Questions and Answers #1 Comparing Two Groups

Questions and Answers Sheet 1 Comparing Two Groups Question #1 Water flows through a water hose at a rate of 𝑄1 = 680𝑐𝑚3 /𝑠 , the diameter of the hose is 𝑑1 = 2.2𝑐𝑚 . A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of 𝑣2 = 9.2𝑚/𝑠 . a) Enter an expression for the cross-sectional area of the hose, 𝐴1 , in terms of its diameter, 𝑑1 b) Calculate the numerical value of 𝐴1 , in square centimeters. c) Enter an expression for the speed of the water in the hose, 𝑣1 , in terms of the volume floe rate 𝑄1 and cross-sectional area 𝐴1 d) Calculate the speed of the water in the hose, 𝑣1 in meters per second. e) Enter an expression for the cross-sectional area of the nozzle, 𝐴2 , in terms of 𝑣1 , 𝑣2 and 𝐴1 f) Calculate the cross-sectional area of the nozzle, 𝐴2 in square centimeters. Answer: 2 a) 𝐴1 = 𝜋 × 𝑑14 2 b) 𝐴1 = 𝜋 × 2.24 = 3.80𝑐𝑚2 𝑄 c) 𝑄1 = 𝐴1 × 𝑣1 ⇒ 𝑣1 = 1 d) 𝑣1 = 𝐴1 680 3.80 =179cm/s =1.79m/s e) Using continuty equation, 𝐴2 × 𝑣2 = 𝐴1 × 𝑣1 𝑣 𝐴2 = 𝐴1 × 1 𝑣2 f) 𝐴2 = 3.80 × = 0.739𝑐𝑚2 1.79 9.2 Question #2 True/False True or False? A clinical trial is conducted to compare an experimental medication to placebo to reduce the symptoms of asthma. One hundred participants are enrolled in the study and randomized to receive either the experimental medication or placebo. The primary outcome is self-reported reduction of symptoms. Among 50 participants who receive the experimental medication, 30 report a reduction of symptoms as compared to 15 participants of 50 assigned to the placebo. The 95% CI for the relative risk of participants reporting a reduction of symptoms between the experimental and placebo groups is between 1.529 and 8.012. Answer: The 95% CI for the relative risk of participants reporting a reduction of symptoms between the experimental and placebo groups is between 1.529 and 8.012.'' Given: 𝑛1 = 50 𝑛2 = 50 𝑋 𝑝1 = 1 ̂ = 30 50 𝑝 ̂2 = = 15 50 𝑛1 = 0.6 𝑋2 𝑛2 = 0.3 Calculating the relative risk (RR): 𝑅𝑅 = = 0.6 ̂1 𝑝 ̂2 𝑝 0.3 =2 Creating the 95% confidence interval for the log of relative risk is: 𝑛1 − 𝑥1 𝑛2 − 𝑥2 𝑥1 𝑥2 ln(𝑅𝑅) ± 𝑍𝛼/2=0.025 √ + 𝑛1 𝑛2 20 15 √ 30 15 ln(2) ± 1.96 × + 50 50 (0.213, 1.173) The 95% confidence interval for the relative risk hence is the antilog of the above values: (1.237, 3.232) Hence the answer is FALSE. Question #3 Iranian researchers studied factors affecting patients’ likelihood of wearing orthodontic appliances, noting that orthodontics is perhaps the area of health care with the highest need for patient cooperation (Behenam & Pooya, 2007). Among their analyses, they compared students in primary school, junior high school, and high school. The data that follow have almost exactly the same means as the researchers found in their study, but with far smaller samples. The score for each student is his or her daily hours of wearing the orthodontic appliance. Did age of students impact how long they wore the orthodontic appliance for? Use alpha of .05. Primary school: 16, 13, 18 (M=15.67, s=2.52) Junior high school: 8, 13, 14, 12 (M=11.75, s=2.63) High school: 20, 15, 16, 18 (M=17.25, s=2.22) Answer: Step 1 ANOVA or Analysis of Variances is one of the most important fields in Statistic. The reason for this is that is goes into the core of analyzing the variation exhibited samples, by breaking down the total variation into various different sources of variation. The most basic use of ANOVA is to test for the difference between the populations for several groups (2 or more). Let us recall that a t-test is used to compare the means of two groups, so then ANOVA is some sort extension that allows to perform comparisons for two or more groups. Step 2 As with any other hypothesis test, ANOVA uses a null and the alternative hypothesis. The null hypothesis is a statement that claims that all population means are equal, and the alternative hypothesis is the hypothesis that not all means are equal (observe that this does NOT imply that all means are unequal, it implies that al least one mean is equal. Question #4 For the following experiment, state the type of test we should use. Assume homogeneity of variance (where applicable), unless something is directly stated (e.g., ''the data are not at all normal'') or otherwise indicated (viz., by the inspection of the data) which would indicate an obvious violation of an assumption. This means you must inspect the data for violations of all assumptions. Assume n is large enough, unless stated. Assume 𝜎 is unknown, unless stated. Brianna works at an infant/toddler center and wants to know whether children at the age of two have a preference for a certain type of toy. She puts four different toys in front of the infant/toddler and sees which toy the child crawls to first. 