Chapter 10 Comparing Two Populations or Treatments Section 10.2

Chapter 10 Comparing Two Populations or Treatments Section 10.2
Comparing Two Means

Comparing Two Means DESCRIBE the shape, center, and variability of the sampling distribution of 𝑥 1 − 𝑥 2 . DETERMINE whether the conditions are met for doing inference about a difference between two means. CONSTRUCT and INTERPRET a confidence interval for a difference between two means. CALCULATE the standardized test statistic and P-value for a test about a difference between two means. PERFORM a significance test about a difference between two means.

The Sampling Distribution of a Difference Between Two Means
The heights of 10-year-old girls can be modeled by a Normal distribution with mean µG = 56.4 inches and standard deviation σG = 2.7 inches.

The heights of 10-year-old girls can be modeled by a Normal distribution with mean µG = 56.4 inches and standard deviation σG = 2.7 inches. The heights of 10-year-old boys can be modeled by a Normal distribution with mean µB = 55.7 inches and standard deviation σB = 3.8 inches.

The heights of 10-year-old girls can be modeled by a Normal distribution with mean µG = 56.4 inches and standard deviation σG = 2.7 inches. The heights of 10-year-old boys can be modeled by a Normal distribution with mean µB = 55.7 inches and standard deviation σB = 3.8 inches. Suppose we take independent SRSs of 12 girls and 8 boys of this age and measure their heights.

What can we say about the difference 𝑥 𝐺 − 𝑥 𝐵 in the sample means?

What can we say about the difference 𝑥 𝐺 − 𝑥 𝐵 in the sample means? Simulated sampling distribution of the difference in sample means 𝑥 𝐺 − 𝑥 𝐵 .

What can we say about the difference 𝑥 𝐺 − 𝑥 𝐵 in the sample means? Simulated sampling distribution of the difference in sample means 𝑥 𝐺 − 𝑥 𝐵 . One set of samples gave 𝑥 𝐺 =56.09 and 𝑥 𝐵 =54.68, resulting in a difference of 𝑥 𝐺 − 𝑥 𝐵 =56.09−54.68=1.41inches

What can we say about the difference 𝑥 𝐺 − 𝑥 𝐵 in the sample means? mean = 𝜇 𝐺 − 𝜇 𝐵 =0.7 inches

What can we say about the difference 𝑥 𝐺 − 𝑥 𝐵 in the sample means? standard deviation = 𝜎 𝑛 𝜎 𝑛 1 = = 1.55 inches mean = 𝜇 𝐺 − 𝜇 𝐵 =0.7 inches

Choose an SRS of size n1 from Population 1 with mean µ1 and standard deviation σ1 and an independent SRS of size n2 from Population 2 with mean µ2 and standard deviation σ2. Then: • The sampling distribution of 𝑥 1 − 𝑥 2 is Normal if both population distributions are Normal. It is approximately Normal if both sample sizes are large (n1 ≥ 30 and n2 ≥ 30) or if one population is Normally distributed and the other sample size is large.

Choose an SRS of size n1 from Population 1 with mean µ1 and standard deviation σ1 and an independent SRS of size n2 from Population 2 with mean µ2 and standard deviation σ2. Then: • The sampling distribution of 𝑥 1 − 𝑥 2 is Normal if both population distributions are Normal. It is approximately Normal if both sample sizes are large (n1 ≥ 30 and n2 ≥ 30) or if one population is Normally distributed and the other sample size is large. • The mean of the sampling distribution of 𝑥 1 − 𝑥 2 is 𝜇 𝑥 1 − 𝑥 2 = 𝜇 1 − 𝜇 2

Choose an SRS of size n1 from Population 1 with mean µ1 and standard deviation σ1 and an independent SRS of size n2 from Population 2 with mean µ2 and standard deviation σ2. Then: • The sampling distribution of 𝑥 1 − 𝑥 2 is Normal if both population distributions are Normal. It is approximately Normal if both sample sizes are large (n1 ≥ 30 and n2 ≥ 30) or if one population is Normally distributed and the other sample size is large. • The mean of the sampling distribution of 𝑥 1 − 𝑥 2 is 𝜇 𝑥 1 − 𝑥 2 = 𝜇 1 − 𝜇 2 • The standard deviation of the sampling distribution of 𝑥 1 − 𝑥 2 is 𝜎 𝑥 1 − 𝑥 2 = 𝜎 𝑛 𝜎 𝑛 2 as long as the 10% condition is met for both samples: n1 < 0.10N1 and n2 < 0.10N2.

