Thus the p-value (since the area must be doubled) is between 0.400 and 0.200. Since 1.040 is between 0.866 and 1.341 the area it cuts off is between 0.200 and 0.100. The area cut off by t = 0.866 is 0.200 and the area cut off by t = 1.341 is 0.100. ![]() Looking in the row of Figure 12.3 "Critical Values of " headed d f = 15, the number 1.040 is between the numbers 0.866 and 1.341, corresponding to t 0.200 and t 0.100. Because the test is two-tailed the observed significance or p-value of the test is the double of the area of the right tail of Student’s t-distribution, with 15 degrees of freedom, that is cut off by the test statistic T = 1.040. The first three steps are identical to those in Note 9.13 "Example 5". Perform the test of Note 9.13 "Example 5" using the p-value approach. ![]() From the row in Figure 12.3 "Critical Values of " with the heading d f = 15 we read off t 0.005 = 2.947. ![]() Since the symbol in H a is “≠” this is a two-tailed test, so there are two critical values, ± t α ∕ 2 = ± t 0.005. Inserting the data and the value D 0 = 0 into the formula for the test statistic gives T = ( x - 1 − x - 2 ) − D 0 s p 2 ( 1 n 1 + 1 n 2 ) = ( 52 − 46 ) − 0 129. Which has Student’s t-distribution with d f = 11 + 6 − 2 = 15 degrees of freedom. Since the samples are independent and at least one is less than 30 the test statistic is T = ( x - 1 − x - 2 ) − D 0 s p 2 ( 1 n 1 + 1 n 2 ) The relevant test is H 0 : μ 1 − μ 2 = 0 vs. Test at the 1% level of significance whether the data provide sufficient evidence to conclude that the mean sales per month of the two designs are different. Refer to Note 9.11 "Example 4" concerning the mean sales per month for the same computer game but sold with two package designs. Because the interval contains both positive and negative values the statement in the context of the problem is that we are 95% confident that the average monthly sales for Design 1 is between 18.3 units higher and 6.3 units lower than the average monthly sales for Design 2. We are 95% confident that the difference in the population means lies in the interval, in the sense that in repeated sampling 95% of all intervals constructed from the sample data in this manner will contain μ 1 − μ 2. From the formula for the pooled sample variance we compute s p 2 = ( n 1 − 1 ) s 1 2 + ( n 2 − 1 ) s 2 2 n 1 + n 2 − 2 = ( 10 ) ( 12 ) 2 + ( 5 ) ( 10 ) 2 15 = 129. 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. A software company markets a new computer game with two experimental packaging designs.
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