STATS mintab express lab, please and thank you in advance! (pictures uploaded a a odd order the fourth picture is the start) you can see the problem numbers 1-20 scattered

To investigate the effect of age on the duration of stay in the ICU, construct a scatterplot of duration of stay in the ICU versus age. Use Graphs>Scatterplot to do this, choose option Single Y Variable Simple, and select ICU as Y variable, and age as X variable. Click OK. Does the scatterplot suggest that there is an association between variables ICU and age? Use Statistics>Regression>Simple Regression, use ICU as the response variable, and age as predictor. Use Minitab's output for regression analysis to answer questions below. 13. What percent of variation in response is explained using the predictor? R-sq= (Enter the number found under Model Summary without % symbol into LON-CAPA). On the output, find the table called Coefficients that lists predictors, coefficients, and their standard errors (between Model Summary and Regression Equation). Check Minitab's calculations of T and p-value for the test if the slope of the population regression equation E(Y)= B. +Byx is zero. The hypotheses are H,:B= 0 vs H,:B, +0. Output for the slope is in the second row of the table (for predictor age). .. = t= Use Minitab's output to identify the sample slope (under Coef) b, and its standard error (under SE Coef) s.e.(b)= 14. Use the sample slope and its standard error to calculate the value of the test statistic b. Does it agree with the value reported under T-value in Minitab's s.e.(b) output? From Minitab's output, the p-value for the test of H.:B1 = 0 vs He:B, 70 is 15. Based on the p-value, what can we conclude about the significance of age as explanatory variable for the duration of stay in the ICU, is it significant or not significant? Is it important to control for age in trial design (e.g., during randomization)? Use Statistics>1-Sample Inference>t and select C1 for the sample to perform a hypothesis test for state 1. Check the field perform hypothesis test, and enter value 16 into the hypothesized mean field. We would like to test H, : u = 16 vs. H: uProbability Distributions>Inverse Cumulative Distribution Function, select l distribution, enter the appropriate degrees of freedom (n - 2), and enter value 0.975. The non-centrality parameter should remain 0. Click OK. 16. The multiplier t*= From Minitab's output for regression (use the Navigator menu on the left to return to the Simple Regression output), the standard error of the sample slope, s.e.(b) = Calculate the margin of error by multiplying* by the standard error. 17. Calculate the margin of error=t* xs.e.(b,) = Keep least 3 decimals. 18. Does the confidence interval b, t*xs.e.(b) contain 0? .. .. To investigate the effect of BMI on the duration of stay in the ICU, use Statistics>Regression>Simple Regression, use ICU as the response variable, and BMI as predictor. 19. For the regression of ICU on BMI, the p-value for the test of H.:B, = 0 vs He:, 0 is 20. Based on the p-value, what can we conclude about the significance of BMI as explanatory variable for the duration of stay in the ICU, is it significant or not significant? Is it important to control for BMI in trial design (e.g., during randomization)? In this lab, we will use some of the statistical methods discussed in Chapters 14 and 15 to analyze hypothetical clinical trial data using Minitab. While these data are not real, they have been generated to reflect what happened in a real clinical trial. For hypothesis testing throughout this lab, we will use 0.05 level of significance. Download the worksheet for this lab, icu_trial.mtw, from LON-CAPA to your computer. A. Suppose that 200 patients who undergo surgery were randomly assigned to one of two groups: group 1 received an experimental treatment following surgery, and group 2 received a conventional treatment following surgery. The duration of stay in the intensive care unit (ICU) following surgery is the outcome (response) variable of interest. The research question is whether the experimental treatment affects the duration of stay in the ICU following surgery. Use File>Open Worksheet to open the worksheet icu_trial.mtw. The first column has a patient id, second column (ICU) has the duration of stay in the ICU in days, column c3 has the group variable (1=experimental, 2=conventional). Columns C4 and C5 contain patients' sex and age (in years). Column C6 has patients' body mass index (BMI). It is calculated as weight in kilograms divided by height in meters squared. We begin the analyses by comparing the two groups to see if there is a difference between the groups with respect to the outcome variable, number of days spent in the ICU following surgery (variable name in the worksheet is ICU). We will test Ho: Hi - U2 = 0 versus Ha: Mi - H2 +0. Subscript 1 refers to the population represented by a sample of patients who received experimental treatment. Subscript 2 refers to the population represented by a sample of patients who received conventional treatment. Use Statistics>2-Sample Inference> t, choose the option Both samples are in one column, select column ICU into samples field, and select column group into sample IDs field. Click on the Options tab, keep the confidence level at 95%, and the alternative hypothesis field should be set to not equal. Do not check assume equal variances. We are conducting a two-tailed test to see if there are differences between the groups with respect to the duration of stay in the ICU following surgery: Ho: Mi - U2 = 0 versus Ha: H1 - H2 +0. Click OK. Use Minitab's output to answer the following questions: 1. How many patients were randomized to the experimental group (group 1)? 2. How many patients were randomized to the conventional group (group 2)? 