Confidence intervals for the difference between two means: Independent samples Section 10.1

Objectives Distinguish between independent and paired samples Construct confidence intervals for the difference between two population means Describe the pooled standard deviation and the known standard deviation methods

Distinguish between independent and paired samples Objective 1 Distinguish between independent and paired samples

Independent Samples vs. Paired Samples A drug company has developed a new drug that is designed to reduce high blood pressure. The researchers wish to design a study to compare the effectiveness of the new drug to that of the old drug. Here is one way in which the study can be designed. Design 1: Two samples of individuals are chosen. One sample is given the old drug and the other sample is given the new drug. After several months, blood pressures of the members of both samples are measured. We compare the blood pressures in the first sample to the blood pressures in the second sample to determine which drug is more effective. In this design, the samples are independent. This means that the observations in one sample do not influence the observations in the other.

Independent Samples vs. Paired Samples A drug company has developed a new drug that is designed to reduce high blood pressure. The researchers wish to design a study to compare the effectiveness of the new drug to that of the old drug. Here is another way in which the study can be designed. Design 2: A single group of individuals is chosen. They are given the old drug for a month, then their blood pressures are measured. They then switch to the new drug for a month, after which their blood pressures are measured again. This produces two samples of measurements, the first one from the old drug and the second one from the new drug. We compare the blood pressures in the first sample to the blood pressures in the second sample to determine which drug is more effective. In this design, the samples are paired. Each observation in one sample can be paired with an observation in the second.

Objective 2 Construct confidence intervals for the difference between two population means

Confidence Intervals for the Difference Between Two Population Means Imagine that we will compare the effectiveness of a new drug designed to reduce blood pressure to the effectiveness of a standard drug. We will draw two independent samples. The first sample will get the new drug, and the second sample will get the standard drug. We can imagine these samples as coming from two populations—a population of patients who take the new drug, and a population of patients who take the standard drug. For each population, there is a mean reduction in blood pressure. We are interested in estimating the difference between the population means. Let 𝜇 1 be the population mean reduction for the new drug, and let 𝜇 2 be the population mean reduction for the standard drug. We wish to construct a confidence interval for the difference 𝜇 1 − 𝜇 2 . It follows that the point estimate for the difference 𝜇 1 − 𝜇 2 is the difference between the sample mean. In other words, the point estimate for 𝜇 1 − 𝜇 2 is 𝑥 1 − 𝑥 2 .

Standard error of 𝑥 1 − 𝑥 2 The sample means 𝑥 1 and 𝑥 2 have variances 𝜎 1 2 𝑛 1 and 𝜎 2 2 𝑛 2 . When the samples are independent, the variance of the difference 𝑥 1 − 𝑥 2 can be shown to be the sum of the variances. In other words, the variance of 𝑥 1 − 𝑥 2 is 𝜎 1 2 𝑛 1 + 𝜎 2 2 𝑛 2 The standard error of 𝑥 1 − 𝑥 2 is the square root of the variance. Since we don’t know the values of 𝜎 1 2 and 𝜎 2 2 , we approximate them with the sample variance 𝑠 1 2 and 𝑠 2 2 . Therefore, the standard error is 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 .

Critical Value and Margin of Error Let 1−𝛼 be the confidence level, expressed as a decimal. The critical value for a confidence interval is 𝑡 𝛼 2 , based on the Student’s 𝑡 distribution. There are two ways to compute the number of degrees of freedom: a simple method that is easier when computing by hand, and a more complicated method that is used by software packages and calculator procedures. The simple method is: Degrees of Freedom = smaller of 𝑛 1 − 1 and 𝑛 2 − 1. The margin of error is obtained by multiplying the critical value and the standard error. The margin of error is 𝑡 𝛼 2 ∙ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 . The confidence interval is constructed by adding and subtracting the margin of error from the point estimate. 𝑥 1 − 𝑥 2 − 𝑡 𝛼 2 ∙ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 < 𝜇 1 − 𝜇 2 < 𝑥 1 − 𝑥 2 + 𝑡 𝛼 2 ∙ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 This method is often referred to as Welch’s method.

