Homework answers / question archive / STA 301 Online Module 12 Activity: Rock, Paper, Scissors Introduction: A truly random game of rock-paper-scissors would result in a statistical tie with each player winning, tying, and losing one-third of the time

STA 301 Online Module 12 Activity: Rock, Paper, Scissors Introduction: A truly random game of rock-paper-scissors would result in a statistical tie with each player winning, tying, and losing one-third of the time

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STA 301 Online Module 12 Activity: Rock, Paper, Scissors Introduction: A truly random game of rock-paper-scissors would result in a statistical tie with each player winning, tying, and losing one-third of the time. However, people are not truly random and thus can be studied and analyzed. While the computer in the applet won’t win all rounds, over time it can exploit a person’s tendencies and patterns to gain an advantage over its human opponent. Research Question: Is there a significant difference between a “gut” method and a “random” method? Data Collection: We will be using the Rock, Paper, Scissors Applet. The link to this applet is posted on the Module 12 page on Canvas. Gut Method: Using the applet, play the Rock, Paper, Scissors game normally 40 times. Random Method: Using random.org to determine whether you play Rock, Paper, or Scissors, play the game 40 more times. The numbers 1, 2, and 3 will correspond to the following moves: ? Let 1 = Rock ? Let 2 = Paper ? Let 3 = Scissors Note: You must refresh the page before playing the game 40 more times using the Random Method 1. Complete the table below to record your data. Win (W) Tie (T) Lose (L) Gut Method 40 Random Method 40 Now that we have two samples, we want to determine if the proportions are similar in the two samples. In other words, is the distribution of Win(W)-Tie(T)-Lose(L) the same for the Gut Method and the Random Method? 2. State the name of the Chi-Square Test that can be used to answer this research question. Then, explain why that test is appropriate. 3. Write the null and alternative hypotheses in context. 4. Compute the expected counts (under the null hypothesis). Win (W) Source: Bob Lochel (Hatboro-Horsham High School) Tie (T) Lose (L) Module 12: Rock, Paper, Scissors Page 1 By Guts Randomly 5. Explain how you know that all assumptions for this hypothesis test are satisfied. In order to receive full credit, you must check all relevant assumptions. 6. Write the complete line of R Code that you used to enter the contingency table into R. 7. Write the complete line of R code that you used to conduct this hypothesis test. 8. Report the value of your test statistic and degrees of freedom. 9. Report the value of your p-value. 10. Write the conclusion to this hypothesis test in context. Source: Bob Lochel (Hatboro-Horsham High School) Module 12: Rock, Paper, Scissors Page 2 STA 301 Online Module 10 Activity: Pulse Lab Consider a person’s heart rate, which can be measured by a pulse. ? To check your pulse at your wrist, place two fingers between the bone and the tendon over your radial artery - which is located on the thumb side of your wrist. When you feel your pulse, count the number of beats in 60 seconds. This is your number of beats per minute. Research Question: Does a person’s position influence their heart rate? More specifically, suppose we wished to answer this question: Are people’s true mean standing pulse rates generally different from their true mean sitting pulse rates? Consider the following two experiments to address this question. 1. The class is randomly distributed into two groups. Half the class will stand and measure their pulse while the other half will sit and measure their pulse. ? This is an example of the independent samples t-test that we covered earlier in Module 10. 2. Each person will measure their pulse while seated and while standing (the order is randomly assigned). ? This is an example of the dependent samples paired t-test that we covered today in class. Exploring Option #2 (Repeated Measures Design): ? How does this option work? ? For each person, calculate ?????????? = ??????? − ????????. Use the collection of differences as your sample ? What are the benefits of using this option? ? Minimizing unknown factors (nuisance variability) ? Overall reduce variability ? What are the potential costs associated with using this option? ? Can be harder to get more data ? Takes more time (observing 2 measurements per individual) ? More expensive Experiment #1 (Completely Randomized Design): Analyzing the Results Please answer Questions #1 - 8 using the data from Experiment #1. 1. Determine whether it is appropriate to use the pooled or non-pooled procedures. You must show all work in order to receive full credit. By looking the 2 variances we can say that we can use pooled procedures because there is no big difference between the 2 variances. By looking the ratio of the 2 variances, the value is 1.168699 which suggest that the difference is small so we can use pooled procedures. Module 10 Activity: Pulse Lab Page 1 2. State and explain whether all necessary assumptions are satisfied in order to conduct the hypothesis test. The samples must be randomly selected. In this case they randomly distribute the persons into the groups. The variables must be normally distributed. The 2 groups have normal distributions by looking their normal q-q plot. The dots are following the straight line in the plot. The sitting pulse and standing pulse groups must be independent. They are independent because they are using different individuals in each group. 