3. For the test of Ho: Mi - U2 = 0 versus Ha: Mi - U2 #0, what is the value of the test statistic?t= 4. Based on the p-value and 0.05 level of significance, what decision is reached? (a) The experimental and conventional treatments produce the same mean number of days in the ICU stay; (b) The experimental and conventional treatments do not produce the same mean number of days in the ICU stay. The data come from the randomized trial, meaning that the differences in ICU stays may be attributable to the experimental versus conventional treatment. The idea is that randomization creates two very similar groups, so that everything for the two groups is the same, except for the treatments they receive. Thus when differences in outcomes are found, they can be attributed to the treatment received as opposed to other factors such as age, sex, or BMI. In practice, when randomization is performed, the groups may or may not be the same at intake on every variable. To verify the success of randomization, we will compare the groups on other variables such as age, sex and BMI to see if the differences in outcome (ICU stay) may be attributable to other factors that were not equalized by the randomization. Use Statistics>2-Sample Inference> t, choose the option Both samples are in one column", select column age into samples field, and select column group into sample IDs field. Click OK. 5. Based on the p-value and 0.05 level of significance, what decision is reached? (a) The populations represented by samples that received experimental versus conventional treatments have the same mean age; (b) The populations represented by samples that received experimental versus conventional treatments do not have the same mean age. Use Statistics>2-Sample Inference> t, choose the option Both samples are in one column", select column BMI into samples field, and select column group into sample IDs field. Click OK. 6. Based on the p-value and 0.05 level of significance, what decision is reached? (a) The populations represented by samples that received experimental versus conventional treatments have the same mean BMI; (b) The populations represented by samples that received experimental versus conventional treatments do not have the same mean BMI; For the categorical variable sex, we will conduct a test of Ho: two populations have the same distribution of a For the categorical variable sex, we will conduct a test of Ho: two populations have the same distribution of a categorical variable versus Ha: not so. Use Statistics>Tables>Cross Tabulation and Chi-Square. Select group as categorical variable for rows, select sex as categorical variable for columns. Click on the Display tab, and check the boxes Percent of row total and Chi-square test for association. Click OK. When submitting your answer to LON-CAPA, enter the 7. Females make up what percent of group 1? number between 0 and 100 without % symbol. 8. Females make up what percent of group 2? number between 0 and 100 without % symbol. When submitting your answer to LON-CAPA, enter the 9. For the chi-square test comparing the distribution of sex in two groups, what is the value of the chi-square test statistic (found under Pearson chi-square on Minitab's output)? 10. What is the p-value? 11. Is the distribution of sex the same in two populations, represented by the samples that received experimental versus conventional treatment? While randomization appears to be successful in this case, often the researchers choose to deliberately control for important factors when designing randomization, so that not everything is left to chance. To choose these important factors, the researchers need to know whether these factors affect the outcome, ICU stay in this case. Let's see if age, BMI and sex affect the duration of stay in the ICU. Use Statistics>2-Sample Inference>t, choose the option Both samples are in one column, select ICU into samples field, and select sex into sample IDs field. Click on the Options tab and check that the alternative is not equal: we are conducting a two-tailed test to see if there are differences in the duration of stay in the ICU between males and females: Ho: Ufemale - Umale = 0 versus Ha: Ufemale - Umale 70. Do not check assume equal variances. The p-value for the test of Ho: Ufemale - Umale = 0) versus Ha: Ufemale - Umale #0 is_ 12. Based on the p-value for the test of Ho: Ufemale - Umale = 0 versus Ha: Ufemale - Umale 70, what decision do we make at .05 level of significance, retain the null hypothesis or reject it? To investigate the effect of age on the duration of stay in the ICU, construct a scatterplot of duration of stay in the ICU versus age. Use Graphs>Scatterplot to do this, choose option Single Y Variable Simple, and select ICU as Y variable, and age as X variable. Click OK. Does the scatterplot suggest that there is an association between variables ICU and age? Use Statistics>Regression>Simple Regression, use ICU as the response variable, and age as predictor. Use Minitab's output for regression analysis to answer questions below. 13. What percent of variation in response is explained using the predictor? R-sq= (Enter the number found under Model Summary without % symbol into LON-CAPA). On the output, find the table called Coefficients that lists predictors, coefficients, and their standard errors (between Model Summary and Regression Equation). Check Minitab's calculations of T and p-value for the test if the slope of the population regression equation E(Y)= B. +Byx is zero. The hypotheses are H,:B= 0 vs H,:B, +0. Output for the slope is in the second row of the table (for predictor age). .. = t= Use Minitab's output to identify the sample slope (under Coef) b, and its standard error (under SE Coef) s.e.(b)= 14. Use the sample slope and its standard error to calculate the value of the test statistic b. Does it agree with the value reported under T-value in Minitab's s.e.