time (seco group 25.7 unscented 41.9 unscented 51.9 unscented 32.2 unscented 64.7 unscented 31.4 unscented 40.1 unscented 43.2 unscented 33.9 unscented 40.4 unscented 58 unscented 61.5 unscented 44.6 unscented 35.3 unscented 37.2 unscented 39.4 unscented 77.4 unscented 52.8 unscented 63.6 unscented 56.6 unscented 58.9 unscented 30.2 scented 56.7 scented 42.4 scented 34.4 scented 44.8 scented 42.9 scented 42.7 scented 24.8 scented 25.1 scented 59.2 scented 42.2 scented 48.4 scented 32 scented 48.1 scented 33.7 scented 42.6 scented 54.9 scented 64.5 scented 43.1 scented 52.8 scented 44.3 scented body temperature 98 99 98.4 98.4 98.8 97.6 98 99.5 98.6 97.2 98.5 98.8 98 98.6 98.5 98.6 97.9 98.8 98.7 97.9 98.4 Problem 1: (21 points total) Is the average body temperature actually less than 98.6 degrees Fahrenheit for healthy human adults? A sample of 21 healthy human adults was taken. The body temperature of each of these 21 individuals is given in the bodytemps.csv data set on Canvas. Use these data to answer the following questions. (R Tutorial 6 may be helpful in answering some of the questions.) 1. State the null and alternative hypotheses in notation. Define the notation used. (Be specific - make sure the population is included when you define the notation.) (3 points) Let is the true average body temperature for healthy human adults. Then the null and alternative hypotheses for this test are, H 0 : The average body temperature is at least 98.6 degrees Fahrenheit for healthy human adults, i.e. 98.6. H a : The average body temperature is less than 98.6 degrees Fahrenheit for healthy human adults, i.e. < 98.6. 2. Why is it appropriate to use the one-sample t-test in this problem? (Do not assess the conditions in answering this question. Rather, refer to the \"statistical bible\" of when to use a ttest.) (2 points) Since the population standard deviation () is not known and the size of the sample is less than 30, it is appropriate to use the one-sample t-test in this problem. 3. There are three conditions that must be satisfied for conclusions from the t-test to be valid to the population of interest. Assess each of these conditions. a. Condition 1: the sample is representative of the population. Note that there is not much information about the make-up of the 21 individuals and how they were selected. Discuss what would make these 21 \"representative\" of the population. (2 points) The sample of these selected 21 adults would be \"representative\" of the population, when the sample constitutes individuals from all the age groups of adults, males and females in the proportion those were in the population. b. Condition 2: the distribution of sample means is normal. You'll need to include a graph to support your assessment of this condition. (2 points) The box-plot of the sample data indicates that the distribution is left-skewed with no outliers present. So there is violation of normality assumption. c. Condition 3: the observations are independent of each other. Discuss what this condition means in the context of the problem and whether or not you believe this condition is true. (2 points) Since the body temperature of one individual does not affect the body temperature of another individual, the sample of body temperatures of 21 healthy human adults are considered as independent of each other. 4. Use R to obtain the t-statistic and p-value. Report the t-statistic (with degrees of freedom) and the p-value here. Do NOT include the rest of the output! (2 points) Test statistic, t (20) = -1.8585 p-value = 0.03894 5. Interpret the p-value in the context of the problem. (Note: this is NOT asking for a conclusion. Think about the definition of the p-value to answer this question.) (2 points) The p-value in this context indicates the probability that test statistic is less than the observed value, i.e. -1.8595, when actually the average body temperature is at least 98.6 degrees Fahrenheit for healthy human adults is 0.03894. 6. State a conclusion in the context of the problem based on the p-value. If you had any concerns that one or more of the conditions wasn't satisfied, mention that here. (3 points) Since p-value < .05, reject the null hypothesis. Therefore, the reported results provide sufficient evidence to support the claim that the average body temperature is less than 98.6 degrees Fahrenheit for healthy human adults. 