by freezing
ptape = the true proportion of warts that are successfully removed by using duct tape ptape = .5 Randomly take one of the sample proportions for the freezing treatment and one of the sample proportions for the duct tape treatment and find the difference. Suppose we repeatedly treated 100 warts using the duct tape method and calculated the proportion of warts that are successfully removed. We would have the sampling distribution of ptape. Suppose we repeatedly treated 100 warts using the traditional freezing treatment and calculated the proportion of warts that are successfully removed. We would have the sampling distribution of pfreeze .6 .5 .1 Doing this repeatedly, we will create the sampling distribution of (pfreeze – ptape) pfreeze - ptape

Properties of the Sampling Distribution of 𝒑 1 - 𝒑 2
Use: Properties of the Sampling Distribution of 𝒑 1 - 𝒑 2 When performing a hypothesis test, we will use the null hypothesis that p1 and p2 are equal. We will not know the common value for p1 and p2. If two random samples are selected independently of one another, the following properties hold: 1. This says that the sampling distribution of 𝑝 1 - 𝑝 2 is centered at p1 – p2 so 𝑝 1 - 𝑝 2 is an unbiased statistic for estimating p1 – p2. Since the value for p1 and p2 are unknown, we will combine 𝑝 1 and 𝑝 2 to estimate the common value of p1 and p2 2. 3. If both n1 and n2 are large (that is, if n1p1 > 10, n1(1 – p1) > 10, n2p2 > 10, and n2(1 – p2) > 10), then 𝑝 1 and 𝑝 2 each have a sampling distribution that is approximately normal, and their difference 𝑝 1 - 𝑝 2 also has a sampling distribution that is approximately normal.

Summary of Large-Sample z Test for p1 – p2 = 0
Null Hypothesis: H0: p1 – p2 = 0 Test Statistic: Alternative Hypothesis: P-value: Ha: p1 – p2 > area to the right of calculated z Ha: p1 – p2 < area to the left of calculated z Ha: p1 – p2 ≠ (area to the right of z) if +z or 2(area to the left of z) if -z Use:

Another Way to Write Hypothesis statements:
Be sure to define both p1 & p2! H0: p1 - p2 = 0 Ha: p1 - p2 > 0 Ha: p1 - p2 < 0 Ha: p1 - p2 ≠ 0 H0: p1 = p2 Ha: p1 > p2 Ha: p1 < p2 Ha: p1 ≠ p2

Summary of Large-Sample z Test for p1 – p2 = 0 Continued
Assumptions: The samples are independently chosen random samples or treatments were assigned at random to individuals or objects. Since p1 and p2 are unknown we must use 𝑝 1and 𝑝 2 to verify that the samples are large enough. Both sample sizes are large n1 𝑝 1 > 10, n1(1 - 𝑝 1) > 10, n2 𝑝 2 > 10, n2(1 - 𝑝 2) > 10

Number with wart successfully removed
Investigators at Madigan Army Medical Center tested using duct tape to remove warts. Patients with warts were randomly assigned to either the duct tape treatment or to the more traditional freezing treatment. Those in the duct tape group wore duct tape over the wart for 6 days, then removed the tape, soaked the area in water, and used an emery board to scrape the area. This process was repeated for a maximum of 2 months or until the wart was gone. The data follows: Do these data suggest that freezing is less successful than duct tape in removing warts? Treatment n Number with wart successfully removed Liquid nitrogen freezing 100 60 Duct tape 104 88

Duct Tape Continued . . . H0: p1 – p2 = 0 Ha: p1 – p2 < 0 Assumptions: 1) Subjects were randomly assigned to the two treatments. Treatment n Number with wart successfully removed Liquid nitrogen freezing 100 60 Duct tape 104 88 Where p1 is the true proportion of warts that would be successfully removed by freezing and p2 is the true proportion of warts that would be successfully removed by duct tape 2) The sample sizes are large enough because: n1 𝑝 1 = 100(.6) = 60 > n1(1 - 𝑝 1) = 100(.4) = 40 > 10 n2 𝑝 2 = 104(.88) = > 10 n2(1 - 𝑝 2) = 104(.12) = 12.48> 10

Duct Tape Continued . . . H0: p1 – p2 = 0 Ha: p1 – p2 < 0 Treatment n Number with wart successfully removed Liquid nitrogen freezing 100 60 Duct tape 104 88 P-value ≈ 0 a = .01 Since the P-value < a, we reject H0. There is convincing evidence to suggest the proportion of warts successfully removed is lower for freezing than for the duct tape treatment.

