InferenceScenarios_2013

# Solution matched pairs t test of two dependent means

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SOLUTION: Matched pairs t-test of two dependent means This is a matched pairs design. The ten differences, Friday 13 th admissions minus Friday 6 th admissions, are: d i = {12, 18, 6, 3, –1, 27, 0, 4, –1, –1} = 6.70; s d = 9.50; n = 10 H 0 : μ d = 0; H a : μ d > 0 Test statistic: t = (6.70–0)/ 9.50/√10 = 2.23, with 9 df. P-value = TDIST(2.23,9,10) = 0.027 Conclusion: At the 5% significance level we reject the null hypothesis. There is sufficient evidence to conclude that there are more emergency admissions on Fridays the 13 th than on the preceding Fridays. If you used a two-sample t-test by mistake, you would get a test statistic of 1.364 and a P-value of 0.0947. In this case you would conclude that there is insufficient evidence of a difference between admission numbers—the opposite conclusion from the matched pairs design! This shows why it is important to keep the experimental design in mind as you choose your test. 5

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Scenario 6. Home Heating A housing survey of single-family homes in two provinces is conducted to determine the proportion of such homes that have gas heat. A sample of 300 homes in Province A shows 185 have gas heat, compare with 75 of 200 in Province B. Compare the proportions of single-family homes with gas heat. SOLUTION: Two-sample z-test of two proportions H o : p 1 = p 2 H a : p 1 ≠ p 2 Use the pooled estimate of proportion (p-hat) for the standard error; p-hat=(185+75)/(300+200) = 0.52 z = (0.617 – 0.375)/√[(0.52)(0.48)(1/300+1/200)] = 5.30 P-value < .001; there is strong evidence that the proportions of single-family homes with gas heat is different between the two provinces. Scenario 7. More on Home Heating To study the effectiveness of wall insulation in saving energy for home heating, the energy consumption (in MWh) for 10 houses was recorded for two winters – one before insulation and one after. Test whether insulation has reduced average energy consumption. House: 1 2 3 4 5 6 7 8 9 10 Before: 12.1 11.0 14.1 13.8 15.5 12.2 12.8 9.9 10.8 12.7 After: 12.0 10.6 13.4 11.2 15.3 13.6 12.6 8.8 9.6 12.4 di: 0.1 0.4 0.7 2.6 0.2 –1.4 0.2 1.1 1.2 0.3 SOLUTION: This is a paired t-test of two dependent means: the same house is used for two measurements. H o : μ d = 0 H a : μ d > 0 Since the question asks whether insulation has reduced consumption this is a one-sided alternative. The mean of the di’s is 0.54 and the SD is 1.016. Then t = 0.54/(1.016/√10) = 1.681. Excel will compute an exact P-value, but if you use the t-tables you can only find lower and upper bounds on the P-value: since t(.05,9)=1.833 and t(.10,9)=1.383 then 0.05 < P-value < 0.10. Hence, there is not enough evidence to conclude that consumption has been reduced. Note that if a two-sided alternative were being tested, the P-value bounds would be: 2x(0.05) < P-value < 2x(0.10) or 0.10 < P-value < 0.20. Note also that this is an observational study and many factors that are uncontrolled could affect the differences. For example, a warmer second winter could confound the insulation effect; there is no control on homeowners changing the thermostat settings, etc.
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• Spring '10
• E.Fowler
• Statistics, Statistical hypothesis testing

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