7. Report AND interpret a 95% confidence interval for the mean body temperature in the population. (3points) The 95% confidence interval for the mean body temperature in the population is (98.155, 98.626). Therefore, we are 95% confident that the true average body temperature for healthy human adults is in between the interval (98.155, 98.626). Problem 2: (14 points total) Can pleasant smells improve learning? Researchers performed an experiment in which 42 subjects were randomly assigned to one of two comparison groups. Subjects assigned to the \"treatment\" group completed a paper-and-pencil maze with the presence of a floral aroma. Subjects assigned to the control group completed a paper-and-pencil maze without the presence of the floral aroma. All subjects were timed as to how long it took them to complete the maze (in seconds). The data are given in thearoma.csv data set on Canvas. Use these data to determine if there is evidence that the floral scent improved the subjects' ability to complete the maze? (That is, is the average time to complete the maze faster in the presence of the floral aroma than without?)Answer these questions to answer the question of interest. (Note: R tutorial 6 may be helpful in answering some of these questions.) 1. What is the response variable? Is it quantitative or categorical? (1 point) The response variable is the \"Time\" and it is a quantitative variable, measured in seconds. 2. State the null and alternative hypotheses in notation and words. Define the notation. Don't forget to include the populations when defining the notation and when writing the hypotheses in words! (3points) Consider two populations in which one population, subjects completed a paper-and-pencil maze with the presence of a floral aroma and in another population without the presence of a floral aroma. Let 1 and 1 are the true average times it took to complete a maze from the above two populations. Then the null and alternative hypotheses for this test are, H 0 : The floral scent hasn't improved the subjects' ability to complete the maze, i.e. 1 = 2. Ha: The floral scent has improved the subjects' ability to complete the maze, i.e. 1 < 2. 3. Obtain and include a properly labeled side-by-side box-and-whisker plot. From the box-andwhisker plots, do you feel there is convincing evidence to indicate the presence of a floral aroma improves subjects' ability to complete the maze, on \"average\"? Explain answer these questions: (2 points - 1point for properly labeled graph and 1 point for explanation) 4. Use R to obtain the test-statistic and p-value. Report the test-statistic with degrees of freedom and the p-value here. Do NOT include the rest of the output! (2 points) Test statistic, t (40) = -1.0229 p-value = 0.1562 5. State a conclusion in the context of the problem based on the p-value. (3 points) Since p-value > .10, fail to reject the null hypothesis. Therefore, the reported results provide no sufficient evidence to support the claim that the floral scent has improved the subjects' ability to complete the maze. 6. Report and interpret a 90% confidence interval for the difference in the mean time to complete the maze between those in the presence of the floral aroma and those not in the presence of the floral aroma. (3 points) The 90% confidence interval for the difference in the mean time to complete the maze between those in the presence of the floral aroma and those not in the presence of the floral aroma is (-10.194, 2.489). Therefore, we are 90% confident that the true difference in the mean time to complete the maze between those in the presence of the floral aroma and those not in the presence of the floral aroma is in between the interval (-10.194, 2.489). Appendix: R-Code for Box-Plot, t-test and confidence interval. Question #1: > > > > > > library(readxl) BODY_TEMPERATURES <- read_excel("Path of the File/BODY TEMPERATURES.xlsx") View(BODY_TEMPERATURES) attach(BODY_TEMPERATURES) boxplot(body_temperature, main = "Box-Plot of Body Temperatures") t.test(body_temperature, mu = 98.6,alternative = "less") One Sample t-test data: body_temperature t = -1.8585, df = 20, p-value = 0.03894 alternative hypothesis: true mean is less than 98.6 95 percent confidence interval: -Inf 98.58492 sample estimates: mean of x 98.39048 > t.test(body_temperature, mu = 98.6) One Sample t-test data: body_temperature t = -1.8585, df = 20, p-value = 0.07788 alternative hypothesis: true mean is not equal to 98.6 95 percent confidence interval: 98.15531 98.62564 sample estimates: mean of x 98.39048 Question #2: > > > > > library(readxl) AROMA <- read_excel("C:/Users/SASTRY/Desktop/AROMA.xlsx") View(AROMA) attach(AROMA) boxplot(time~group,main = "Box-Plot of Times by AROMA Group") > t.test(time~group,var.equal=TRUE, alternative = "less") Two Sample t-test data: time by group t = -1.0229, df = 40, p-value = 0.1562 alternative hypothesis: true difference in means is less than 0 95 percent confidence interval: -Inf 2.489097 sample estimates: mean in group scented mean in group unscented 43.32381 47.17619 > t.test(time~group,var.equal=TRUE, conf.level = 0.9) Two Sample t-test data: time by group t = -1.0229, df = 40, p-value = 0.3125 alternative hypothesis: true difference in means is not equal to 0 90 percent confidence interval: -10.193859 2.489097 sample estimates: mean in group scented mean in group unscented 43.32381 47.1761 Problem 1: (21 points total) Is the average body temperature actually less than 98.6 degrees Fahrenheit for healthy human adults? A sample of 21 healthy human adults was taken. The body temperature of each of these 21 individuals is given in the bodytemps.csv data set on Canvas. Use these data to answer the following questions. (R Tutorial 6 may be helpful in answering some of the questions.) 