Problem: A fast-food restaurant uses an automated filling machine to pour its soft drinks. The machine has different settings for small, medium, and large drink cups. According to the machine’s manufacturer, when the large setting is chosen, the amount of liquid L dispensed by the machine follows a Normal distribution with mean 27 ounces and standard deviation 0.8 ounce. When the medium setting is chosen, the amount of liquid M dispensed follows a Normal distribution with mean 17 ounces and standard deviation 0.5 ounce. To test the manufacturer’s claim, the restaurant manager measures the amount of liquid in each of 20 cups filled using the large setting and 25 cups filled using the medium setting. Let 𝑥 𝐿 − 𝑥 𝑀 be the difference in the sample mean amount of liquid under the two settings. What is the shape of the sampling distribution of 𝑥 𝐿 − 𝑥 𝑀 ? Why? Find the mean of the sampling distribution. Calculate and interpret the standard deviation of the sampling distribution. RichLegg/Getty Images

Problem: … the amount of liquid L dispensed by the machine follows a Normal distribution with mean 27 ounces and standard deviation 0.8 ounce. When the medium setting is chosen, the amount of liquid M dispensed follows a Normal distribution with mean 17 ounces and standard deviation 0.5 ounce. …the restaurant manager measures the amount of liquid in each of 20 cups filled using the large setting and 25 cups filled using the medium setting. Let 𝑥 𝐿 − 𝑥 𝑀 be the difference in the sample mean amount of liquid under the two settings. What is the shape of the sampling distribution of 𝑥 𝐿 − 𝑥 𝑀 ? Why? Find the mean of the sampling distribution. Calculate and interpret the standard deviation of the sampling distribution. RichLegg/Getty Images

Problem: … the amount of liquid L dispensed by the machine follows a Normal distribution with mean 27 ounces and standard deviation 0.8 ounce. When the medium setting is chosen, the amount of liquid M dispensed follows a Normal distribution with mean 17 ounces and standard deviation 0.5 ounce. …the restaurant manager measures the amount of liquid in each of 20 cups filled using the large setting and 25 cups filled using the medium setting. Let 𝑥 𝐿 − 𝑥 𝑀 be the difference in the sample mean amount of liquid under the two settings. What is the shape of the sampling distribution of 𝑥 𝐿 − 𝑥 𝑀 ? Why? Find the mean of the sampling distribution. Calculate and interpret the standard deviation of the sampling distribution. RichLegg/Getty Images (a) Normal, because both population distributions are Normal.

Problem: … the amount of liquid L dispensed by the machine follows a Normal distribution with mean 27 ounces and standard deviation 0.8 ounce. When the medium setting is chosen, the amount of liquid M dispensed follows a Normal distribution with mean 17 ounces and standard deviation 0.5 ounce. …the restaurant manager measures the amount of liquid in each of 20 cups filled using the large setting and 25 cups filled using the medium setting. Let 𝑥 𝐿 − 𝑥 𝑀 be the difference in the sample mean amount of liquid under the two settings. What is the shape of the sampling distribution of 𝑥 𝐿 − 𝑥 𝑀 ? Why? Find the mean of the sampling distribution. Calculate and interpret the standard deviation of the sampling distribution. RichLegg/Getty Images (b) 𝜇 𝑥 𝐿 − 𝑥 𝑀 = 27 – 17 = 10 ounces.

Problem: … the amount of liquid L dispensed by the machine follows a Normal distribution with mean 27 ounces and standard deviation 0.8 ounce. When the medium setting is chosen, the amount of liquid M dispensed follows a Normal distribution with mean 17 ounces and standard deviation 0.5 ounce. …the restaurant manager measures the amount of liquid in each of 20 cups filled using the large setting and 25 cups filled using the medium setting. Let 𝑥 𝐿 − 𝑥 𝑀 be the difference in the sample mean amount of liquid under the two settings. What is the shape of the sampling distribution of 𝑥 𝐿 − 𝑥 𝑀 ? Why? Find the mean of the sampling distribution. Calculate and interpret the standard deviation of the sampling distribution. RichLegg/Getty Images (c) 𝜎 𝑥 𝐿 − 𝑥 𝑀 = = ounce. Note: We do not have to check the 10% condition because we are not sampling without replacement from a finite population.