3. Define your parameters (in context). Let mu sitting be the population mean pulse for those who were sitting. Module 10 Activity: Pulse Lab Page 2 Let mu standing be the population mean pulse for those who were standing. 4. Write the null and alternative hypotheses (in context). Ho: people’s true mean standing pulse rates are equal to their true mean sitting pulse rates. Ha: people’s true mean standing pulse rates are generally different from their true mean sitting pulse rates. 5. Report the value of the test statistic and degrees of freedom. Test statistic = -2.7044 Df = 120 6. Report the p-value. P value = 0.007839 7. Write the conclusion in context. As p value is lower than alpha 0.05, we can reject Ho and conclude that people’s true mean standing pulse rates are generally different from their true mean sitting pulse rates. 8. Construct and interpret a 95% confidence interval for the true mean difference between standing and sitting pulse rates. The 95% confidence interval for the true mean difference between standing and sitting pulse rates is (-9.7396 , -1.5062). we can say that we are 95% confident that the true mean difference between standing and sitting pulse rates is between -9.7396 and -1.5062 Experiment #2 (Repeated Measures Design): Analyzing the Results Please answer Questions #9 - 16 using the data from Experiment #2. 9. Explain how you know that it is appropriate to use the paired procedures. We know it’s a paired procedure because we are using the same individuals in both groups. 10. State and explain whether all necessary assumptions are satisfied in order to conduct the hypothesis test. Module 10 Activity: Pulse Lab Page 3 The samples were randomly selected. They said that the indivuals were randomly selected The samples are dependent. Both groups are using the same participants so they are dependent. The samples are normally distributed. By looking the normal q-q plot for the mean difference pulse rate we can see that the dots are following the straight line so it is normally distributed. 11. Define your parameters (in context). Let mu the difference in mean pulse rate between sitting and standing group 12. Write the null and alternative hypotheses (in context). Ho: the difference in mean pulse rate between sitting and standing group is equal 0 Ha: the difference in mean pulse rate between sitting and standing group is different than 0 13. Report the value of the test statistic and degrees of freedom. The test statistic is -14.153 and df = 121 14. Report the p-value. The p value is < 2.2e-16, that is a significant small number Module 10 Activity: Pulse Lab Page 4 15. Write the conclusion in context. As the p value is lower than alpha 0.05 we can reject Ho and conclude that the difference in mean pulse rate between sitting and standing group is different than 0. In context, this suggest that people’s true mean standing pulse rates are generally different from their true mean sitting pulse rates. 16. Construct and interpret a 95% confidence interval for the true mean difference between standing and sitting pulse rates. The 95% confidence interval for the true mean difference between standing and sitting pulse rates is (-10.3617 , -7.8186). We can say that we are 95% confident that the true mean difference between standing and sitting pulse rate is between -10.3617 and -7.8186 Comparing the Results: Please answer Questions #17 - 18 to compare the results of Experiment #1 (Completely Randomized Design) and Experiment #2 (Repeated Measures Design). 17. Compare the results of both experimental designs. Did you get the same result for both experiments? Does this match what you expected to see? Please explain. Yes we got the same results for both experiments, both experiments suggested that there is a significant difference between standing and sitting pulse rate. Yes I expected to see that both results were equal because we are using a significant number of participants in both experiments. 18. Which experiment do you think has the better design? Explain your reasoning. I consider the repeated measures design has a better design because we are using the same individuals in each group, which reduces variability within groups by fixing the confounding variables. Module 10 Activity: Pulse Lab Page 5 Module 12: Inferences for Count Data Module 12 Overview: ? We will study inferential methods that can be used with count data for one or more samples. We will focus on the Two-Proportions z-test and z-interval, Chi-Square Goodness of Fit Test, the Chi-Square Test for Homogeneity, and the Chi-Square Test for Independence. Module 12 Outline: ? Part 1: Review of Chi-Square Curves ? Part 2: Working With One Sample (Chi-Square Goodness of Fit Test) ? Part 3: Working With Two or More Samples ? Two-Proportions z-test ? Two-Proportions z-interval ? Reminders ? Chi-Square Test of Homogeneity ? Part 4: Chi-Square Test of Independence ? Part 5: Your Turn Part 1: Review of Chi-Square Curves: ? Total area under the curve = ______ ? ______________ skewed ? Determined by ______________________________________ ? Mean = ________ ? Always __________________ (all chi-square values are ≥ 0) Image Retrieved from: https://sahilmohnani.wordpress.com/2013/06/03/the-chi-square-distribution/ Part 2: Working With One Sample (Chi-Square Goodness-of-Fit Test): ? Purpose: Used to test the _________________________ of a ___________________________ variable (i.e. to see whether