1. State the null and alternative hypotheses in notation. Define the notation used. (Be specific - make sure the population is included when you define the notation.) (3 points) Let is the true average body temperature for healthy human adults. Then the null and alternative hypotheses for this test are, : The average body temperature is at least 98.6 degrees Fahrenheit for healthy human adults, i.e. 98.6. : The average body temperature is less than 98.6 degrees Fahrenheit for healthy human adults, i.e. < 98.6. 2. Why is it appropriate to use the one-sample t-test in this problem? (Do not assess the conditions in answering this question. Rather, refer to the \"statistical bible\" of when to use a ttest.) (2 points) Since the population standard deviation () is not known and the size of the sample is less than 30, it is appropriate to use the one-sample t-test in this problem. 3. There are three conditions that must be satisfied for conclusions from the t-test to be valid to the population of interest. Assess each of these conditions. a. Condition 1: the sample is representative of the population. Note that there is not much information about the make-up of the 21 individuals and how they were selected. Discuss what would make these 21 \"representative\" of the population. (2 points) The sample of these selected 21 adults would be \"representative\" of the population, when the sample constitutes individuals from all the age groups of adults, males and females in the proportion those were in the population. b. Condition 2: the distribution of sample means is normal. You'll need to include a graph to support your assessment of this condition. (2 points) The box-plot of the sample data indicates that the distribution is left-skewed with no outliers present. So there is violation of normality assumption. c. Condition 3: the observations are independent of each other. Discuss what this condition means in the context of the problem and whether or not you believe this condition is true. (2 points) Since the body temperature of one individual does not affect the body temperature of another individual, the sample of body temperatures of 21 healthy human adults are considered as independent of each other. 4. Use R to obtain the t-statistic and p-value. Report the t-statistic (with degrees of freedom) and the p-value here. Do NOT include the rest of the output! (2 points) Test statistic, t (20) = -1.8585 p-value = 0.03894 5. Interpret the p-value in the context of the problem. (Note: this is NOT asking for a conclusion. Think about the definition of the p-value to answer this question.) (2 points) The p-value in this context indicates the probability that test statistic is less than the observed value, i.e. -1.8595, when actually the average body temperature is at least 98.6 degrees Fahrenheit for healthy human adults is 0.03894. 6. State a conclusion in the context of the problem based on the p-value. If you had any concerns that one or more of the conditions wasn't satisfied, mention that here. (3 points) Since p-value < .05, reject the null hypothesis. Therefore, the reported results provide sufficient evidence to support the claim that the average body temperature is less than 98.6 degrees Fahrenheit for healthy human adults. 7. Report AND interpret a 95% confidence interval for the mean body temperature in the population. (3points) The 95% confidence interval for the mean body temperature in the population is (98.155, 98.626). Therefore, we are 95% confident that the true average body temperature for healthy human adults is in between the interval (98.155, 98.626). Problem 2: (14 points total) Can pleasant smells improve learning? Researchers performed an experiment in which 42 subjects were randomly assigned to one of two comparison groups. Subjects assigned to the \"treatment\" group completed a paper-and-pencil maze with the presence of a floral aroma. Subjects assigned to the control group completed a paper-and-pencil maze without the presence of the floral aroma. All subjects were timed as to how long it took them to complete the maze (in seconds). The data are given in thearoma.csv data set on Canvas. Use these data to determine if there is evidence that the floral scent improved the subjects' ability to complete the maze? (That is, is the average time to complete the maze faster in the presence of the floral aroma than without?)Answer these questions to answer the question of interest. (Note: R tutorial 6 may be helpful in answering some of these questions.) 