Problem: … the amount of liquid L dispensed by the machine follows a Normal distribution with mean 27 ounces and standard deviation 0.8 ounce. When the medium setting is chosen, the amount of liquid M dispensed follows a Normal distribution with mean 17 ounces and standard deviation 0.5 ounce. …the restaurant manager measures the amount of liquid in each of 20 cups filled using the large setting and 25 cups filled using the medium setting. Let 𝑥 𝐿 − 𝑥 𝑀 be the difference in the sample mean amount of liquid under the two settings. What is the shape of the sampling distribution of 𝑥 𝐿 − 𝑥 𝑀 ? Why? Find the mean of the sampling distribution. Calculate and interpret the standard deviation of the sampling distribution. RichLegg/Getty Images (c) The difference (Large cup – Medium cup) in the sample mean amounts of liquid typically varies by about 0.2 ounce from the true difference in means of 10 ounces..

Confidence Intervals for µ1 – µ2
Conditions for Constructing a Confidence Interval About a Difference in Means • Random: The data come from two independent random samples or from two groups in a randomized experiment.

Conditions for Constructing a Confidence Interval About a Difference in Means • Random: The data come from two independent random samples or from two groups in a randomized experiment. ◦ 10%: When sampling without replacement, n1 < 0.10N1 and n2 < 0.10N2.

Conditions for Constructing a Confidence Interval About a Difference in Means • Random: The data come from two independent random samples or from two groups in a randomized experiment. ◦ 10%: When sampling without replacement, n1 < 0.10N1 and n2 < 0.10N2. • Large Counts: Normal/Large Sample: For each sample, the corresponding population distribution (or the true distribution of response to the treatment) is Normal or the sample size is large (n ≥ 30). For each sample, if the population (treatment) distribution has unknown shape and n < 30, a graph of the sample data shows no strong skewness or outliers.

Problem: A college student wants to compare the cost of one- and two-bedroom apartments near campus. She collects the following data on monthly rents (in dollars) for a random sample of 10 apartments of each type. Let µ1 = the true mean monthly rent of all one-bedroom apartments near campus and µ2 = the true mean monthly rent of all two-bedroom apartments near campus. Check if the conditions for calculating a confidence interval for µ1 – µ2 are met. Justin Sullivan/Getty Images

Problem: Let µ1 = the true mean monthly rent of all one-bedroom apartments near campus and µ2 = the true mean monthly rent of all two-bedroom apartments near campus. Check if the conditions for calculating a confidence interval for µ1 – µ2 are met. Justin Sullivan/Getty Images

Problem: Let µ1 = the true mean monthly rent of all one-bedroom apartments near campus and µ2 = the true mean monthly rent of all two-bedroom apartments near campus. Check if the conditions for calculating a confidence interval for µ1 – µ2 are met. Justin Sullivan/Getty Images Random: Independent random samples of 10 one-bedroom apartments and 10 two-bedroom apartments near campus. ✓

Problem: Let µ1 = the true mean monthly rent of all one-bedroom apartments near campus and µ2 = the true mean monthly rent of all two-bedroom apartments near campus. Check if the conditions for calculating a confidence interval for µ1 – µ2 are met. Justin Sullivan/Getty Images 10%: We can assume that 10 < 10% of all one-bedroom apartments near campus and that 10 < 10% of all two-bedroom apartments near campus. ✓

Problem: Let µ1 = the true mean monthly rent of all one-bedroom apartments near campus and µ2 = the true mean monthly rent of all two-bedroom apartments near campus. Check if the conditions for calculating a confidence interval for µ1 – µ2 are met. Justin Sullivan/Getty Images Normal/Large Sample? The sample sizes are small, but the dotplots don’t show any outliers or strong skewness. ✓

statistic ± (critical value) · (standard deviation of statistic)

statistic ± (critical value) · (standard deviation of statistic) 𝑥 1 − 𝑥 2 ±(critical value)∙ (standard deviation of statistic)

statistic ± (critical value) · (standard deviation of statistic) 𝑥 1 − 𝑥 2 ±(critical value)∙ (standard deviation of statistic) 𝜎 𝑥 1 − 𝑥 2 = 𝜎 𝑛 𝜎 𝑛 2

statistic ± (critical value) · (standard deviation of statistic) 𝑥 1 − 𝑥 2 ±(critical value)∙ (standard deviation of statistic) The formula for the standard deviation of 𝑥 1 − 𝑥 2 uses σ1 and σ2, which are usually unknown. 𝜎 𝑥 1 − 𝑥 2 = 𝜎 𝑛 𝜎 𝑛 2