1) a one-sample t-test for means 2) none of the other tests 3) an independent samples t-test with heterogeneity of variance 4) a related samples t-test 5) and independent samples t-test with homogeneity of variance 6) a chi-squared goodness-of-fit test 7) a z-test for the difference between proportions 8) a one sample z-test for proportions 9) a one-sample z test for means Answer: Step 1 Given: The assumptions that can be made are: 1) Homogeneity of variances. 2) Validity of parametric test. 3) The size of the sample(n) is large. 4) The population standard deviation is unknown. Explanation of incorrect options: In general, a t-test and a z-test cannot be performed for the given study as both these tests are used for comparing the means of the two populations, and in the given scenario, the data is based on the counts and the options available for the children. Therefore, a t-test or a z-test of any kind like two-sample (heterogeneity and homogeneity), one sample cannot be performed. Step 2 Explanation to correct answer: The test that will be performed is a ''Chi-squared goodness of fit test''. In the given scenario, the variable in the given study is categorical, and the data is based on the preference of children for a certain type of toy; this means there is a goodness of fit that is to be tested. Therefore, it is the only test from the given options that will be used for the given study. Hence, the correct option is: chi-squared goodness of fit. Question #5 Researchers compared the effectiveness of a mood induction technique (exposure to depressing film footage) across 3 different groups of 11 participants each: 1) 0 minutes exposure (Control Group), 2) 10 minutes exposure, & 3) 20 minutes exposure. The response variable is the score on the Beck Depression Inventory (higher score = more depressed mood). M 1 = 7 M 2 = 11 M 3 = 12 SS1 = 56 SS 2 = 96 SS3 = 62 n1 = 11 n2 = 11 n3 = 11 SS-total =368 Which of the hypotheses is listed incorrectly? Select all that apply a) 𝐻0: 𝜇1 > 𝜇2 > 𝜇3 b) 𝐻0: 𝜇1 = 𝜇2 = 𝜇3 c) 𝐻1: 𝜇1 ≠ 𝜇2 ≠ 𝜇3 d) 𝐻1: 𝜇1 < 𝜇2 < 𝜇3 Answer: Step 1 It is basically a comparison of the statistical measures of the data with one sample(population) or two samples(populations) or more than two samples(populations). There are two hypotheses namely NULL and ALTERNATIVE. The tail of the hypothesis will be decided from the alternative hypothesis. If it is '<' then left tail; '>' then right tail; ′ ≠≠′ then two-tail. Whereas the null hypothesis is almost '=0' in most of the cases. Step 2 Here a total of three groups so here we use Anova. The correct hypotheses are, b) 𝐻0: 𝜇1 = 𝜇2 = 𝜇3 c) 𝐻1: 𝜇1 ≠ 𝜇2 ≠ 𝜇3 The incorrect hypotheses are, a) 𝐻0: 𝜇1 > 𝜇2 > 𝜇3 d) 𝐻1: 𝜇1 < 𝜇2 < 𝜇3 These two hypotheses are incorrect because in the ANOVA we check there is a significant difference between treatments or not. No significant difference means (𝜇1 = 𝜇2 = 𝜇3) and significant difference means (𝜇1 ≠ 𝜇2 ≠ 𝜇3) So, these two options are incorrect. Question #6 A study was performed comparing different nonpharmacologic treatments for people with high-normal diastolic blood pressure (DBP) (80-89 mm Hg). One of the modes of treatment studied was stress management. People were randomly assigned to a stress management intervention (SMI) group or a control group. Participants randomized to SMI were given instructions in a group setting conserning different techniques for stress management and met periodically over a 1-month period. Participants randomized to the control group were advised to pursue their normal lifestyles and were told that their blood pressure would be closely monitored and that their physician would be notified of any consistent elevation. The results for the SMI group (n=242) at the end of the study (18 months) were as follows: Mean (change) =-5.53mm Hg (follow-up - baseline), sd (change) =6.48mm Hg. How much power did the study have for detecting a significant difference between groups (using a twosided test with a 5% level of significance) if the true effect of the SMI intervention is to reduce mean DBP by 2mm Hg more than the control group and the standard deviation of change within a group is 6mm Hg? Answer: From the ''Table 3-The normal distribution'' in Appendix, the value of 𝑧0.975 is 1.96. The value of power is obtained below: Power = 𝜙 −𝑧1−𝑎 + 2 2 2 𝜎 𝜎 √ 1+ 2 𝑛1 𝑛2 ( = 𝜙 −1.96 + (|𝜇2 −𝜇1 |) ) (2) 62 62 √ + ( 242 320) = 𝜙(1.953) [Use Table 3 - The normal