Statistical Hypothesis Testing and Confidence Intervals: A Critical Thinking Approach

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Statistical Hypothesis Testing and Confidence Intervals: A Critical ThinkingApproachPart IJackson, Sherri L. (2012). Research methods and statistics: A critical 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 ata 6-monthcheckup 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·==-1.590689589d. What is zcv?Answer: zcv=-1.644853627e. Should Hobe rejected? What should the researcher conclude?Answer: TheHₒ should 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)*)=1.5+1.644853627*=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?Answer:Critical t: ±2.0687 at DF= 23Critical t: ±2.1604 at DF= 13Type II error.

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6. A researcher Hypothesizes that individuals who listen to classical music will score differently from thegeneral population on a test of special ability,μ∂= 58. A random sample of 14 individuals who listen toclassical music is given the same test. Their scores on the test are52, 59, 63, 65, 58, 55, 62, 63, 53, 59,57, 61, 60, 59.a.Is this a one-or two-tailed test?b. What are Hoand Hafor this study?c. Compute tobt.d. What is tcv?e. Should Hobe rejected? What should the researcher conclude?f. Determine the 95% confidence interval for the population mean, based on the sample mean.Answer:a.Two tailed test.b.Null H0: μ=58 against the alternative H1:μ.c.tobt=52.34.d.tcv=2.16.e.Since the calculated t=52.35 which is greater than tcv=2.16 therefore we reject the nullhypothesis that μwhich meansindividuals who listen to classical music will scoredifferently from the general population on a test of special ability.f.95% confidence interval for the population mean is(56.80, 61.20)8. A researcher believes that the percentage of people who exercise in California is greater than thenational exercise rate. The national rate is 20%. The researcher gathers a random sample of 120individuals who live in California and finds that the number who exercise regularly is 31 out of 120.a. What is x2obt?Answer:ObservedFrequenciesProportionsExpectedFrequenciesChi-Square (O-E)^2/O310.2242.042890.8960.510Total12011202.552Chi-Square Value2.552p-value0.4660Therefore x2obt=2.552.b. What is df for this test?Answer: The df for this test will be =(2-1)=1.c. What is x2cv?Answer: x2cv=3.841 at 5% level of significance.d.What conclusion should be drawnfrom these results?Answer: since 2.552<3.841 therefore we fail to reject null hypothesis for this test. The conclusion shouldbe drawn from these result is that the percentage of people who exercise in California is at most thenational exercise rate, pJackson S. L. pp 273–275. Evenexercises.

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2. A student is interested in whether students who study with music playing devote as much attentionto their studies as do students who study under quiet conditions (he believes that studying under quietconditions leads to better attention). He randomly assigns participants to either the music or no-musiccondition and has them read and study the same passage of information for the same amount of time.Subjects are given the same 10-item test on the material. Their scores appear next. Scores on the testrepresent interval-ratio data and are normally distributed.Music No Music6105967576666788659a. What statistical test should be used to analyze these data?Answer:Since, the same sample of participants is used in both the cases; we’ll apply paired t-test fordifference of means to analyze the given data.b. IdentifyH0andHafor this study.Answer:H0: No music does not have any effect on scores of students, μX = μYHA: No music will increase the scores of students, μX<μYc. Conduct the appropriate analysis.Answer:Let X be the scores of students with music. And Y be the scores without musicMusic (X)No Music (Y)d77016821572167108910880164Under H0, the test statistics is:No. of observations == 9

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