statistic ± (critical value) · (standard deviation of statistic) 𝑥 1 − 𝑥 2 ±(critical value)∙ (standard deviation of statistic) The formula for the standard deviation of 𝑥 1 − 𝑥 2 uses σ1 and σ2, which are usually unknown. 𝜎 𝑥 1 − 𝑥 2 = 𝜎 𝑛 𝜎 𝑛 2 𝑆𝐸 𝑝 1 − 𝑝 2 = 𝑠 𝑛 𝑠 𝑛 2

statistic ± (critical value) · (standard deviation of statistic) 𝑥 1 − 𝑥 2 ±(critical value)∙ (standard deviation of statistic) The formula for the standard deviation of 𝑥 1 − 𝑥 2 uses σ1 and σ2, which are usually unknown. 𝜎 𝑥 1 − 𝑥 2 = 𝜎 𝑛 𝜎 𝑛 2 𝑆𝐸 𝑥 1 − 𝑥 2 = 𝑠 𝑛 𝑠 𝑛 2 We use 𝑠 1 and 𝑠 2 to estimate σ1 and σ2. We call this the standard error of 𝑥 1 − 𝑥 2 .

Two-Sample t Interval for a Difference Between Two Means When the conditions are met, a C% confidence interval for µ1 – µ2 is 𝑥 1 − 𝑥 2 ± 𝑡 ∗ 𝑠 𝑛 𝑠 𝑛 2 where t* is the critical value with C% of the area between –t* and t* for the t distribution with degrees of freedom using either Option 1 (technology) or Option 2 (the smaller of n1 – 1 and n2 – 1).

Problem: The Wade Tract Preserve in Georgia is an old- growth forest of longleaf pines that has survived in a relatively undisturbed state for hundreds of years. One question of interest to foresters who study the area is “How do the sizes of longleaf pine trees in the northern and southern halves of the forest compare?” To find out, researchers took random samples of 30 trees from each half and measured the diameter at breast height (DBH) in centimeters. Here are summary statistics and comparative boxplots of the data: Danita Delimont/Getty Images

Problem: (a) Based on the graph and numerical summaries, write a few sentences comparing the sizes of longleaf pine trees in the two halves of the forest.

Problem: (a) Based on the graph and numerical summaries, write a few sentences comparing the sizes of longleaf pine trees in the two halves of the forest. (a) Shape: The distribution of DBH in the northern sample appears skewed to the right, while the distribution of DBH in the southern sample appears skewed to the left. Outliers: No outliers are present in either sample.

Problem: (a) Based on the graph and numerical summaries, write a few sentences comparing the sizes of longleaf pine trees in the two halves of the forest. (a) Center: It appears that trees in the southern half of the forest have larger diameters. The mean and median DBH for the southern sample are much larger than the corresponding values for the northern sample. Variability: There is more variability in the DBH of the northern longleaf pines. The range, IQR, and standard deviation are all larger for the northern sample.

Problem: (b) Construct and interpret a 90% confidence interval for the difference in the mean DBH of longleaf pines in the northern and southern halves of the Wade Tract Preserve.

Problem: (b) Construct and interpret a 90% confidence interval for the difference in the mean DBH of longleaf pines in the northern and southern halves of the Wade Tract Preserve. (b) STATE 90% CI for µ1 – µ2, where µ1 = the true mean DBH of all trees in the southern half of the forest. µ2 = the true mean DBH of all trees in the northern half of the forest.

Problem: (b) Construct and interpret a 90% confidence interval for the difference in the mean DBH of longleaf pines in the northern and southern halves of the Wade Tract Preserve. (b) PLAN Two-sample t interval for µ1 – µ2 • Random: Independent random samples of 30 trees each from the northern and southern halves of the forest. ✓ 10%: Assume 30 < 10% of all trees in the northern half of the forest and 30 < 10% of all trees in the southern half of the forest. • Normal/Large Sample: n1 = 30 ≥ 30 and n2 = 30 ≥ 30. ✓

Problem: (b) Construct and interpret a 90% confidence interval for the difference in the mean DBH of longleaf pines in the northern and southern halves of the Wade Tract Preserve. (b) DO 𝑥 1 =34.53, 𝑠 1 =14.26, 𝑛 1 =30 𝑥 2 =23.70, 𝑠 2 =17.50, 𝑛 2 =30 Using technology 2-SampTInt gives (3.9362, ) using df =