distribution'' in Appendix] =0.975 Thus, the value of power is 0.975. Question #7 A teacher sees students straggling into her class at highly diverse times. She secretly records when each person comes into her class. Everyone in the class at the bell gets a score of zero. Then, as each late person comes in, the number of seconds after the bell that the person arrives is recorded. Number of seconds late, then, is each person's individual score. The scores ranged from 0 to 600 (for people who were 10 minutes late). The teacher also has each person's exam average over the semester. She wants some estimate of the degree of relationship between a person's lateness and his or her exam scores in her class. Which statistic should she use? a. Z-test b. One Sample t-test c. Independent groups t-test d. Dependent groups t-test e. One Factor ANOVA f. Two factor ANOVA g. t-test for r > 0 h. Chi-square test i. Phi Coefficient j. Cramer’s Phi k. Point Biserial r L. Eta-squared m. r-squared or r Answer: Step 1 Introduction: The lateness scores of the arriving students varied between 0 and 600; these scores are measured in seconds, and is the number of seconds that a student arrives late for class. The lateness score can be any value between 0 and 600, and the value 0 is absolute, implying that the person was not late. Thus, the lateness score is a continuous variable, measured on the ratio scale. The exam average of a student over the semester can take any value between 0 and the maximum average marks in the exam, with the value 0, which implies that the person did not get any score on an average throughout the semester, being absolute. Thus, the lateness score is a continuous variable, measured on the ratio scale. The teacher wants to estimate the degree of relationship between two continuous, ratio scale variablethe lateness and exam scores of the student. Step 2 Explanation: The z-test is used when one wishes to test if the true population mean of a single population equals a given constant value, when the population standard deviation is known. The one-sample t-test is used for the same purpose, but when the population standard deviation is unknown. But here, two different populations are to be compared. Independent groups t-test is used when the means for the same or comparable variables for two independent populations are to be compared. Dependent groups t-test is used when the means for the same or comparable variables for two dependent populations are to be compared. But here, two completely different variables- lateness score (time) and exam scores are to be compared. One factor ANOVA is used when three or more categories of a single factor are to be compared with respect to their means, provided the data are independent and normally distributed, the groups are independent, and the populations are homoscedastic (same variance). Two factor ANOVA is used when three or more categories of one factor, and three or more categories of a different factor are to be compared separately with respect to their means, provided the data satisfy similar conditions. However, two continuous, and not categorical variables are to be compared here. The t-test for r > 0 would be used if one wishes to test for the presence of significant positive correlation, r between two continuous variables. However, the teacher here wants to test the degree of relationship between the two variables, and not just positive linear relationship. Further, the teacher is likely to want to test if higher lateness is associated with lower exam scores on average, and not to test if higher lateness is associated with higher exam scores on average. The chi-square test of association is useful for testing for the presence of association between two categorical variables. Although the two variables here are continuous, it is possible for the teacher to divide each of the lateness scores exam average scores into several classes or categories, and then use chi-square test of association to estimate the degree of relationship between the variables. Phi coefficient is used to measure the association between two binary variables. However, each variable here is non-binary. Even if it is possible to classify each variable into exactly two categories, much information regarding the original data would be lost in such a process, which is not desirable. Cramer’s phi is usually useful for measuring the association between two nominal or categorical variables. Although the variables here are ratio scale, like in case of chi-square test, it is possible for the teacher to divide each of