BTM8106-week2 complete work
BTM8106-week2 complete work
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Date Created: 11/05/15
BTM810682 1. Jackson evennumbered Chapter Exercises (pp. 220221; 273275) Chapter 8/pp. 220221 Part I Jackson, Sherri L. (2012). Research methods and statistics: A critical thinking approach. Fourth ed. pp220-221. 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 and H for this study? o a Answer: Null hypothesis H :0The Mean number of cavities with the new toothpaste ≥ 1.73. (µ ≥ 1.73) against the alternative H :1The Mean number of cavities with the new toothpaste < 1.73. (µ < 1.73). c. Compute z obt. 1.5−1.73 Answer: Zₒbt·= 1.12/√60 =-1.590689589 d. What is z cv Answer: z cv1.644853627 e. Should H oe rejected? What should the researcher conclude? Answer: The Hₒ 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 the 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 √sample¿¿ 1.12 samplestandardedviation )=1.5+1.644853627* √60 = 1.738 ¿ 4. Henry performed a two-tailed test for an experiment in which N=24. He could not find his table of t critical values, but he remembered the t acvdf = 13. He decided to compare his t obtwith this cv. Is he more likely to make a Type I error or a Type II error in this situation? 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 classical music is given the same test. 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 ond H far this study? c. Compute t obt. d. What is tcv e. Should H oe 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 1µ ≠58 . c. tobt2.34. d. tcv2.16. e. Since the calculated t=52.35 which is greater than t =2cv6 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 x2obt Answer: Observed Proportion Expected Chi-Square Frequencies s Frequencies (O-E)^2/O 31 0.2 24 2.042 89 0.8 96 0.510 Total 120 1 120 2.552 Chi-Square 2.552 Value p-value 0.4660 Therefore x2obt2.552. b. What is df for this test? Answer: The2df for this test will be =(2-1)=1. c. What is xcv Answer: x 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 ≤0.20 who exercise in California is at most the national exercise rate, p Jackson S. L. pp 273 – 275. Even exercises. Chapter 10/pp. 273275 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 and H for this study. 0 a Answer: H0: No music does not have any effect on scores of students, μX = μY HA: No music will increase the scores of students, μX < μY c. Conduct the appropriate analysis. Answer: Let X be the scores of students with music. And Y be the scores without music Music No Music (d−d´) (X) (Y) d 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 , the test statistics is: 0 ´ t= d t S√n (n−2) No. of observations = n = 9 ´ 1 d= n ∑ d=1 1 2 4 S = ∑ d−d´) = =0.8 n−1 5 d 1 Therefore, t= = =2.738 S√n √0.8√ 6 Tabulated t0.05for 5 d.f. for single-tail test is 2.015. d. Should H 0 be rejected? What should the researcher conclude? Answer: Since calculated t > tabulated , H0 is rejected at 5% level of significance, i.e., the difference between the scores is significant. Hence, we conclude that studying under quiet conditions leads to better attention and thus, the higher scores. e. If significant, compute and interpret the effect size. Answer: The effect size is given by: ´−´x 7.667−6.667 ES= S = =1.118 √ 0.8 An effect size of 1.118 means that the average scores in the experimental (or No Music) group is 1.118 standard deviations above the average score in the control (Music) group. f. If significant, draw a graph representing the data. Answer: The graph representing the data is as follows: g. Determine the 95% confidence interval. Answer: The 95% confidence interval for difference mean is given by: ´ S 0.8 d±t(n−). n=1±2.5705∗ √ 6 =1±0.93864 √ Thus the confidence interval for difference mean is [0.06136,1.93864] 4. The researcher in exercise 2 decides to conduct the same study using a with-in participants design to control for differences in cognitive ability. He selects a random sample of subjects and has them study different material of equal difficulty in both the music and no-music conditions. The study is completely counter balanced to control for order effects. The data appear next. As before, they are measured on an interval-ratio scale and are normally distributed; he believes that studying under quiet conditions will lead to better performance. Music No Music 7 7 6 8 5 7 6 7 8 9 8 8 a. What statistical test should be used to analyze these data? Answer: To analyze the data we need to use paired t test. b. Identify H0 and H a for this study. Answer: H0: Studying under the quite condition and music condition have the same attention ie, d =0 H1: Studying under quite condition leads better attention i.d, µ <0 c. Conduct the appropriate analysis. Answer: The output of the above said paired t test is below: t-Test: Paired Two Sample for Means Music No Music 6.6666 7.666666 Mean 67 667 1.4666 0.666666 Variance 67 667 Observations 6 6 Pearson Correlation 0.6742 Hypothesized Mean Difference 0 df 5 - 2.7386 t Stat 1 0.0204 P(T<=t) one-tail 3 2.0150 t Critical one-tail 48 0.0408 P(T<=t) two-tail 59 2.5705 t Critical two-tail 82 d. Should H 0 be rejected? What should the researcher conclude? Answer: Test statistics: t=-2.739 P-value of the tst=0.020. Since the P-value of the test=0.020<0.05 at 5% level of significance therefore we reject the null hypothesis. The conclusion from these test is that the studying under quite condition leads better attention. e. If significant, compute and interpret the effect size. Answer: The effect size is given by: y−´x 7.667−6.667 ES= S = =1.118 √0.8 An effect size of 1.118 means that the average scores in the experimental (or No Music) group is 1.118 standard deviations above the average score in the control (Music) group. 