a frequency distribution fits a specific pattern). ? We are often testing particular proportion values. ? We compare our data (______________________ counts) to a set of specified proportion values (________________________ counts). ? We are working with only ______________________ and ______________________. ? Hypotheses: ? H0: The variable _________ the specified distribution. (i.e. there is no difference or no change) ? HA: The variable ________________________________ the specified distribution. ? Assumptions: ? Data obtained by ____________________________________ ? All ____________________________________ ≥ 5 ? Test Statistic: χ 2 =Σ (Observed count − Expected count)2 Expected count with df = ______________________________ Module 12 Notes Page 1 ? Works by comparing the ___________________________ to the _______________________________ ? How to compute the expected counts: ? If all the expected frequencies are equal, the expected frequency E can be calculated by using E = n / k where n is the total number of observations and k is the number of categories. ? If all the expected frequencies are not equal, then the expected frequency E can be calculated by E = n • p, where n is the total number of observations and p is the probability for that category. Example: Suppose as a market analyst you wished to see whether consumers have any preference among five flavors of a new fruit soda. A random sample of 100 people provided these data: Cherry Strawberry Orange Lime Grape 32 28 16 14 10 At α = 0.05 , can it be concluded that consumers show a preference for flavors of the fruit soda? Compute the Expected Counts: ? If there were no preference, you would expect each flavor to be selected with _____________________. In this case, the equal frequency is _______________________________________________________. Frequency Observed Cherry Strawberry Orange Lime Grape 32 28 16 14 10 Expected Hypotheses: Assumptions: ? Data obtained by randomization - ________________________________________________that we have a random sample. ? All expected counts ____________________________________ Test Statistic: ? If the null hypothesis were true, meaning that customers show no preference for flavors of the fruit soda, we would expect all of the expected counts to be ____________________________. ? We need to compare ? 32 to ________ ? 28 to ________ ? 16 to ________ ? 14 to ________ ? 10 to ________ Module 12 Notes Page 2 χ2 = with df = P-Value: Conclusion: R Code and Output: > chisq.test(c(32,28,16,14,10), p=c(0.2, 0.2, 0.2, 0.2, 0.2)) Chi-squared test for given probabilities data: c(32, 28, 16, 14, 10) X-squared = 18, df = 4, p-value = 0.001234 Part 3: Working With Two or More Samples ? We will now begin working with __________________ samples when we have categorical data. ? All of these methods involve comparing the __________________________________________ to make inferences about the unknown ______________________________________________. Two-Proportion z-Test: ? Purpose: To test the hypothesis that ______________________________________________ are equal. Example Text: In a study to estimate the proportion of residents in a certain city and its suburbs who favor construction of a nuclear power plant, it is found (based on a random sample) that 63 of 100 urban residents favor the construction, while 59 of 125 suburban residents are in favor. Example: Is there a significant difference between the proportion of urban and suburban residents who favor construction of a nuclear power plant? Use α = 0.05. Assumptions: ? Random samples - We are ______________________________________ that both samples are random. Module 12 Notes Page 3 ? Independent samples - We ______________ that the samples of urban and suburban residents are independent. ? Large enough sample sizes - At least 10 successes and 10 failures in each sample to ensure normality of the _____________________________________________________________by _____________ ? For the urban sample we have ______ successes and ______ failures ? For the suburban sample we have ______ successes and ______ failures Define Parameters: Hypotheses: Before we can use R to solve the problem, we need to do some preliminary work. First, we can set up a contingency table. In Favor Not in Favor Urban 63 37 Suburban 59 66 Creating a Contingency Table in R: > table=rbind(c(63,37),c(59,66)) > table [,1] [,2] [1,] 63 37 [2,] 59 66 R Code and Output: > prop.test(table) 2-sample test for equality of proportions with continuity correction data: table X-squared = 4.969, df = 1, p-value = 0.02581 Module 12 Notes Page 4 alternative hypothesis: two.sided 95 percent confidence interval: 0.02010755 0.2958924 5sample estimates: prop 1 prop 2 0.630 0.472 Test Statistic: P-Value: Conclusion: Two-Proportion z-Interval: ? Purpose: To construct a confidence interval for the difference between two population proportions. Your Turn: Construct and interpret a 95% confidence interval for the true difference between the proportion of urban and suburban residents who favor construction of a nuclear power plant. Assumptions: Already checked earlier. 