the lateness scores exam average scores into several classes or categories, and then use Cramer’s phi to estimate the degree of relationship between the variables. Point biserial correlation or point biserial r is a measure of association between a continuous, and a binary variable. However, both variables in this study are continuous. The quantity eta-squared calculates the effect size of a test, by measuring the standardized difference of the sample mean from the population mean, or between two sample means of comparable quantities. However, the two means here are not directly comparable. The quantity r, also called Pearson correlation coefficient is a measure of strength of linear association between any two continuous, ratio scale variables, provided some inherent conditions are satisfied. The quantity r-squared or 𝑟 2 is simply the square of r, which gives the proportion (or percentage, when multiplied with 100) of variation in the dependent variable that is being explained by the independent variable, when it is possible to determine one of the variables in the study to be dependent on the other one. Here, as both the variables are continuous, ratio scale variables, r or 𝑟 2 can be used to measure the degree of association between them, if the researcher is specifically interested in determining the degree of linear association, and if it is possible to determine one of the variables as dependent on the other. Logically speaking, exam score of a student might be considered to be dependent on the level of lateness to class, at least from the teacher’s point of view. Thus, among the given tests, it would be the most suitable to use r-squared or r; additionally, with some modifications to the available data, chi-square test or Cramer’s phi might also be used. Question #8 Explain whether homogeneity of variance is satisfied for this analysis, and be clear on how you decided this. We saw data from Everitt that showed that girls receiving cognitive behavior therapy gained weight over the course of that therapy. However, it is possible that they just gained weight because they got older. One way to control for this is to look at the amount of weight gained by the cognitive therapy group (n=29) in contrast with the amount gained by girls in a Control group (n=26), who received no therapy. The data on weight gain for the two groups is shown below. Control −0.5 −9.3 −5.4 12.3 −2.0 −10.2 −12.2 11.6 −7.1 6.2 −0.2 −9.2 8.3 Cognitive Therapy 3.3 1.7 −9.1 11.3 0.7 2.1 0.0 −0.1 −1.4 −1.0 −0.7 1.4 −10.6 −3.5 −0.3 −4.6 14.9 −3.7 −6.7 3.5 −0.8 2.8 17.1 2.4 0.3 −7.6 12.6 1.8 1.6 1.9 3.7 11.7 3.9 15.9 6.1 0.1 −10.2 1.1 15.4 −4.0 −0.7 20.9 Mean −0.45 3.01 St. Dev 7.99 7.31 Variance 63.82 53.41 Run the appropriate test to compare the group means. What would you conclude? Answer: Step 1 Homogeneity of Variance: Homogeneity of variance (homoscedasticity) is a crucial assumption shared by many parametric statistical methods. This assumption requires that the variance within each population be equal for all populations (two or more, counting on the method). Step 2 In this case, we can run the ANOVA because there are two different groups are there and the mean and variance are known so we have to test the means of these two groups. Our null hypothesis is there is a significant difference between two groups which means, the two means are are equal and the alternative is there is no significant difference between two groups which means, the two means are not equal. The measures are given in the problem. In these measures, the population variance are not equal. So, we have to check the homoscedasticity by using Levene's test this is nothing but test of homogeneity of variance. After, using this test we get the result Source SS df MS F statistic Between treatments 12.25 1 12.249 0.5569 Within treatments 1165.75 53 21.995 Total 1177.998 54 Therefore, the p value is 0.459. Thus, the calculated value is greater than the critical value so we prove the homogeneity is met in this problem. Step 3 Answer: In this case, the homogeneity is met. Question #9 Trucks are used in many configurations. Three common configurations are tractors with 1, 2, and 3 trailers. Accident statistics for these configurations in 2002 are listed below. Fatal U.S. Truck Accidents in 2002 by Configuration Truck Configuration Tractor, 1 trailer Tractor, 2 trailers Tractor, 3 trailers Number of Registered Vehicles Number of Accidents 150, 000 2,889 5,100 154 200 1 Assume these statistics represent simple random samples for each configuration. Which of the following would