6. Researchers at a food company are interested in how a new spaghetti sauce made from green tomatoes (and green in color) will compare to their traditional red spaghetti sauce. They are worried that the green color will adversely affect the tastiness scores. They randomly assign subjects to either the green or red sauce condition. Participants indicate the tastiness of the sauce on a 10-point scale. Tastiness scores tend to be skewed. The scores follow. Red Sauce Green Sauce 7 4 6 5 9 6 10 8 6 7 7 6 8 9 a. What statistical test should be used to analyze these data? As here the sample size is small and the data is not symmetric so we can’t use any parametric test thus we have to use a nonparametric test and in this case the best test would be Wilcoxon rank-sum test for tied case. b. Identify H0 and H a for this study. The null and alternative hypotheses are, H0: The two samples come from same population i.e. both the samples have same median Ha: The two samples come from different population i.e. both the samples have different medians c. Conduct the appropriate analysis. Using Minitab the obtained output is given below, Mann-Whitney Test and CI: Red, Green N Median Red 7 7.000 Green 7 6.000 Point estimate for ETA1-ETA2 is 0.000 95.9 Percent CI for ETA1-ETA2 is (-3.000,2.001) W = 55.0 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.7983 The test is significant at 0.7947 (adjusted for ties) d. Should H 0 be rejected? What should the researcher conclude? As we can see that the p-value for the test is 0.7983 so we are failing to reject the null hypothesis and concluding that the effects of both the sauce is same. 8. You notice in your introductory psychology class that more women tend to sit up front, and more men sit in the back. To determine whether this difference is significant you collect data on the seating preferences for the students in your class. The data follow. Men Women Front of the Room 15 27 Back of the Room 32 19 a. What is x 2obt? Hence x 2obt6.732323 b. What is df for this test? The df for this test is =(2-1)*(2-1)=1 c. What is x 2cv ? The X cv=3.84 at 5% level of significance. d. What conclusion should be drawn from these results? Answer: With regard to sitting in the front or back, there is an association between men and women. That is, more women tend to sit up front, and more men sit in the back. This difference is significant, Part 2 (use citations as required) 1. What are degrees of freedom? How are the calculated? Degrees of freedom can be described as the number of scores that are free to vary. For example, suppose your friend tossed three dice, and the total score added up to 12. If your friend told you that he rolled a 3 on the first die and a 5 on the second, then you know that the third die must be a 4 (otherwise, the total would not add up to 12). In this example, 2 die are free to vary while the third is not. Therefore, there are 2 degrees of freedom. In many situations, the degrees of freedom are equal to the number of observations minus one. Thus, if the sample size were 20, there would be 20 observations; and the degrees of freedom would be 20 minus 1 or 19. 2. What do inferential statistics allow you to infer? Inferential statistics allow us to draw inferences about a larger population from data of a sample. Inferential statistics are used to: Determine if the difference observed between two sets of scores (groups) is significant. Determine whether an odds ratio or relative risk ratio shows a significant association between two variables, and, if there is, the strength of that association between variables. 3. What is the General Linear Model (GLM)? Why does it matter? The General Linear Model (GLM) underlies most of the statistical analyses that are used in applied and social research. The General Linear Model (GLM) is mathematically identical to a multiple regression analysis but stresses its suitability for both multiple qualitative and multiple quantitative variables. The GLM is suited to implement any parametric statistical test with one dependent variable, including any factorial ANOVA design as well as designs with a mixture of qualitative and quantitative variables (covariance analysis, ANCOVA. 4. Compare and contrast parametric and nonparametric statistics. Why and in what types of cases would you use one over the other? When information about population is completely known then we use parametric test and when the information about population is unknown or little bit of information is known for the population then we use non-parametric tst. 5. Why is it important to pay attention to the assumptions of the statistical test? What are your options if your dependent variable scores are not normally distributed? The most commonly applied statistical assumptions are: i) Independence of observations from each other: This assumption is a common error ii) Independence of observational error from potential confounding effects iii) Exact or approximate normality of observations: The assumption of normality is often erroneous, because many populations are not normal. However, it is standard practice to assume that the sample mean from a random sample is normal, because of the central-limit theorem. Part 3 Part II introduces you to a debate in the field of education between those who support Null Hypothesis Significance Testing (NHST) and those who argue that NHST is poorly suited to most of the questions educators are interested in. Jackson (2012) and Trochim and Donnelly (2006) pretty much follow this model. Northcentral follows it. But, as the authors of the readings for Part II argue, using statistical analyses based on this model may yield very misleading results. You may or may not propose a study that uses alternative models of data analysis and presentation of findings (e.g., confidence intervals and effect sizes) or supplements NHST with another model. In any case, by learning about alternatives to NHST, you will better understand it and the culture of the field of education. Answer the following questions: 1. What does p = .05 mean? What are some misconceptions about the meaning of p =.05? Why are they wrong? Should all research adhere to the p = .05 standard for significance? Why or why not? Answer: p = 0.05 means signifance level of the given sample size. P= 0.05 is a threshold given by fisher used to check the significance or non-significance of a sample size. Fisher does not explain the 0.05 clearly like from where it came what are the concepts behind the value of p. So people will remain in confusion about the choice of the value of 0.05. But it was assumed that 0.05 is just an arbitrary number fisher assumed which is used to check the significance level. No it is not necessary to adhere all research to the value 0.05 as it is not proved its origin hence we can make some other arbitrary constant for the same with proper reasoning. 