![]() From Figure 12.3 "Critical Values of ", in the row with the heading df = 11 + 6 − 2 = 15 we read that t 0.025 = 2.131. To apply the formula for the confidence interval, we must find t α ∕ 2. In words, we estimate that the average monthly sales for Design 1 is 6 units more per month than the average monthly sales for Design 2. The point estimate of μ 1 − μ 2 is x - 1 − x - 2 = 52 − 46 = 6 ![]() Construct a point estimate and a 95% confidence interval for the difference in average monthly sales between the two package designs. Design 2 is sent to 6 stores their average sales the first month is 46 units with sample standard deviation 10 units. Design 1 is sent to 11 stores their average sales the first month is 52 units with sample standard deviation 12 units. Ī software company markets a new computer game with two experimental packaging designs. Since the symbol in H a is “>” this is a right-tailed test, so there is a single critical value, z α = z 0.01, which from the last line in Figure 12.3 "Critical Values of " we read off as 2.326. Since the samples are independent and both are large the test statistic is Z = ( x - 1 − x - 2 ) − D 0 s 1 2 n 1 + s 2 2 n 2 The test is therefore H 0 : μ 1 − μ 2 = 0 vs. To say that the mean customer satisfaction for Company 1 is higher than that for Company 2 means that μ 1 > μ 2, which in terms of their difference is μ 1 − μ 2 > 0. If the mean satisfaction levels μ 1 and μ 2 are the same then μ 1 = μ 2, but we always express the null hypothesis in terms of the difference between μ 1 and μ 2, hence H 0 is μ 1 − μ 2 = 0. Test at the 1% level of significance whether the data provide sufficient evidence to conclude that Company 1 has a higher mean satisfaction rating than does Company 2. Refer to Note 9.4 "Example 1" concerning the mean satisfaction levels of customers of two competing cable television companies. In the context of the problem we say we are 99% confident that the average level of customer satisfaction for Company 1 is between 0.15 and 0.39 points higher, on this five-point scale, than that for Company 2. We are 99% confident that the difference in the population means lies in the interval, in the sense that in repeated sampling 99% of all intervals constructed from the sample data in this manner will contain μ 1 − μ 2. From Figure 12.3 "Critical Values of " we read directly that z 0.005 = 2.576. To apply the formula for the confidence interval, proceed exactly as was done in Chapter 7 "Estimation". In words, we estimate that the average customer satisfaction level for Company 1 is 0.27 points higher on this five-point scale than it is for Company 2. Without reference to the first sample we draw a sample from Population 2 and label its sample statistics with the subscript 2.Ĭonstruct a point estimate and a 99% confidence interval for μ 1 − μ 2, the difference in average satisfaction levels of customers of the two companies as measured on this five-point scale. We draw a random sample from Population 1 and label the sample statistics it yields with the subscript 1. We arbitrarily label one population as Population 1 and the other as Population 2, and subscript the parameters with the numbers 1 and 2 to tell them apart. Each population has a mean and a standard deviation. Figure 9.1 "Independent Sampling from Two Populations" illustrates the conceptual framework of our investigation in this and the next section. Suppose we wish to compare the means of two distinct populations. To learn how to perform a test of hypotheses concerning the difference between the means of two distinct populations using large, independent samples.To learn how to construct a confidence interval for the difference in the means of two distinct populations using large, independent samples.To understand the logical framework for estimating the difference between the means of two distinct populations and performing tests of hypotheses concerning those means.
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