be the most, appropriate test to use to determine whether the populations for each configuration have the same rate of fatal accidents? A) One-sample proportion z-test. B) Two-sample proportion z-test C) Chi-square test for independence D) Chi-square goodness-of-fit test E) Chi-square test for homogeneity of proportions Answer: Step 1 Given: The data is provided for the U.S Truck Accidents in 2002 by configuration. The data is: Truck Configuration Tractor, 1 trailer Tractor, 2 trailers Tractor, 3 trailers Number of Registered Vehicles Number of Accidents 150, 000 2,889 5,100 154 200 1 The researcher wants to determine whether the populations for each configuration have same rate of fatal accidents. Explanation of incorrect options: (A) The test ''One-sample proportion z-test'' will not be used, as the samples in the given scenario are three, and the test is performed for only one sample. Therefore, this option is incorrect. Step 2 (B) The test ''two-sample proportion z-test'' will not be used as this is applied in the scenario where specific characteristics of two groups or populations are compared, and the test is performed for two samples. Therefore, this option is incorrect. (C) The test '' Chi-square test for independence'' is used to test whether the two variables are independent or not, and in the given scenario, the researcher does not want to test the independence between the truck configurations. Therefore, this option is incorrect. (D) The test '' Chi-square goodness of fit test'' is used to determine whether the variable studied in the test follows a specific distribution or not. In the given scenario, the researcher does not want to test if the truck configuration follows a specified distribution. Therefore, this is an incorrect option. Explanation to correct option: (E) The test '' Chi-square test for homogeneity of proportion'' is used when more than three variables are involved and test whether the sample data belong to the same population or not. In the given scenario, it can be determined whether the populations for each configuration have the same rate of fatal accidents by identifying if the sample data of truck configuration came from the same population. Hence, the researcher can test his claim by using the '' Chi-square test for homogeneity of proportion''. Thus, option (E) is correct. Question #10 To find: The 95% confidence interval for the comparison. Answer: Calculation: The study is conducted to see that whether a sample of children of different age groups consumed an adequate amount of calcium or not. In the study, samples are considered from two independent populations (different age groups). First, the proportion of children aged 5 to 10 year who met the calcium requirement is defined by the formula: Count 𝑝1 = ̂ Sample size Substitute the values in the above formula: Count 𝑝1 = ̂ Sample size 861 = 1055 =0.8161 Now, the proportion of children aged 11 to 13 years who met the calcium requirement is defined by the formula: Count Sample size Substitute the values in the above formula: Count 𝑝 ̂2 = 𝑝 ̂2 = Sample size = 417 974 =0.4281 The 95% confidence interval is defined by the formula: 𝐶𝐼 = (𝑝1 − 𝑝2 ) ± 𝑧𝛼/2 × √ 𝑝1 (1 − 𝑝1 ) 𝑝2 (1 − 𝑝2 ) + 𝑛1 𝑛2 Substitute the values in the above formula: 𝐶𝐼 = (𝑝1 − 𝑝2 ) ± 𝑧𝛼/2 × √ 𝑝1 (1 − 𝑝1 ) 𝑝2 (1 − 𝑝2 ) + 𝑛1 𝑛2 = (0.8161 − 0.4281) ± 1.96 × √0.8161(1 − 0.8161)1055 + 0.4281(1−0.4281) 974 = 0.388 ± 1.96 × 0.01983 = 0.388 ± 0.03886 = (0.3492,0.4268) Hence, the required confidence interval is (0.3492, 0.4268). Question #11 GMU uses a robot food delivery service which now has been utilized in the City of Fairfax. One of the potential benefits of this service is to help the busiest students eat breakfast. Research has shown that about 80% of college students skip breakfast due to busy schedules and other reasons. Initial data were collected from a random sample of 595 Mason students who utilize the robot food delivery service and are presented in StatCrunch. Show the full calculation of the sample proportion by including the number of students who skipped breakfast, the total number of students sampled, and the value of the sample proportion. Present this sample proportion as a decimal rounded to four decimal places. 