Problem: (b) Construct and interpret a 90% confidence interval for the difference in the mean DBH of longleaf pines in the northern and southern halves of the Wade Tract Preserve. (b) DO 𝑥 1 =34.53, 𝑠 1 =14.26, 𝑛 1 =30 𝑥 2 =23.70, 𝑠 2 =17.50, 𝑛 2 =30 Using df = 29 and Table B, t*= −23.70 ± =10.83±7.00=(3.83, 17.83)

Problem: (b) Construct and interpret a 90% confidence interval for the difference in the mean DBH of longleaf pines in the northern and southern halves of the Wade Tract Preserve. (b) CONCLUDE We are 90% confident that the interval from to centimeters captures µ1 – µ2 = the difference in the true mean DBH of all the southern trees and the true mean DBH of all the northern trees.

The TI-83/84 can be used to construct a confidence interval for µ1 – µ2. For Pooled, choose “No” for inference about µ1 – µ2.

The TI-83/84 can be used to construct a confidence interval for µ1 – µ2. Using data in lists from the apartments example.

AP® Exam Tip The formula for the two-sample t interval for µ1 – µ2 often leads to calculation errors by students. Also, the interval produced by technology is narrower than the one calculated using the conservative method. As a result, your teacher may recommend using the calculator’s 2-SampTInt feature to compute the confidence interval. Be sure to name the procedure (two-sample t interval for µ1 – µ2) in the “Plan” step and give the interval (3.9362, ) and df (55.728) in the “Do” step.

Significance Tests for µ1 – µ2
H0: µ1 – µ2 = hypothesized value

H0: µ1 – µ2 = hypothesized value H0: µ1 – µ2 = 0 H0: µ1 = µ2

H0: µ1 – µ2 = hypothesized value H0: µ1 – µ2 = 0 H0: µ1 = µ2 Conditions for Performing a Significance Test About a Difference in Means • Random: The data come from two independent random samples or from two groups in a randomized experiment. ◦ 10%: When sampling without replacement, n1 < 0.10N1 and n2 < 0.10N2. • Normal/Large Sample: For each sample, the corresponding population distribution (or the true distribution of response to the treatment) is Normal or the sample size is large (n ≥ 30). For each sample, if the population (treatment) distribution has unknown shape and n < 30, a graph of the sample data shows no strong skewness or outliers.

Problem: Has the mean number of hours Americans work in a week changed? One of the questions on the General Social Survey (GSS) asked respondents how many hours they work each week. Responses from random samples of employed Americans were recorded for 1975 and 2014—they are summarized here. Do these data give convincing evidence at the α = 0.05 significance level of a difference in the true mean number of work hours per week for Americans in 1975 and 2014? (a) State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest. (b) Check if the conditions for performing the test are met. Bill Varie/Getty Images

Problem: (a) State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest. (b) Check if the conditions for performing the test are met. Bill Varie/Getty Images

Problem: (a) State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest. (b) Check if the conditions for performing the test are met. Bill Varie/Getty Images (a) H0: µ2014 – µ1975 = 0 Ha: µ2014 – µ1975 ≠ 0 µ1 = the true mean hours worked per week by Americans in 1975. µ2 = the true mean hours worked per week by Americans in 2014.

Problem: (a) State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest. (b) Check if the conditions for performing the test are met. Bill Varie/Getty Images (b) • Random? Independent random samples of 764 working Americans in 1975 and 1501 working Americans in ✓ º 10%: 764 < 10% of all working Americans in 1975; 1501 < 10% of all working Americans in ✓ • Normal/Large Sample? n1975 = 764 ≥ 30 and n2014 = 1501 ≥ 30.

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑧𝑒𝑑 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 −𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑧𝑒𝑑 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 −𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 𝑡= 𝑥 1 − 𝑥 2 −0 𝑠 𝑛 𝑠 𝑛 2

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑧𝑒𝑑 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 −𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 𝑡= 𝑥 1 − 𝑥 2 −0 𝑠 𝑛 𝑠 𝑛 2 When the Normal/Large Sample condition is met, we can find the P-value using the t distribution with degrees of freedom given by Option 1 (technology) or Option 2 (df = smaller of n1 – 1 and n2 – 1).

