Statistical Analysis of Experimental Research: Hypothesis Testing and Confidence Intervals

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Statistical Analysis of Experimental Research: Hypothesis Testing andConfidence IntervalsPart I1.Jackson even-numbered Chapter exercises (pp. 220-221; 273-275)Jackson, Sherri L. (2012). Research methods and statistics: Acritical thinking approach.Fourth ed.pp220-221.Evenexercises.2. The producers of a new toothpaste claim that itprevents more cavities than other brands oftoothpaste. A random sample of 60 people usedthe new toothpaste for 6 months. The meannumber ofcavities at their next checkup is 1.5. Inthe general population, the mean number of cavities at a 6-month checkup is 1.73 (∂= 1.12).a. Is this a one-or two-tailed test?Answer: one tailed test.b. What are Hoand Hafor this study?Answer:Null hypothesis H0: The Mean number of cavities with the newtoothpaste≥1.73. (μ≥1.73)against thealternative H1:The Mean number of cavities with the new toothpaste < 1.73. (μ < 1.73).c. Compute zobt.Answer:Zₒbt·=1.5−1.731.12/√60=-1.590689589d. What is zcv?Answer: zcv=-1.644853627e. Should Hobe rejected? What should the researcher conclude?Answer: TheHₒshould be rejected if the value of the test is less than the critical value of 0.05.The nullhypothesis should not be rejected since the calculated value of the test statistic is not less than thecritical value. The researcher does not have enough samples to conclude that the new tooth paste willprevent cavities. If there were enough samples the researcher could support the claim that the newtooth paste would prevent cavities compare to other brands.From this researcher can conclude that theMean number of cavities with the newtoothpaste≥1.73.f. Determine the 95% confidence interval for the population mean, based on the sample mean.Answer:95% confidence interval for the population mean is the upper bound so the 95% confidence interval isthe upper bound = sample mean +(-zcv)*௦௔௠௣௟௘௦𝑡௔𝑛𝑑௔௥𝑑௘𝑑௩௜௔𝑡௜௢𝑛ඥ௦௔௠௣௟௘௦௜௭௘)=1.5+1.644853627*1.12√60=1.7384.Henry performed a two-tailed test for an experiment in which N=24. He could not find his table of tcritical values, but he remembered the tcvat df = 13.He decided to compare his tobtwith thistcv. Is hemore likely to make a Type I error or a Type II error in this situation?

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