COMM
InferenceScenarios_2013

# E the margin of error in a confidence interval

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point estimate only and ignore sampling error (i.e., the margin of error in a confidence interval). Suppose that a simple random sample of 51 stores this month shows mean sales of 52 units of a small appliance with standard deviation 10 units. During the same month last year, a sample of 75 stores gave mean sales of 48 units with standard deviation 9 units. An increase from 48 to 52 is a rise of 8.33% so the manager is happy because “Sales are up 8.33%,” he claims. SOLUTION: Two-sample t-test of two independent means Use the version with the pooled estimate for the variance. Since the sample standard deviations are so close to one another (10 and 9), we assume that the samples come from populations with equal variances. H 0 : µ 1 − µ 2 = 0, H a : µ 1 − µ 2 ≠ 0, where µ 1 and µ 2 are the average sales of this year and last year, respectively. Test statistic: t = (52–48) / 9.416√[1/51+1/75] = 2.341, with 124 df. P-value = TDIST(2.341,124,2) = 0.0208 At a 5% significance level, there is sufficient evidence to conclude that the change in sales from last year to this is statistically significant. (But not at the 1% sig. level!) Note: With the unpooled version, the test statistic = 2.294. Explanation to Manager: The data come from only a sample of stores, not all stores in the chain. It is probable that the true average sales—from all stores—changed by some amount other than the 4 units, or 8.33%, observed in the sample. *** 95% confidence interval for the difference in the mean number of units sold at all retail stores: Pooled SD = 9.416 95% CI = (52-48) ± 1.979 x 9.416√[1/51+1/75] = 4 ± 3.4 or (0.6,7.4) Note: The unpooled version would give an interval of (0.5,7.5) Since 0 is not within the 95% confidence interval from part, a two-sided test of H 0 , µ 1 − µ 2 = 0 at the 5% level of significance will necessarily be rejected. The difference between the mean sales year over year is significantly different from 0. c) Explain in language that the marketing manager can understand why he cannot be certain that sales went up by 8.33%. 4

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Scenario 5. Hospital Admissions on Friday the 13th This question is based on a research study quoted on CBC Radio 1 on Friday, February 13, 2009. Is Friday the 13 th really unlucky? Do hospital emergency wards see more admissions (that is, acutely injured or sick patients) on a Friday the 13 th than they see, for example, on each previous Friday? A British study examined the number of emergency admissions on various Fridays the 13 th and on the previous Fridays the 6 th between 1989 and 1995. Here are their results. Admissions on Admissions on Friday the 6 th Friday the 13 th October 1989 29 41 July 1990 22 40 September 1991 40 46 December 1991 36 39 March 1992 15 14 November 1992 16 43 August 1993 19 19 May 1994 38 42 January 1995 28 27 October 1995 45 44 Assess the evidence that, in the long run, there are more emergency admissions on Fridays the 13 th than there are on the preceding Fridays by conducting a hypothesis test at the 5% level.
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• Spring '10
• E.Fowler
• Statistics, Statistical hypothesis testing

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