95% CI: Interpretation: Reminders: ? The decision of which sample is Sample 1 and which sample is Sample 2 is completely arbitrary and must be made prior to constructing the confidence interval. ? However, pay close attention to the order of subtraction. ? It is important when interpreting the confidence interval. ? Interpreting the confidence interval for the difference of proportions: ? If 0 is in the CI, it is plausible that p1 = p2 , p1 > p2 , or p1 < p2 ? If 0 is NOT in the CI ? If all values in the CI are positive, infer that p1 > p2 ? If all values in the CI are negative, infer that p1 < p2 ? The magnitude of the values in the CI tells you how large any true difference is ? If values are near 0, the true difference may be relatively small in practical terms. Module 12 Notes Page 5 Chi-Square Test of Homogeneity: ? Purpose: To compare the distributions of a variable of two or more populations. ? __________________________ means that each population has ________________ distribution for the variable of interest, whereas _________________________________ means that the populations have _____________________ distributions for the variable of interest. ? We are working with __________________________________ and ______________________. ? Hypotheses: ? H0: The populations are ______________________________ with respect to the variable ? HA: The populations are ______________________________ with respect to the variable ? Assumptions: ? Data obtained by randomization ? All expected counts ≥ 5 ? _________________________________ samples ? Test Statistic: χ 2 =Σ (Observed count − Expected count)2 Expected count with df = _______________________________ ? Expected count = ? Works by comparing the ____________________________ to the ________________________ ? Note: When there are only 2 categories, we can use the prop.test ( ) function in R. For a general contingency table, the test is implemented in the chisq.test ( ) function in R. Example: A survey was conducted in Indiana, Kentucky, and Ohio to determine the attitude of voters concerning school busing. A simple random sample of 200 voters from each of these states yielded the following results: Voter Attitude Support Not Support Undecided Indiana 82 97 21 Kentucky 107 66 27 Ohio 93 74 33 At α = 0.05 , is the true proportion of voters within each category the same for each of the three states? Hypotheses: Module 12 Notes Page 6 Assumptions: ? Data obtained by randomization - _______________________________ that we have a random sample. ? Independent samples - We __________________ that the samples are all independent. ? Expected counts: Voter Attitude Support Not Support Undecided Total Indiana 200 Kentucky 200 Ohio 200 Total 282 237 81 600 We can see from the table that all expected counts are ___________. R Code and Output: > survey=rbind(c(82,97,21), c(107,66,27), c(93,74,33)) > survey [,1] [,2] [,3] [1,] 82 97 21 [2,] 107 66 27 [3,] 93 74 33 > chisq.test(survey) Pearson's Chi-squared test data: survey X-squared = 12.564, df = 4, p-value = 0.01361 Test Statistic (with degrees of freedom): P-Value: Conclusion: Part 4: Chi-Square Test of Independence: ? Purpose: To determine whether an ______________________ exists between two categorical variables. ? We are working with ______________________ and ______________________. Module 12 Notes Page 7 ? Hypotheses: ? H0: The two variables are ____________________________________________. ? HA: The two variables are ____________________________________________. ? Assumptions: ? Data obtained by randomization ? All expected counts ≥ 5 ? Test Statistic: χ 2 =Σ (Observed count − Expected count)2 Expected count with df = (# rows − 1) ? (# columns − 1) ? column total ? Expected count = row total grand total ? Works by comparing the observed counts to the expected counts Example: In an experiment to study the relationship between hypertension (high blood pressure) and smoking habits, the following data were collected for a random sample of 180 individuals. Nonsmokers Moderate Smoker Heavy Smoker Hypertension 21 36 30 No Hypertension 48 26 19 Total Total Source: Walpole, R.E., Myers, R.H., Myers, S.L., & Ye, K. (2012). Probability & Statistics for Engineers & Scientists. Boston, MA: Pearson. Is the presence or absence of hypertension independent of smoking habits? Or, equivalently, is there a ________________________________ between smoking habits and high blood pressure? Use α = 0.05 . Hypotheses: Assumptions: ? Data obtained by randomization - Told in the problem that we have a random sample. ? Expected counts: Nonsmokers Moderate Smoker Heavy Smoker Hypertension No Hypertension We can see from the table that all expected counts are _____________. Module 12 Notes Page 8 R Code and Output: > table=rbind(c(21,36,30), c(48,26,19)) > table [,1] [,2] [,3] [1,] 21 36 30 [2,] 48 26 19 > chisq.test(table) Pearson's Chi-squared test data: table X-squared = 14.464, df = 2, p-value = 0.0007232 Test Statistic (with degrees of freedom): P-Value: Conclusion: Part 5: Your Turn: Your Turn: In a recent year, the most popular colors for light trucks were white, 30%; black, 17%; red, 14%; silver, 12%; gray, 11%; blue, 8%; and other, 8%. A survey of randomly selected light truck owners in a particular area revealed the following. At α = 0.05 , do the proportions differ from those stated? White Black Red Silver Gray Blue Other 45 32 30 30 22 15 6 State the name of the Chi-Square test that you will be using: ________________________________________ Enter the expected counts into the table below. Frequency White Black Red Silver Gray Blue Other Observed 45 32 30 30 22 15 6 Expected State and discuss the appropriate assumptions. Module 12 Notes Page 9 Write the null and alternative hypotheses (in context). Write the line(s) of RStudio code that is/are needed to conduct the appropriate Chi-Square Test. Test Statistic (with degrees of freedom): P-Value: Conclusion: Module 12 Notes Page 10