1. What is the response variable? Is it quantitative or categorical? (1 point) The response variable is the \"Time\" and it is a quantitative variable, measured in seconds. 2. State the null and alternative hypotheses in notation and words. Define the notation. Don't forget to include the populations when defining the notation and when writing the hypotheses in words! (3points) Consider two populations in which one population, subjects completed a paper-and-pencil maze with the presence of a floral aroma and in another population without the presence of a floral aroma. Let 1 and 1 are the true average times it took to complete a maze from the above two populations. Then the null and alternative hypotheses for this test are, : The floral scent hasn't improved the subjects' ability to complete the maze, i.e. 1 = 2. : The floral scent has improved the subjects' ability to complete the maze, i.e. 1 < 2. 3. Obtain and include a properly labeled side-by-side box-and-whisker plot. From the box-andwhisker plots, do you feel there is convincing evidence to indicate the presence of a floral aroma improves subjects' ability to complete the maze, on \"average\"? Explain answer these questions: (2 points - 1point for properly labeled graph and 1 point for explanation) 4. Use R to obtain the test-statistic and p-value. Report the test-statistic with degrees of freedom and the p-value here. Do NOT include the rest of the output! (2 points) Test statistic, t (40) = -1.0229 p-value = 0.1562 5. State a conclusion in the context of the problem based on the p-value. (3 points) Since p-value > .10, fail to reject the null hypothesis. Therefore, the reported results provide no sufficient evidence to support the claim that the floral scent has improved the subjects' ability to complete the maze. 6. Report and interpret a 90% confidence interval for the difference in the mean time to complete the maze between those in the presence of the floral aroma and those not in the presence of the floral aroma. (3 points) The 90% confidence interval for the difference in the mean time to complete the maze between those in the presence of the floral aroma and those not in the presence of the floral aroma is (-10.194, 2.489). Therefore, we are 90% confident that the true difference in the mean time to complete the maze between those in the presence of the floral aroma and those not in the presence of the floral aroma is in between the interval (-10.194, 2.489). Appendix: R-Code for Box-Plot, t-test and confidence interval. Question #1: > > > > > > library(readxl) BODY_TEMPERATURES <- read_excel("Path of the File/BODY TEMPERATURES.xlsx") View(BODY_TEMPERATURES) attach(BODY_TEMPERATURES) boxplot(body_temperature, main = "Box-Plot of Body Temperatures") t.test(body_temperature, mu = 98.6,alternative = "less") One Sample t-test data: body_temperature t = -1.8585, df = 20, p-value = 0.03894 alternative hypothesis: true mean is less than 98.6 95 percent confidence interval: -Inf 98.58492 sample estimates: mean of x 98.39048 > t.test(body_temperature, mu = 98.6) One Sample t-test data: body_temperature t = -1.8585, df = 20, p-value = 0.07788 alternative hypothesis: true mean is not equal to 98.6 95 percent confidence interval: 98.15531 98.62564 sample estimates: mean of x 98.39048 Question #2: > > > > > library(readxl) AROMA <- read_excel("C:/Users/SASTRY/Desktop/AROMA.xlsx") View(AROMA) attach(AROMA) boxplot(time~group,main = "Box-Plot of Times by AROMA Group") > t.test(time~group,var.equal=TRUE, alternative = "less") Two Sample t-test data: time by group t = -1.0229, df = 40, p-value = 0.1562 alternative hypothesis: true difference in means is less than 0 95 percent confidence interval: -Inf 2.489097 sample estimates: mean in group scented mean in group unscented 43.32381 47.17619 > t.test(time~group,var.equal=TRUE, conf.level = 0.9) Two Sample t-test data: time by group t = -1.0229, df = 40, p-value = 0.3125 alternative hypothesis: true difference in means is not equal to 0 90 percent confidence interval: -10.193859 2.489097 sample estimates: mean in group scented mean in group unscented 43.32381 47.1761 # Question 1 library(readxl) BODY_TEMPERATURES <- read_excel("Path of the File/BODY TEMPERATURES.xlsx") View(BODY_TEMPERATURES) attach(BODY_TEMPERATURES) boxplot(body_temperature, main = "Box-Plot of Body Temperatures") # Box-plot of the sample data t.test(body_temperature, mu = 98.6,alternative = "less") # Fot left-tailed t-test t.test(body_temperature, mu = 98.6) # For 95% Confidence Interval # Question 2 library(readxl) AROMA <- read_excel("Path of the File/AROMA.xlsx") View(AROMA) attach(AROMA) boxplot(time~group,main = "Box-Plot of Times by AROMA Group") # Side-by-side Box-plot of the sample data t.test(time~group,var.equal=TRUE, alternative = "less") # Fot left-tailed t-test t.test(time~group,var.equal=TRUE, conf.level = 0.9) # For 90% Confidence Interval