Problem: Refer to the preceding example. The table summarizes data on hours worked per week from the independent random samples of working Americans in 1975 and 2014. (a) Explain why the sample results give some evidence for the alternative hypothesis. (b) Calculate the standardized test statistic and P-value. (c) What conclusion would you make using α = 0.05?

Problem: Refer to the preceding example. The table summarizes data on hours worked per week from the independent random samples of working Americans in 1975 and 2014. (a) Explain why the sample results give some evidence for the alternative hypothesis. (b) Calculate the standardized test statistic and P-value. (c) What conclusion would you make using α = 0.05? (a) The observed difference in the sample proportions is 𝑥 − 𝑥 =41.91 −38.97=2.94 which gives some evidence in favor of Ha: µ2014 – µ1975 ≠ 0 because ≠ 0.

Problem: Refer to the preceding example. The table summarizes data on hours worked per week from the independent random samples of working Americans in 1975 and 2014. (a) Explain why the sample results give some evidence for the alternative hypothesis. (b) Calculate the standardized test statistic and P-value. (c) What conclusion would you make using α = 0.05? (b) 𝑡= −38.97 − = =4.88

Problem: Refer to the preceding example. The table summarizes data on hours worked per week from the independent random samples of working Americans in 1975 and 2014. (a) Explain why the sample results give some evidence for the alternative hypothesis. (b) Calculate the standardized test statistic and P-value. (c) What conclusion would you make using α = 0.05? (b) Option 1: 2-SampTTest gives P-value = using df = Option 2: Using df = 100 and Table B, P-value < 2(0.0005) = 0.001

Problem: Refer to the preceding example. The table summarizes data on hours worked per week from the independent random samples of working Americans in 1975 and 2014. (a) Explain why the sample results give some evidence for the alternative hypothesis. (b) Calculate the standardized test statistic and P-value. (c) What conclusion would you make using α = 0.05? (c) Because the P-value of < α = 0.05, we reject H0. There is convincing evidence of a difference in the true mean hours worked by Americans in 1975 and 2014.

Putting It All Together: Two-Sample t Test for µ1 – µ2
Two-Sample t Test for the Difference Between Two Means Suppose the conditions are met. To test the hypothesis H0: µ1 – µ2 = 0, compute the standardized test statistic, 𝑡= 𝑥 1 − 𝑥 2 −0 𝑠 𝑛 𝑠 𝑛 2 Find the P-value by calculating the probability of getting a t statistic this large or larger in the direction specified by the alternative hypothesis Ha. Use the t distribution with degrees of freedom approximated by Option 1 (technology) or Option 2 (the smaller of n1 – 1 and n2 – 1).

Problem: Does increasing the amount of calcium in our diet reduce blood pressure? Examination of a large sample of people revealed a relationship between calcium intake and blood pressure. Such observational studies do not establish causation. Researchers therefore designed a randomized comparative experiment. The subjects were 21 healthy men who volunteered to take part in the experiment. They were randomly assigned to two groups: 10 of the men received a calcium supplement for 12 weeks, while the control group of 11 men received a placebo pill that looked identical. The experiment was double-blind. The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative number. Here are the data: rdrgraphe/Shutterstock

Problem: The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative number. Here are the data: (a) Do the data provide convincing evidence that a calcium supplement reduces blood pressure more than a placebo, on average, for subjects like the ones in this study? rdrgraphe/Shutterstock

Problem: The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative number. Here are the data: (a) Do the data provide convincing evidence that a calcium supplement reduces blood pressure more than a placebo, on average, for subjects like the ones in this study? rdrgraphe/Shutterstock (a) STATE H0: µC – µP = 0 Ha: µC – µP > 0 µC = the true mean decrease in systolic blood pressure for healthy men like the ones in this study who take a calcium supplement µP = the true mean decrease in systolic blood pressure for healthy men like the ones in this study who take a placebo We’ll use α = 0.05

Problem: The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative number. Here are the data: (a) Do the data provide convincing evidence that a calcium supplement reduces blood pressure more than a placebo, on average, for subjects like the ones in this study? rdrgraphe/Shutterstock (a) PLAN Two-sample t test for µ1 – µ2 • Random: The 21 subjects were randomly assigned to the calcium or placebo treatments.✓ • Normal/Large Sample: The sample sizes are small, but the dotplots show no strong skewness and no outliers.✓

