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In the last few years, many research studies have shown that the purported benefits of hormone replacement therapy (HRT) do not exist, and in fact, that hormone replacement therapy actually increases the risk of several serious diseases. A four-year experiment involving 4472 women was conducted at 35 medical centres. Half of the women took placebos and half took a prescription drug, a widely prescribed type of hormone replacement therapy. There were x, = 43 cases of dementia in the hormone group and x2 = 22 in the placebo group. Is there sufficient evidence to indicate that the risk of dementia is higher for patients using the prescription drug? Test at the 1% level of significance. (Round your answers to two decimal places.) 1-2. Null and alternative hypotheses: O Ho: (P1 - P2) * 0 versus Ha: (P1 - P2) = 0 O Ho: (P1 - P2) = 0 versus Ha: (P1 - P2) * 0 O Ho: (P1 - P2) = 0 versus Ha: (P1 - P2) > 0 O Ho: (P1 - P2) = 0 versus Ha: (P1 - P2) 0 3. Test statistic: Z = 4. Rejection region: If the test is one-tailed, enter NONE for the unused region. 5. Conclusion: O Ho is not rejected. There is insufficient evidence to indicate that the risk of dementia is higher for patients using the prescription drug. O Ho is not rejected. There is sufficient evidence to indicate that the risk of dementia is higher for patients using the prescription drug. O Ho is rejected. There is sufficient evidence to indicate that the risk of dementia is higher for patients using the prescription drug. O Ho is rejected. There is insufficient evidence to indicate that the risk of dementia is higher for patients using the prescription drug. Need Help? Read ItHere are the red blood cell counts (in 10 cells per microliter) of a healthy person measured on each of 15 days. 5.5 5.1 5.0 5.3 5.5 5.4 5.3 5.3 5.1 5.2 5.2 4.8 5.5 5.1 5 .3 LO USE SALT Find a 95% confidence interval estimate of , the true mean red blood cell count (in 10 cells per microliter) for this person during the period of testing. (Round your answers to three decimal places.) 10 cells per microliter to 10 cells per microliter You may need to use the appropriate appendix table to answer this question. Need Help? Read It [-/2 Points] DETAILS MENDSTATC4 10.E.025. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Calculate s , the pooled estimator for 2, in these cases. (Round your answers to three decimal places.) n, = 11, n, = 4, s, 2 = 3.1, s, 0 O Ho: (/41 \"2) 0 H0: (#1 \"2) = 0 versus Ha: (u1 [42) 0 O HO: (#1 142) = 0 versus Ha: (141 142) t 0 O Ho: ( M1 - 12) = 0 versus Ha: (1 1 - 12) * 0 O Ho: (1 1 - M2) = 0 versus Ha: (M1 - 12) 0 OH: (M1 - 12 ) = 0 versus Ha: (1 1 - 12) = 0 (b) Do the data provide sufficient evidence to indicate that the oral antiplaque rinse is effective? Test using a = 0.05. (Round your answers to three decimal places.) Test statistic: t = Rejection region: If the test is one-tailed, enter NONE for the unused region. t > Conclusion: O Ho is not rejected. There is insufficient evidence to indicate that the oral antiplaque rise is effective. O Ho is rejected. There is insufficient evidence to indicate that the oral antiplaque rise is effective. O Ho is not rejected. There is sufficient evidence to indicate that the oral antiplaque rise is effective. O Ho is rejected. There is sufficient evidence to indicate that the oral antiplaque rise is effective. (c) Find the approximate p-value for the test.Industrial wastes and sewage dumped into our rivers and streams absorb oxygen and thereby reduce the amount of dissolved oxygen available for fish and other forms of aquatic life. One state agency requires a minimum of 5 parts per million (ppm) of dissolved oxygen in order for the oxygen content to be sufficient to support aquatic life. A pollution control inspector suspected that a river community was releasing amounts of semitreated sewage into a river. To check his theory, he drew five randomly selected specimens of river water at a location above the town, and another five below. The dissolved oxygen readings (in parts per million) are as follows. Above Town 4.8 5.3 5.1 4.9 5.2 Below Town 5.1 4.8 4.9 4.7 4.8 (a) Do the data provide sufficient evidence to indicate that the mean oxygen content below the town is less than the mean oxygen content above? Test using a = 0.05. (Use , for the population mean for the above town location and u, for the population mean for the below town location.) State the null and alternative hypotheses. O Ho : ( H1 - M2 ) = 0 versus Ha: (M1 - 12 ) $ 0 O HO : ( 1 1 - 1 2 ) 0 O Ho : ( 1 1 - 12 ) = 0 versus Ha : ( 1 1 - 12 ) = 0 O Ho: (M1 - 12 ) = 0 versus Ha: (1 1 - 12) > 0 O HO : ( M 1 - 1 2 ) = 0 versus Ha : ( 1 1 - 1 2 ) State the conclusion. O Ho is rejected. There is insufficient evidence to indicate that the mean content of oxygen below town is less than the mean content above town. O Ho is rejected. There is sufficient evidence to indicate that the mean content of oxygen below town is less than the mean content above town. O Ho is not rejected. There is sufficient evidence to indicate that the mean content of oxygen below town is less than the mean content above town. O Ho is not rejected. There is insufficient evidence to indicate that the mean content of oxygen below town is less than the mean content above town. (b) Suppose you prefer estimation as a method of inference. Estimate the difference in the mean dissolved oxygen contents (in ppm) for locations above and below the town. Use a 95% confidence interval. (Use "1 - M2. Round your answers to three decimal places.) ppm to ppm You may need to use the appropriate appendix table or technology to answer this question.Chronic anterior compartment syndrome is a condition characterized by exercise-induced pain in the lower leg. Swelling and impaired nerve and muscle function also accompany this pain, which is relieved by rest. Researchers conducted an experiment involving 10 healthy runners and 10 healthy cyclists to determine whether there are significant differences in pressure measurements within the anterior muscle compartment for runners and cyclists. In addition to the compartment pressures, the level of creatine phosphokinase (CPK) in blood samples, a measure of muscle damage, was determined for each of n, = 10 runners and n2 = 10 cyclists before and after exercise. The data summary-CPK values in units/liter-is as follows. Runners Cyclists Standard Standard Condition Mean Deviation Mean Deviation Before Exercise 252.69 115.49 175.3 60.65 After Exercise 284.79 132.68 179.4 64.55 Difference 32.1 21.04 4,1 6.85 (a) Test for a significant difference in mean CPK values for runners and cyclists before exercise under the assumption that of 2 # 62 ; use a = 0.05. State the null and alternative hypotheses. O HO : ( M 1 - 12 ) = 0 versus Ha : ( 1 1 - 12 ) > 0 O Ho: ( M1 - M2 ) = 0 versus Ha: (1 1 - 12) = 0 O Ho: (M1 - M2) = 0 versus Ha: (1 1 - 12) * 0 O Ho : ( 1 1 - 1 2 ) 2.10 X t 0 O Ho: (M1 - H2 ) 2.143 X t 056.207 * units/L to 154.57 X units/L (c) Test for a significant difference in mean CPK values for runners before and after exercise. (Use M, for after exercise and u2 for before exercise.) State the null and alternative hypotheses O HO: (1 1 - 12 ) * 0 versus Ha: (1 1 - 12 ) = 0 O Ho: (M1 - M2) = 0 versus Ha: (1 1 - 12) * 0 O HO : ( M 1 - 12 ) = 0 versus Ha: (1 1 - 12 ) > 0 O Ho : ( M 1 - 12 ) t