SmalleyTylerStatsHomeworkWeek6

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Jan 9, 2024

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STA 144 Week 6 Homework Work on the following problem set from chapter 11 1. Using the data in the file named Chapter 11 Data Set 2 (in the appendix), test the research hypothesis at the .05 level of significance that boys raise their hand in class more often than girls. Do this practice problem by using SPSS or by hand. What is your conclusion regarding the research hypothesis? Remember to first decide whether this is a one or two-tailed test. a. It seems like in the first test, the value of p is just barely at about the significance level and does not warrant completely rejecting the null hypothesis as it is extremely close and perhaps further testing would be required because it cannot be said with certainty that one group is better at raising their hands than the other. However on the second test, the p value does have a significant enough higher value that might allow us to assume in that instance that boys raise their hands more often. 2. Using the same data set (Chapter 11 Data Set 2), test the research hypothesis at the . 01 level of significance that there is a difference between boys and girls in the number of times they raise their hand in class. Do this practice problem by using SPSS or by hand. What is your conclusion regarding the research hypothesis? You used the same data for this problem as for Question 1, but you have a different hypothesis (one is directional and the other is nondirectional). How do the results differ and why? a. The conclusion shows that there is definitely a difference among boys and girls when raising their hands during class. This is made clear from two different aspects. The first being that the significance level was made smaller and therefore easier to attain a higher value and therefore show a promising result. Also, making it a two tailed test means that there is not effort to prove a bias towards any one way and all that needs to be shown is that there is in fact a difference. 3. Time for some tedious, by-hand practice just to see if you can get the numbers right. Using the following information, calculate the t-test statistics by hand. a. X1 = 62 X2 = 60 n1 = 10 n2 = 10 s1 = 2.45 s2 = 3.16,,,,, T = 1.58173
b. X1 = 158 X2 = 157.4 n1 = 22 n2 = 26 s1 = 2.06 s2 = 2.59…. T = 0.876629 c. X1 = 200 X2 = 198 n1 = 17 n2 = 17 s1 = 2.45 s2 = 2.35… T = 2.55437 3. Using the results you got from Question 3, and a level of significance of .05, what are the two-tailed critical values associated with each? Would the null hypothesis be rejected? A – Critical Value is 2.1098, the null hypothesis would not be rejected B – Critical Value is 2.0141, the null hypothesis would not be rejected C – Critical Value is 2.0395, the null hypothesis would be rejected 4. Here’s a good one to think about. A public health researcher tested the hypothesis that providing new car buyers with child safety seats will also act as an incentive for parents to take other measures to protect their children (such as driving more safely, child-proofing the home, etc.). Dr. L counted all the occurrences of safe behaviors in the cars and homes of the parents who accepted the seats versus those who did not. The findings? A significant difference at the .013 level. Another researcher did exactly the same study, and for our purposes, let’s assume that everything was the same—same type of sample, same outcome measures, same car seats, and so on. Dr. R’s results were marginally significant (remember that from Chapter 9?) at the .051 level. Whose results do you trust more and why? a. I would trust Dr. R’s research results more as it appears that his significance level would have been harder to beat and by achieving it, would show that there is indeed some form of correlation between the habits of the driving and the implementation of the child safety seats within the car.