1. Frequency table results for Breakfast? Count = 595 Breakfast? Frequency Relative Frequency 0 142 0.23865546 Using 𝛼 = 0.05 , is there sufficient evidence to 1 453 0.76134454 conclude that less than 80% GMU students who utilize the food delivery robots skip breakfast? Conduct a full hypothesis test by following the steps below. Answer: Step 1 Hypothesis Testing: It is basically a comparison of the statistical measures of the data with one sample(population) or two samples(populations) or more than two samples(populations). There are two hypotheses namely NULL and ALTERNATIVE. The tail of the hypothesis will be decided from the alternative hypothesis. If it is '<' then left tail; '>' then right tail; ′ ≠≠′ then two-tail. Whereas the null hypothesis is almost ' =0 ' in most of the cases. One-Sample Z-Test: It is the hypothesis testing done for comparing the mean of the data to a specific value. As the name says it is a one-sample z-test, the data of one sample is considered for the complete hypothesis testing and the test statistics will be found from that data. Step 2 Here, the level of significance is 𝛼 = 0.05 , The hypothesis statement is 𝐻0 : 𝑝̂ = 𝑃𝑣𝑠𝐻1 : 𝑝̂ < 𝑃 Now, before calculating the test statistics, the standard deviation of the population is calculated using the given population proportion. Here, the proportion of the sample is, 453 𝑝̂ = = 0.761344538 ≅ 0.7613 595 The standard deviation of the population using the population proportion is, 𝜎=√ =√ =√ =√ 𝑃×(1−𝑃) 𝑛 0.80×(1−0.80) 50 0.80×0.20 50 0.16 50 = √0.0032 = 0.056568542 ≅ 0.0566 Step 3 The test statistics is, 𝑝̂ − 𝑃 𝑍= 𝜎 0.7613−0.80 = = 0.0566 −0.0389 0.0566 = −0.683745583 Now, the p-value is calculated by the normal distribution for the probability of the Z-score is less than 0.683745583. That is, 𝑃(𝑍 < −0.683745583) = 0.24707 Therefore the p-value for the test statistics is approximately 0.24707. Step 4 Conclusion: As the p-value of the hypothesis test statistics is greater than 0.05, that is the level of significance of the test; there is no sufficient evidence to reject the null hypothesis. So, the GMU students who are all using the food delivery robots skip breakfast is 80% . Question #12 For each of the following scenarios, select the test (from the choices below), that is most appropriate for analyzing the data. a) Z-test b) One-sample t-test c) Independent t-test d) Dependent t-test e) 1-Way ANOVA f) Regression/Correlation g) Chi-Square 1. Does caffeine help test performance? 90 participants are randomly assigned to one of 4 groups (no caffeine, small dose, medium dose, and large dose) and then asked to complete a GRE math section. 2. Do seniors study for fewer hours per week than freshmen? 3. Is the average SAT math score of Hunter College students higher than the national average (𝜇 = 500, 𝜎 = 100)? Answer: Step 1 1. e) 1-Way ANOVA A 1-way ANOVA is used to compare the means of more than two independents groups in order to find out the means of the groups are significantly different from each other. A 1-way ANOVA is used for the given scenario because the levels of caffeine are more than two. That is, (no caffeine, small does, medium does and large dose). After the test, the scores will be collected for each group and then a 1-way ANOVA test will be carried out to find whether caffeine really helps in test performance by comparing the means of four groups. A significant P-value results in the rejection of hypothesis and concludes that the caffeine helps in the test performance at a given level of significance. Thus, 1-way ANOVA suits the given experiment. Step 2 2. c) Independent t-test: This test can be used to determine whether there is a significant difference between the mean of two independent groups or samples. Independent sample t-test can be used for the given scenario since there are two different groups (seniors and freshmen) and the test aims to compare the mean number of hours of study per week between seniors and freshmen and the test is carried out by comparing the two means. A significant Pvalue concludes that seniors study for fewer hours per week than freshmen. Step 3 3. b) One-sample t-test: A one sample t test is used to find out whether a sample mean with the hypothesized population mean. Here, the average SAT math score obtained from the sample will be tested using the population mean SAT math score 500. A significant P-value results in the rejection of hypothesis and concludes that the average SAT math score of Hunter College students is higher than the national average at a given level of significance. Question #13 Gender gap Candidates for political office realize that different levels of support among men and women may be a crucial factor in determining the outcome of an election. One candidate finds that 52% of 473 men polled say they will vote for him, but only 45% of the 522 women in the poll express support. a) Write a 95% confidence interval for the