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Statistical Analysis of Experimental Research: Hypothesis Testing and Confidence Intervals

Application of hypothesis testing in experimental research studies.

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    Statistical Analysis of Experimental Research: Hypothesis Testing and Confidence Intervals Part I 1. Jackson even - numbered Chapter exercises (pp. 220 - 221; 273 - 275) Jackson, Sherri L. (201 2 ). Research methods and statistics: A critical thinking approach . Fourth ed. p p220 - 221 . Even exercises . 2. The producers of a new toothpaste claim that it prevents more cavities than other brands of toothpaste. A random sample of 60 people used the new toothpaste for 6 months. The mean number of cavities at their next checkup is 1.5. In the 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 H o and H a for this study? Answer: Null hypothesis H 0 : T he Mean number of cavities with the new toothpaste ≥ 1.73. (μ ≥ 1.73) against the alternative H 1 : T he Mean number of cavities with the new toothpaste < 1.73. (μ < 1.73) . c. Compute z obt . Answer: Z ₒ bt · = 1 . 5 − 1 . 73 1 . 12 / √ 60 = - 1.590689589 d. What is z cv ? Answer: z cv = - 1.644853627 e. Should H o be rejected? What should the researcher conclude? Answer: The H ₒ should be rejected if the value of the test is less than the critical value of 0.05 . The null hypothesis should not be rejected since the calculated value of the test statistic is not less than the critical value. The researcher does not have enough samples to conclude that the new tooth paste will prevent cavities. If there were enough samples the researcher could support the claim that the new tooth paste would prevent cavities compare to other brands. From this researcher can conclude that t he Mean number of cavities with the new toothpaste ≥ 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 is the upper bound = sample mean + ( - z cv ) * ௦௔௠௣௟௘ ௦𝑡௔𝑛𝑑௔௥𝑑 ௘𝑑௩௜௔𝑡௜௢𝑛 ඥ ௦௔௠௣௟௘ ௦௜௭௘ )=1.5+ 1.644853627 * 1 . 12 √ 60 = 1.738 4. Henry performed a two - t ailed test for an experiment in which N=24. He could not find his table of t critical values, but he remembered the t cv at df = 13. He decided to compare his t obt with this t cv . Is he more likely to make a Type I error or a Type II error in this situation?

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    Answer: Critical t: ±2.0687 at DF= 23 Critical t: ±2.1604 at DF= 13 Type II error. 6. A researcher Hypothesizes that individuals who listen to classical music will score differently from the general population on a test of special ability, μ ∂ = 58. A random sample of 14 individuals who listen to cla ssical music is given the same t es t . Their scores on the test are 52, 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 H o and H a for this study? c. Compute t obt . d. What is t cv ? e. Should H o be 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 H 0 : μ=58 against the alternative H 1 :μ ≠ 58 . c. t obt = 52.34. d. t cv =2.16. e. Since the calculated t=52.35 which is greater than t cv =2.16 therefore we reject the null hypothesis that μ ≠ 58 which means individuals who listen to classical music will score differently 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 the national exercise rate. The national rate is 20% . The researcher gathers a random sample of 120 individuals who live in California and finds that the number who exercise regularly is 31 out of 120. a. What is x 2 obt ? Answer: Observed Frequencies Proportions Expected Frequencies Chi - Square (O - E)^2/O 31 0.2 24 2.042 89 0.8 96 0.510 Total 120 1 120 2.552 Chi - Square Value 2.552 p - value 0.4660 Therefore x 2 obt =2.552. b. What is df for this test? Answer: The df for this test will be =(2 - 1)=1.

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    c. What is x 2 cv ? Answer: x 2 cv =3.841 at 5% level of significance. d. What conclusion should be drawn from these results? Answer: since 2.552<3.841 therefore we fail to reject null hypothesis for this test. The conclusion should be drawn from these result is that the percentage of people who exercise in California is at most the national exercise rate, p ≤ 0 . 20 Jackson S. L. pp 273 – 275. Even exercises . 2. A student is interested in whether students who study with music playing devote as much attention to their studies as do students who study under quiet conditions (he believes that studying under quiet conditions leads to better attention). He randomly assigns participants to either the music or no - music condition 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 test represent interval - ratio data and are normally distributed. Music No Music 6 10 5 9 6 7 5 7 6 6 6 6 7 8 8 6 5 9 a. 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 for difference of means to analyze the given data. b. Identify H 0 and H a for this study. Answer: H0: No music does not have any effect on scores of students, μ X = μ Y HA: No music will increase the scores of students, μ X < μ Y c. Conduct the appropriate analysis. Answer: Let X be the scores of students with music. And Y be the scores without music Music (X) No Music (Y) d ൫ 𝑑 − 𝑑 ̅ ൯ 2 7 7 0 1 6 8 2 1 5 7 2 1 6 7 1 0 8 9 1 0 8 8 0 1 6 4 Under H 0 , the test statistics is:

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