Problem: The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative number. Here are the data: (a) Do the data provide convincing evidence that a calcium supplement reduces blood pressure more than a placebo, on average, for subjects like the ones in this study? rdrgraphe/Shutterstock (a) DO 𝑥 𝐶 =5.000, 𝑠 𝐶 =8.743, 𝑛 𝐶 =10 𝑥 𝑃 =−0.273, 𝑠 𝑃 =5.901, 𝑛 𝑃 =11 𝑡= − − − − = =1.604

Problem: The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative number. Here are the data: (a) Do the data provide convincing evidence that a calcium supplement reduces blood pressure more than a placebo, on average, for subjects like the ones in this study? rdrgraphe/Shutterstock (a) DO Option 1: 2-SampTTest gives P-value = using df = 15.59 Option 2: Using df = 9 and Table B, 0.05 < P-value < 0.10

Problem: The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative number. Here are the data: (a) Do the data provide convincing evidence that a calcium supplement reduces blood pressure more than a placebo, on average, for subjects like the ones in this study? rdrgraphe/Shutterstock (a) CONCLUDE Because the P-value of > α = 0.05, we fail to reject H0. The experiment does not provide convincing evidence that the true mean decrease in systolic blood pressure is higher for men like these who take calcium than for men like these who take a placebo.

Problem: The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative number. Here are the data: (b) Interpret the P-value you got in part (a) in the context of this experiment. rdrgraphe/Shutterstock

Problem: The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative number. Here are the data: (b) Interpret the P-value you got in part (a) in the context of this experiment. rdrgraphe/Shutterstock (b) Assuming µC – µP = 0 is true, there is a probability of getting a difference (Calcium – Placebo) in mean blood pressure reduction for the two groups of or greater just by the chance involved in the random assignment.

The TI-83/84 can be used to perform significance tests for µ1 – µ2. For Pooled, choose “No” for inference about µ1 – µ2.

The TI-83/84 can be used to perform significance tests for µ1 – µ2. Using data in lists from the work week example.

AP® Exam Tip The formula for the two-sample t statistic for a test about µ1 – µ2 often leads to calculation errors by students. Also, the P-value from technology is smaller and more accurate than the one obtained using the conservative method. As a result, your teacher may recommend using the calculator’s 2-SampTTest feature to perform calculations. Be sure to name the procedure (two-sample t interval for µ1 – µ2) in the “Plan” step and to report the standardized test statistic (t ), P-value (0.0644), and df (15.59) in the “Do” step.

A Word about “Pooled” Most software offers a choice of two-sample t procedures. One is often labeled “unequal” variances; the other, “equal” variances.

A Word about “Pooled” Most software offers a choice of two-sample t procedures. One is often labeled “unequal” variances; the other, “equal” variances. The unequal variance procedures use our formula for the two-sample t interval and test. This interval and test are valid whether or not the population variances are equal.

A Word about “Pooled” Most software offers a choice of two-sample t procedures. One is often labeled “unequal” variances; the other, “equal” variances. The unequal variance procedures use our formula for the two-sample t interval and test. This interval and test are valid whether or not the population variances are equal. The equal variance procedures assumes the two population distributions have the same variance. This procedure combines (the statistical term is pools) the two sample variances to estimate the common population variance. The resulting statistic is called the pooled two-sample t statistic.

A Word about “Pooled” Most software offers a choice of two-sample t procedures. One is often labeled “unequal” variances; the other, “equal” variances. The unequal variance procedures use our formula for the two-sample t interval and test. This interval and test are valid whether or not the population variances are equal. The equal variance procedures assumes the two population distributions have the same variance. This procedure combines (the statistical term is pools) the two sample variances to estimate the common population variance. The resulting statistic is called the pooled two-sample t statistic. Our advice: Never use the pooled t procedures if you have technology that will carry out Option 1.

Section Summary DESCRIBE the shape, center, and variability of the sampling distribution of 𝑥 1 − 𝑥 2 . DETERMINE whether the conditions are met for doing inference about a difference between two means. CONSTRUCT and INTERPRET a confidence interval for a difference between two means. CALCULATE the standardized test statistic and P-value for a test about a difference between two means. PERFORM a significance test about a difference between two means.

Assignment 10.2 SG (Handout)
Tomorrow we will check homework and review for 10.2 Quiz for about 10 min then take the quiz.

Chapter 10 Comparing Two Populations or Treatments Section 10.2

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Chapter 10 Comparing Two Populations or Treatments Section 10.2

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