5. For this question, you have to do two analyses. Using Chapter 11 Data Set 5 (in the appendix), use SPSS to compute the t score for the difference between two groups on the test variable. Then, use Chapter 11 Data Set 6 and do the same. Notice that there is a difference between the two t values, even though the mean scores for each group are the same. What is the source of this difference, and why—with the same sample size—did the t value differ? 6. Independent Samples Levene's Test for Equality of Variances F Sig. t df Sign One-Sided p Score1 Equal variances assumed .122 .736 -.218 8 .41 Equal variances not assumed -.218 7.982 .41 Independent Samples Levene's Test for Equality of Variances F Sig. t df Sign One-Sided p Score2 Equal variances assumed .275 .607 -.327 18 .37 Equal variances not assumed -.327 17.959 .37 The source of the difference would be with the sample size as a greater number of people involved within the test will allow for a greater range of differing scores making the sections of each curve possibly more rounded or elongated and therefore the t values will differ between the two groups as there are more people that change the deviation of the scores as well. 7. One Sample T Test Practice with SPSS
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We want to test if students at our school score differently on a grammar test than the national population of readers (where μ = 89). We take a sample of ten (n=10) readers whose grammar reading scores are given below. Use this as a sample to do the other questions below. 72 67 59 76 93 90 75 81 71 93 1. State the RESEARCH Hypothesis or H1 a. The students at our school score differently on a grammar test than the national population of readers. 2. State the NULL Hypothesis or Ho a. H0: There is no difference in the scores of students at our school and the national population of readers 3. Identify H1 as one or two-tailed a. This is a two-tailed test 4. Specify alpha level (level of significance) a. Alpha level = .05 5. Specify Degrees of Freedom (n – 1) a. Degrees of freedom = 9 (note that SPSS will also give you this) 6. Identify Critical Value—(t crit ) from the t table (in the appendix) a. Critical value = +/- 2.262 7. Calculate t-score—( tobt ) using SPSS (this has been done for you and the results are given below: t obt= -3.118 One-Sample Statistics N Mean Std. Deviation Std. Error Mean Grammar 10 77.70 11.461 3.624 One-Sample Test Test Value = 89 t df Sig. (2-tailed) Mean Difference 95% Confidence Interval of the Difference Lower Upper Grammar -3.118 9 .012 -11.300 -19.50 -3.10
8. State decision (rejection) rule –use the diagram of the normal distribution curve a. If the calculated t is beyond the t crit, then the NULL is rejected b. If the calculated t not beyond the crit, then there is a FAILURE to reject the NULL If t obt is beyond 2.262 in the right tail or beyond -2.262 in the left tail, we have a rejection of the null hypothesis and we can accept the research hypothesis. 9. State the meaning of the results, explain the outcome, and draw a conclusion. T obt is beyond t crit in the left tail; therefore, reject the null hypothesis; the reading scores of children in our school are lower than the scores of the national population. They are significantly lower. Note: SPSS will always calculate significance on a two-tailed test. If you have a one-tailed test you will need to take t crit from the tables. 8. The population mean on a national scholastic achievement test is 100 with a standard deviation of 30. The students in Mr. Smart’s class got the following scores: 127 121 123 128 118 126 120 130 128 119 127 125 Using the criterion of 0.05 in the upper tail only, determine if Mr. Smart’s class is representative of the population. 1. State the RESEARCH Hypothesis or H1 a. Mr. Smart’s classroom tests differently on their tests than the national average 2. State the NULL Hypothesis or Ho a. There is no difference between nationwide scores and Mr. Smart’s classroom 3. Identify H1 as one or two-tailed a. This is a two tailed test. 4. Specify alpha level (level of significance ) a. The level of significance is 0.05 5. Specify Degrees of Freedom (n – 1) a. Df = 11 6. Identify Critical Value—(t crit ) from the t table (in the appendix ) a. Critical value = 1.7959 7. Calculate t-score—( tobt ) using SPSS (this has been done for you and the following printout shows the analysis of the data. a. 21.33
8. State decision (rejection) rule –use the diagram of the normal distribution curve a. If the calculated t is beyond the t crit, then the NULL is rejected a. The t is beyond the critical value and the null hypothesis is rejected b. If the calculated t not beyond the cril, then there is a FAILURE to reject the NULL a. The null hypothesis is rejected 9. State the meaning of the results, explain the outcome, and draw a conclusion. a. Based on the results there is indeed a difference between how Mr. Smart’s class scored on the test and the rest of the nation. It would appear that they were able to perform better according to the incredibly positive number that far exceeded the critical value. 9. Independent Samples T test Practice with SPSS We investigate the effects of sensitivity training on a policeman’s effectiveness at resolving domestic disputes (comparing independent samples of policemen who had or had not completed the training). The dependent variable was their ability to successful resolve domestic disputes. The following scores were obtained: No course Course 11 13 14 16 10 14 12 17 8 11 15 14 12 15 13 18 9 12 11 11 a. What are the independent and dependent variables? a. The dependent variables are that of scores they received for resolving conflict and the independent variable is whether or not they had completed the courses. b. State the null and alternative hypotheses. a. Null Hypothesis = there is no difference in a policeman’s ability to resolve conflict in regards to sensitivity training
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b. Research Hypothesis = there may be a difference in how a policeman is able to resolve conflict based on whether or not they completed sensitivity training. c. Using an alpha = 0.05, what is t ( crit )? a. t = 1.8331 d. Using SPSS, calculate t ( obt ) a. e. What should you conclude?