percent of male voters who may vote for this candidate. Interpret your interval. b) Write and interpret a 95% confidence interval for the percent of female voters who may vote for him. c) Do the intervals for males and females overlap? What do you think this means about the gender gap? d) Find a 95% confidence interval for the difference in the proportions of males and females who will vote for this candidate. Interpret your interval. e) Does this interval contain zero? What does that mean? f) Why do the results in parts c and e seem contradictory? If we want to see if there is a gender gap amongvoters with respect to this candidate, which is the correct approach? Why? Answer: Step 1 Hey, since there are multiple subparts posted, we will answer first three subparts. If you want any specific subpart to be answered then please submit that subpart only or specify the subpart in your message. Given info: One survey was done on gender gap in voting. It was seen that 52% of 473 men support him but 45% of 522 women support him. Step 2 a) Calculation: For finding the confidence interval using z-intervals the assumption and conditions should met. The assumption and conditions for inference: (i) Randomization condition: Though it is not given that the sample is random but it can be assumed that the poll was random. (ii) Independent group assumption: The two groups are independent for different gender. (iii) Success/failure condition: Here for men 𝑛𝑝ˆ = 246 & 𝑛𝑞ˆ = 473 − 246 = 227 and for women 𝑛𝑝ˆ = 235& 𝑛𝑞ˆ = 522 − 235 = 287 All are greater than 10. Hence, all conditions are satisfied for the inference. Step 3 Confidence interval: Software procedure: Step by step procedure to obtain the confidence interval using the MINITAB software: Choose Stat > Basic Statistics > 1 Proportion. Choose Summarized data. In Number of events, enter 246 In Number of trials, enter 473 Click Options. Under Alternative, and choose not equal. Click OK in each dialog box. Output using the MINITAB software is given below: Test and Cl for One Proportion Method p: event proportion Normal approximation method is used for this analysis. Descriptive Statistics N Event Sample p 95%CI for p 473 246 0.520085 (0.475, 0.565) Test Null hypothesis 𝐻0 : 𝑝 = 0.5 Alternative hypothesis 𝐻1 : p ≠ 0.5 From the MINITAB output, the 95% confidence interval is (0.475, 0.565) Interpretation: The 95% confidence interval for the percent of male voters who may vote for this candidate is (0.475, 0.565) means that there is 95% confident that the percentage of men who will vote him lies between 47.5% and 45.5%. Step 4 (b) Confidence interval: Software procedure: Step by step procedure to obtain the confidence interval using the MINITAB software: Choose Stat > Basic Statistics > 1 Proportion. Choose Summarized data. In Number of events, enter 235 In Number of trials, enter 522 Click Options. Under Alternative, and choose not equal. Click OK in each dialog box. Output using the MINITAB software is given below: Test and CI for One Proportion Method p: event proportion Normal approximation method is used for this analysis. Descriptive Statistics N Event Sample p 95%CI for p 522 235 0.450192 (0.4075, 0.4929) From the MINITAB output, the 95% confidence interval is (0.4075, 0.4929) Interpretation: The 95% confidence interval for the percent of female voters who may vote for this candidate is (0.4070, 0.4929) means that there is 95% confident that the percentage of men who will vote him lies between 40.75% and 49.29% . Step 5 (c) From part a, the 95% confidence interval for the percent of male voters who may vote for this candidate is (0.475, 0.565) From part a, the 95% confidence interval for the percent of female voters who may vote for this candidate is (0.4070, 0.4929) From the two outputs, it can be said that the confidence intervals overlap. Hence, there is no evidence to support that the proportion of male and female will differ for voting him. Question #14 Use the following listed arrival delay times (minutes) for American Airline flights from New York to Los Angeles. Negative values correspond to flights that arrived early. Also shown are the SPSS results for analysis of variance. Assume that we plan to use a 0.05 significance level to test the claim that the different flights have the same mean arrival delay time. SPSS ANOVA a. What characteristic of the data above indicates that we should use one-way analysis of variance? b. If the objective is to test the claim that the three flights have the same mean arrival delay time, why is the method referred to as analysis of