2. Compare and contrast the concepts of effect size and statistical significance. Answer: Effect size is a simple way of quantifying the difference between two groups that has many advantages over the use of tests of statistical significance alone. Effect size emphasizes the size of the difference rather than confounding this with sample size. If effect size comes under the purview of 0.05 then it is said to be significant and acceptable but it is goes beyond the purview of 0.05 then the effect size or the sample size is said to be non significant. 3. What is the difference between a statistically significant result and a clinically or “real world” significant result? Give examples of both. Answer: Statistically significant results are based on various assumptions while the clinically significant results are due to a logical conclusion. For example in financial sector performance of the company is analyzed in advance on the basis of statistical calculation of previous years while in reality check it could differ from the statistical result. Like GDP growth of US is 2.3 predicted in advance but in real calculation it was found to be +1.0. 4. What is NHST? Describe the assumptions of the model. Answer: Null Hypothesis Significance Testing (NHST) is a type of statistical techniques or methods which are used to check or calculate the effect of certain factor on our observation. Assumptions used in NHST In the null hypotheses involves the absence of factors such as selection and drift etc. These null hypotheses does not based on reality. Null hypotheses is formulated on a well-informed hypotheses based on tested a priori assumptions. Hypotheses allow the analysis and reconstruction of models. 5. Describe and explain three criticisms of NHST. Answer: For NHST, the two independent dimensions of measurement are (1) the strength of an effect, measured using the distance of a point estimate from zero; and (2) the uncertainty we have about the effect’s true strength, measured using something like the expected variance of our measurement device. These two dimensions are reduced into a single p-value in a way that discards much of the meaning of the original data. In addition to that nobody knows the origin of 0.05 in NHST and why only 0.05 is assumed so it is like an assumption assumed to check result. 6. Describe and explain two alternatives to NHST. What do their proponents consider to be their advantages? Answer: Today there is need to replace the NHST with real life statistical calculations. There are various alternatives available to NHST; Two alternatives to NHST Power Analysis In statistical power analysis it gives you a longterm probability but given that the population effect size, alpha, and sample size, of rejecting the null hypothesis, given that the null hypothesis is false. The influence of sample size on significance has long been understood. It is better than NHST as their results and probability are for long term. PlotPlusErrorBar Procedure In this method a bar graph is used to predict results. This method is reliable as it tracks the data from the various no of years. This procedure is also known as PPE. 7. Which type of analysis would best answer the research question you stated in Activity 1? Justify your answer. Answer: You are a researcher interested in addressing the question: does smiling cause mood rise (i.e. become more positive)? Sketch betweenparticipants, withinparticipants, and matched participants designs that address this question and discuss the advantages and disadvantages of each to yielding that help you answer the question. Describe and discuss each design in 45 sentences. To research whether smiling has an impact on mood changing a researcher can employ different designs to reach his or her findings. Therefore, the impact of a smile on mood changing will be the hypothesis of the research and a researcher can use betweenparticipants, withinparticipants and matchedparticipants design. These designs are applied differently but they all give the same findings. Betweenparticipants design is a number of respondents are divided into two. One group tests whether a smile changes the mood (McKenney & Reeves, 2012). They smile to different people and the findings are recorded. Another group does not smile to people and the findings are recorded. After the sample period is over the findings from both respondents give the answer where smiling is a mood changer. Withinparticipants design is the process by which similar respondents are used to get findings from a research study. In this study to find out if smiling causes mood to change, the same respondents are used to get findings. The study is done in two stages. First, the respondents smile to people to see whether their mood will change (McKenney & Reeves, 2012). Then after some time the same respondents do not smile and see whether that action will have an impact on the mood of a person. The findings are also recorded. Matchedparticipants design is the process by which participants are matched with respondents. In this case, the participants that smile are matched with ones that prefer to be smiled at. From the findings of all three designs, it will be found that smiling will have an impact on the mood of a person. When a respondent smiles to a person he or she changes their mood. It was found that smiling definitely changes moods. The advantages of betweenparticipants are that it gives the study variety and it takes a short time as the study is done concurrently by all participants. The disadvantage is that it uses more money because more people are needed. The advantage of withinparticipants is it costs less as the same participants are used. A disadvantage is that it takes longer as the same participants take part in two stages. The advantage of matched participants is that it takes the best out of withinparticipants and betweenparticipants. Here Since we have to check the significance level of the three designs so from present scenario nothing is better than NHST in predicting significance of events. It can be easily predicted that which one is better or whether mood changes or not.
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