Student Retention Example
A group of college students were asked what they thought the “issue of the day”. Without a pause the class almost to a person said “student retention”. The class then went out and obtained a random sample (questionable) and asked the question, “Do you plan on returning next year?” The responses along with the gender of the person responding are summarized in the following table. Test to see if the proportion of students planning on returning is the same for both genders at the 0.05 level of significance.

Student Retention Example Continued
p1 = true proportion of males who plan on returning p2 = true proportion of females who plan on returning n1 = number of males surveyed n2 = number of females surveyed 𝑝 1 = sample proportion of males who plan on returning 𝑝 2 = sample proportion of females who plan on returning Null hypothesis: H0: p1 – p2 = 0 Alternate hypothesis: Ha: p1 – p2  0

Significance level:  = 0.05 Test statistic: Assumptions: The two samples are independently chosen random samples. Furthermore, the sample sizes are large enough since n1 p1 = 211  10, n1(1- p1) = 64  10 n2p2 = 141  10, n2(1- p2) = 41  10

Calculations:

P-value: The P-value for this test is 2 times the area under the z curve to the left of the computed z = P-value = 2(0.4247) = Conclusion: Since P-value = > 0.05 = , the hypothesis H0 is not rejected at significance level 0.05. There is no evidence that the return rate is different for males and females..

Washing Machine Example
A consumer agency spokesman stated that he thought that the proportion of households having a washing machine was higher for suburban households then for urban households. To test to see if that statement was correct at the 0.05 level of significance, a reporter randomly selected a number of households in both suburban and urban environments and obtained the following data.

Washing Machine Example Continued
p1 = proportion of suburban households having washing machines p2 = proportion of urban households having washing machines p1 - p2 is the difference between the proportions of suburban households and urban households that have washing machines. H0: p1 - p2 = 0 Ha: p1 - p2 > 0

Significance level:  = 0.05 Test statistic: Assumptions: The two samples are independently chosen random samples. Furthermore, the sample sizes are large enough since n1 p1 = 243  10, n1(1- p1) = 57  10 n2p2 = 181  10, n2(1- p2) = 69  10

Calculations:

P-value: The P-value for this test is the area under the z curve to the right of the computed z = 2.39. The P-value = = Conclusion: Since P-value = < 0.05 = a, the hypothesis H0 is rejected at significance level There is sufficient evidence at the 0.05 level of significance that the proportion of suburban households that have washers is more than the proportion of urban households that have washers.

A Large-Sample Confidence Interval for p1 – p2
When The samples are independently chosen random samples or treatments were assigned at random to individuals or objects a large-sample confidence interval for p1 – p2 is calculated by: 2) Both sample sizes are large n1 𝑝 1 > 10, n1(1 - 𝑝 1) > 10, n2 𝑝 2 > 10, n2(1 - 𝑝 2) > 10

The article “Freedom of What
The article “Freedom of What?” (Associated Press, February 1, 2005) described a study in which high school students and high school teachers were asked whether they agreed with the following statement: “Students should be allowed to report controversial issues in their student newspapers without the approval of school authorities.” It was reported that 58% of students surveyed and 39% of teachers surveyed agreed with the statement. The two samples – 10,000 high school students and 8000 high school teachers – were selected from schools across the country. Compute a 90% confidence interval for the difference in proportion of students who agreed with the statement and the proportion of teachers who agreed with the statement.

Newspaper Problem Continued . . .
p1 = p2 = .39 Based on this confidence interval, does there appear to be a significant difference in proportion of students who agreed with the statement and the proportion of teachers who agreed with the statement? Explain. 1) Assume that it is reasonable to regard these two samples as being independently selected and representative of the populations of interest. 2) Both sample sizes are large enough n1p1 = 10000(.58) > 10, n1(1 – p1) = 10000(.42) > 10, n2p2 = 8000(.39) > 10, n2(1 – p2) = 8000(.61) > 10 We are 90% confident that the difference in proportion of students who agreed with the statement and the proportion of teachers who agreed with the statement is between .178 and .202.

Survey Example A student assignment called for the students to survey both male and female students (independently and randomly chosen) to see if the proportions that approve of the College’s new drug and alcohol policy. A student went and randomly selected 200 male students and 100 female students and obtained the data summarized below. Use this data to obtain a 90% confidence interval estimate for the difference of the proportions of female and male students that approve of the new policy.

Survey Example Continued
For a 90% confidence interval the z value to use is This value is obtained from the bottom row of the table of t critical values (Table III). We use p1 to be the female’s sample approval proportion and p2 as the male’s sample approval proportion. We are 90% confident that the proportion of females that approve of the policy exceeds the proportion of males that approve of the policy by somewhere between and

Comparing Two Populations or Treatments

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Comparing Two Populations or Treatments

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