STA 144 Week 6 Homework

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Jan 9, 2024

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STA 144 HOMEWORK 6 STA 144 Homework 6 Zihao Luke. Gao College of behavior and social sciences, Cal Baptist University Dr. McKinney Dec 9th, 2023

STA 144 HOMEWORK 6 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. This is a two tailed test and the p value is at .044 which is below the value of .05 which is the significance level so we fail to reject the hypothesis. 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?

STA 144 HOMEWORK 6 This is a two tailed test and since the p value is at .044 so we reject the null hypothesis at the significance level of .01. The difference from question one is that the significance level went from .05 to .01 which makes easier to reject the null hypothesis than question one. 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 Degree of freedom is 18, pooled sd is 2.8275, t = 1.582 b. X1 = 158 X2 = 157.4 n1 = 22 n2 = 26 s1 = 2.06 s2 = 2.59 Degree of freedom is 46, Pooled so is 2.3628, t = 8.77 c. X1 = 200 X2 = 198 n1 = 17 n2 = 17 s1 = 2.45 s2 = 2.35 Degree of freedom is 32, pooled so is 2.40, t = 2.43 4. 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. Two tailed value is 2.10, fail to reject the null B. Two tailed value is 2.01, fail to reject the null C. The two tailed value is 2.04, fail to reject the null 5. 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?

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STA 144 HOMEWORK 6 Dr. L’s .013 significant level is more significant than dr. R’s at .05 which means that we have more statistical confidence in his research results. 6. 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? The chance in the t value is due to the difference in sample size. Set 5 had a value of 0 and set 6 had 20. 7. One Sample T Test Practice with SPSS 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. 72675976939075817193 1. State the RESEARCH Hypothesis or H1

STA 144 HOMEWORK 6 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 critfrom the tables. 8. The population mean m 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: 127121123128118126120130128119127125 Using the criterion of 0.05 in the upper tail only, determine if Mr. Smart’s class is representative of the population.

STA 144 HOMEWORK 6 1. State the RESEARCH Hypothesis or H1 The population mean NSA score is higher than 100. 2. State the NULL Hypothesis or Ho The population mean NSA score is the same as the national mean score of 100. 3. Identify H1 as one or two-tailed It is a two tailed test. 4. Specify alpha level (level of significance) The level of significance is at .05. 5. Specify Degrees of Freedom (n – 1) The degree of freedom is 11. 6. Identify Critical Value — (t crit ) from the t table (in the appendix) The critical value is 1.80 7. Calculate t-score — ( tobt ) using SPSS (this has been done for you and the following printout shows the analysis of the data. 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 cril, then there is a FAILURE to reject the NULL The t value is 21.003 and we reject the null. 9. State the meaning of the results, explain the outcome, and draw a conclusion. The results are significant and the mean of the sample population score is higher than 100. 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

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STA 144 HOMEWORK 6 10 14 12 17 8 11 15 14 12 15 13 18 9 12 11 11 a. What are the independent and dependent variables? b. State the null and alternative hypotheses. c. Using an alpha = 0.05, what is t ( crit )? d. Using SPSS, calculate t ( obt ) e. What should you conclude? I had problem with this question 

STA 144 Week 6 Homework

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