variance? Flight 1 −32 −25 −26 −6 5 −15 −17 −36 Flight 19 −5 −32 −13 −9 −19 49 −30 −23 Flight 21 −23 28 103 −19 −5 −46 13 −3 Sum of Squares df Mean Square F Sig. Between Groups 2575.000 2 1297.500 1.334 .285 Within Groups 20271.500 21 955.310 Total 22846.500 23 Answer: Step 1 a) The data shows the arrival delay times (minutes) for American Airline flights from New York to Los Angeles and the results for analysis of variance are obtained from the SPSS software. Step 2 The characteristic of the data indicates the arrival delay times of the all three flights are considered in to a single factor. b) Analysis of variance is used to compare more than two groups of means. That is, testing the equality of three or more samples from populations with equal means, which is based different estimates of a common population variance. Here, the analysis of variance is used to compare the three groups of means; this is based on two different estimates of a common population variance Question #15 It appears that there is some truth to the old adage “That which doesn’t kill us makes us stronger.” Seery, Holman, and Silver (2010) found that individuals with some history of adversity report better mental health and higher well-being compared to people with little or no history of adversity. In an attempt to examine this phenomenon, a researcher surveys a group of college students to determine the negative life events that they experienced in the past 5 years and their current feeling of well-being. For n = 16 participants with 2 or fewer negative experiences, the average well-being score is M = 42 with SS = 398, and for n = 16 participants with 5 to 10 negative experiences the average score is M = 48.6 with SS = 370 . 1. Is there a significant difference between the two populations represented by these two samples? Use a two-tailed test with 𝛼 = .01 [use the 4-step procedure]. [Remember to assess whether the assumption of homogeneity of variances is satisfied or not, 𝛼 = .01 for homogeneity test]. 2. Compute Cohen’s d to measure the size of the effect. 3. Report your hypothesis test results and cohen’s d in APA format. Answer: Step 1 Given Information: It appears that there is some truth to the old adage “That which doesn’t kill us makes us stronger.” Seery, Holman, and Silver (2010) found that individuals with some history of adversity report better mental health and higher well-being compared to people with little or no history of adversity. In an attempt to examine this phenomenon, a researcher surveys a group of college students to determine the negative life events that they experienced in the past 5 years and their current feeling of well-being. Participants with 2 or fewer negative experiences: 𝑋1 = 42 SS=398 n=16 𝑠1 = √ =√ 𝑆𝑆 𝑛−1 398 16 − 1 𝑠1 = 5.151 Participants with 5 to 10 negative experiences: 𝑋2 = 48.6 SS=370 n=16 𝑠2 = √ =√ 𝑆𝑆 𝑛−1 370 16−1 𝑠2 = 4.967 Step 2 Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: there is no significant difference between the two populations represented by these two samples i.e. 𝜇1 = 𝜇2 Ha: there is a significant difference between the two populations represented by these two samples i.e 𝜇1 ≠ 𝜇2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. Rejection Region Based on the information provided, the significance level is 𝛼 = 0.01 , and the degrees of freedom are df=30. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this two-tailed test is tc=2.75 , for 𝛼 = 0.01 and df=30 . The rejection region for this two-tailed test is 𝑅 = {𝑡: ∣ 𝑡 ∣> 2.75} . Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: 𝑋1 − 𝑋2 𝑡= (𝑛 − 1)𝑠12 + (𝑛2 − 1)𝑠22 1 1 √ 1 ( + ) 𝑛1 + 𝑛2 − 2 𝑛1 𝑛2 𝑡= 42 − 48.6 = −3.689 2 + (16 − 1)4.9672 1 (16 − 1)5.151 1 √ ( + ) 16 + 16 − 2 16 16 Decision about the null hypothesis Since it is observed that ∣ 𝑡 ∣= 3.689 > 𝑡𝑐 = 2.75 , it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0.0009, and since p=0.0009<0.01, it is concluded that the null hypothesis is rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that there is a significant difference between the two populations represented by these two samples, at the 0.01 significance level. Step 3 Cohen's d using t-statistic: For the independent samples T-test, Cohen's d is determined by calculating the mean difference between your two groups, and then dividing the result by the pooled standard deviation. Cohen's 𝑑 = (𝑀2 − 𝑀1)/𝑆𝐷𝑝𝑜𝑜𝑙𝑒𝑑 where: (𝑆𝐷12 + 𝑆𝐷22 )⁄ 𝑆𝐷𝑝𝑜𝑜𝑙𝑒𝑑 = √( 2) Pluggin all the values in stated formula , (48.6 − 42)⁄ Cohen's 𝑑 = 5.059836 = 1.30439. Since , the cohen's d is equal to 1.304 . It shows a large effect size.