Assumptions Welch’s Method requires some assumptions: Assumptions: We have simple random samples from two populations. The samples are independent of one another. Each sample size is large (𝑛 > 30), or its population is approximately normal.

Example A drug company has developed a new drug that is designed to reduce high blood pressure. To test the drug, a sample of 15 patients is recruited to take the drug. Their systolic blood pressures are reduced by an average of 28.3 mmHg, with a standard deviation of 12.0 mmHg. In addition, another sample of 20 patients takes a standard drug. The blood pressures in this group are reduced by an average of 17.1 mmHg with a standard deviation of 9.0 mmHg. Assume that blood pressure reductions are approximately normally distributed. Find a 95% confidence interval for the difference between the population mean reduction for the new drug and that of the standard drug. Solution: We first check the assumptions. We have two independent random samples, and the populations are approximately normally distributed. The assumptions are satisfied. We summarize the information: New Drug Standard Drug Sample mean 𝑥 1 =28.3 𝑥 2 =17.1 Sample stand dev. 𝑠 1 =12 𝑠 2 =9 Sample size 𝑛 1 =15 𝑛 2 =20 Population Mean 𝜇 1 (unknown) 𝜇 2 (unknown)

Solution The point estimate is: 𝑥 1 − 𝑥 2 =28.3−17.1=11.2 The degrees of freedom is the smaller of 𝑛 1 − 1 = 14 and 𝑛 2 − 1 = 19, which is 14. Using Table A.3, degrees of freedom of 14 and 95% confidence level, we obtain a critical value of 𝑡 𝛼 2 =2.145. The standard error is 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 = 12.0 2 15 + 9.0 2 20 =3.6946. The margin of error is 𝑡 𝛼 2 ∙ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 =2.145∙ 12.0 2 15 + 9.0 2 20 =7.925. The 95% confidence interval is 𝑥 1 − 𝑥 2 - 𝑡 𝛼 2 ∙ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 < 𝜇 1 − 𝜇 2 < 𝑥 1 − 𝑥 2 + 𝑡 𝛼 2 ∙ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 11.2 – 7.925 < 𝜇 1 − 𝜇 2 < 11.2 + 7.925 3.3 < 𝜇 1 − 𝜇 2 < 19.1 We are 95% confident that the new drug provides a greater reduction in systolic blood pressure, and that the improvement due to the new drug is between 3.3 and 19.1 mmHg. Remember: 𝑥 1 =28.3 𝑥 2 =17.1 𝑠 1 =12 𝑠 2 =9 𝑛 1 =15 𝑛 2 =20

Technology and Degrees of Freedom If you construct a confidence interval for the difference between two means with technology, you will get a somewhat different answer than you will get using the method we have presented here. The reason is that computers and calculators compute the number of degrees of freedom differently, using a more accurate but rather complicated formula. Most computers and calculators compute the degrees of freedom as follows: Degrees of freedom = 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 2 𝑠 1 2 𝑛 1 2 𝑛 1 −1 + 𝑠 2 2 𝑛 1 2 𝑛 1 −1

Confidence Intervals on the TI-84 PLUS The 2-SampTInt command constructs confidence intervals for the difference of the means of two independent samples. This command is accessed by pressing STAT and highlighting the TESTS menu. If the summary statistics are given the Stats option should be selected for the input option. If the raw sample data are given, the Data option should be selected.

Example (TI-84 PLUS) A drug company has developed a new drug that is designed to reduce high blood pressure. To test the drug, a sample of 15 patients is recruited to take the drug. Their systolic blood pressures are reduced by an average of 28.3 mmHg, with a standard deviation of 12.0 mmHg. In addition, another sample of 20 patients takes a standard drug. The blood pressures in this group are reduced by an average of 17.1 mmHg with a standard deviation of 9.0 mmHg. Assume that blood pressure reductions are approximately normally distributed. Find a 95% confidence interval for the difference between the population mean reduction for the new drug and that of the standard drug. Solution: We first check the assumptions. We have two independent random samples, and the populations are approximately normally distributed. The assumptions are satisfied. We summarize the information: New Drug Standard Drug Sample mean 𝑥 1 =28.3 𝑥 2 =17.1 Sample stand dev. 𝑠 1 =12 𝑠 2 =9 Sample size 𝑛 1 =15 𝑛 2 =20 Population Mean 𝜇 1 (unknown) 𝜇 2 (unknown)

Example (TI-84 PLUS) Solution: We press STAT and highlight the TESTS menu and select 2-SampTInt. Select Stats as the input method and enter the following values: Enter 0.95 as the confidence level and select No for the pooled option. Select Calculate. The confidence interval is (3.5912, 18.809). The results differ somewhat from those when calculating the confidence interval by hand, because the degrees of freedom has been calculated by the more accurate formula. 𝑥 1 =28.3 𝑥 2 =17.1 𝑠 1 =12 𝑠 2 =9 𝑛 1 =15 𝑛 2 =20

Objective 3 Describe the pooled standard deviation and the known standard deviation methods

Alternate Methods In most situations in practice, Welch’s method is the method of choice for constructing confidence intervals for the difference between two means with independent samples. There are two other methods that are sometimes used. They are often not the best to use in practice, however, so we will always use Welch’s method. Alternate Methods: Using the pooled standard deviation when two population variances 𝜎 1 2 and 𝜎 2 2 are known to be equal. Using 𝑧 𝛼 2 when the population variances 𝜎 1 and 𝜎 2 are known.

Confidence Intervals Using Pooled Standard Deviation When the two population variances, 𝜎 1 2 and 𝜎 1 2 , are known to be equal, there is an alternate method for computing a confidence interval. This alternate method was widely used in the past, and is still an option in many forms of technology. However, it is not recommended. The major problem with this method is that the assumption that the population variances are equal is very strict. The method can be quite unreliable if it is used when the population variances are not equal. Step 1: Compute the pooled standard deviation, 𝑠 𝑝 , as follows: 𝑠 𝑝 = 𝑛 1 −1 𝑠 1 2 + 𝑛 2 −1 𝑠 2 2 𝑛 1 + 𝑛 2 −2 Step 2: Compute the degrees of freedom: Degrees of freedom = 𝑛 1 + 𝑛 2 −2 A level 100(1 - 𝛼)% confidence interval is 𝑥 1 − 𝑥 2 - 𝑡 𝛼 2 ∙ 𝑠 𝑝 1 𝑛 1 + 1 𝑛 2 < 𝜇 1 − 𝜇 2 < 𝑥 1 − 𝑥 2 + 𝑡 𝛼 2 ∙ 𝑠 𝑝 1 𝑛 1 + 1 𝑛 2

Confidence Intervals When 𝜎 1 and 𝜎 2 are Known A level 100(1 − 𝛼)% confidence interval when 𝜎 1 and 𝜎 2 are known is given by 𝑥 1 − 𝑥 2 − 𝑧 𝛼 2 𝜎 1 2 𝑛 1 + 𝜎 2 2 𝑛 2 < 𝜇 1 − 𝜇 2 < 𝑥 1 − 𝑥 2 + 𝑧 𝛼 2 𝜎 1 2 𝑛 1 + 𝜎 2 2 𝑛 2 Note that this method is the same as Welch’s except that the sample standard deviations 𝑠 1 and 𝑠 2 are replaced with the population standard deviations 𝜎 1 and 𝜎 2 , and 𝑡 𝛼 2 is replaced with 𝑧 𝛼 2 .

You Should Know… The difference between independent and paired samples How to construct confidence intervals for the difference between two population means The pooled standard deviation method The known standard deviation method