(b) output? From Minitab's output, the p-value for the test of H.:B1 = 0 vs He:B, 70 is 15. Based on the p-value, what can we conclude about the significance of age as explanatory variable for the duration of stay in the ICU, is it significant or not significant? Is it important to control for age in trial design (e.g., during randomization)? Use Statistics>1-Sample Inference>t and select C1 for the sample to perform a hypothesis test for state 1. Check the field perform hypothesis test, and enter value 16 into the hypothesized mean field. We would like to test H, : u = 16 vs. H: uProbability Distributions>Inverse Cumulative Distribution Function, select l distribution, enter the appropriate degrees of freedom (n - 2), and enter value 0.975. The non-centrality parameter should remain 0. Click OK. 16. The multiplier t*= From Minitab's output for regression (use the Navigator menu on the left to return to the Simple Regression output), the standard error of the sample slope, s.e.(b) = Calculate the margin of error by multiplying* by the standard error. 17. Calculate the margin of error=t* xs.e.(b,) = Keep least 3 decimals. 18. Does the confidence interval b, t*xs.e.(b) contain 0? .. .. To investigate the effect of BMI on the duration of stay in the ICU, use Statistics>Regression>Simple Regression, use ICU as the response variable, and BMI as predictor. 19. For the regression of ICU on BMI, the p-value for the test of H.:B, = 0 vs He:, 0 is 20. Based on the p-value, what can we conclude about the significance of BMI as explanatory variable for the duration of stay in the ICU, is it significant or not significant? Is it important to control for BMI in trial design (e.g., during randomization)? In this lab, we will use some of the statistical methods discussed in Chapters 14 and 15 to analyze hypothetical clinical trial data using Minitab. While these data are not real, they have been generated to reflect what happened in a real clinical trial. For hypothesis testing throughout this lab, we will use 0.05 level of significance. Download the worksheet for this lab, icu_trial.mtw, from LON-CAPA to your computer. A. Suppose that 200 patients who undergo surgery were randomly assigned to one of two groups: group 1 received an experimental treatment following surgery, and group 2 received a conventional treatment following surgery. The duration of stay in the intensive care unit (ICU) following surgery is the outcome (response) variable of interest. The research question is whether the experimental treatment affects the duration of stay in the ICU following surgery. Use File>Open Worksheet to open the worksheet icu_trial.mtw. The first column has a patient id, second column (ICU) has the duration of stay in the ICU in days, column c3 has the group variable (1=experimental, 2=conventional). Columns C4 and C5 contain patients' sex and age (in years). Column C6 has patients' body mass index (BMI). It is calculated as weight in kilograms divided by height in meters squared. We begin the analyses by comparing the two groups to see if there is a difference between the groups with respect to the outcome variable, number of days spent in the ICU following surgery (variable name in the worksheet is ICU). We will test Ho: Hi - U2 = 0 versus Ha: Mi - H2 +0. Subscript 1 refers to the population represented by a sample of patients who received experimental treatment. Subscript 2 refers to the population represented by a sample of patients who received conventional treatment. Use Statistics>2-Sample Inference> t, choose the option Both samples are in one column, select column ICU into samples field, and select column group into sample IDs field. Click on the Options tab, keep the confidence level at 95%, and the alternative hypothesis field should be set to not equal. Do not check assume equal variances. We are conducting a two-tailed test to see if there are differences between the groups with respect to the duration of stay in the ICU following surgery: Ho: Mi - U2 = 0 versus Ha: H1 - H2 +0. Click OK. Use Minitab's output to answer the following questions: 1. How many patients were randomized to the experimental group (group 1)? 2. How many patients were randomized to the conventional group (group 2)? 3. For the test of Ho: Mi - U2 = 0 versus Ha: Mi - U2 #0, what is the value of the test statistic?t= 4. Based on the p-value and 0.05 level of significance, what decision is reached? (a) The experimental and conventional treatments produce the same mean number of days in the ICU stay; (b) The experimental and conventional treatments do not produce the same mean number of days in the ICU stay. The data come from the randomized trial, meaning that the differences in ICU stays may be attributable to the experimental versus conventional treatment. The idea is that randomization creates two very similar groups, so that everything for the two groups is the same, except for the treatments they receive. Thus when differences in outcomes are found, they can be attributed to the treatment received as opposed to other factors such as age, sex, or BMI. In practice, when randomization is performed, the groups may or may not be the same at intake on every variable. To verify the success of randomization, we will compare the groups on other variables such as age, sex and BMI to see if the differences in outcome (ICU stay) may be attributable to other factors that were not equalized by the randomization. Use Statistics>2-Sample Inference> t, choose the option Both samples are in one column", select column age into samples field, and select column group into sample IDs field. Click OK. 5. Based on the p-value and 0.05 level of significance, what decision is reached? (a) The populations represented by samples that received experimental versus conventional treatments have the same mean age; (b) The populations represented by samples that received experimental versus conventional treatments do not have the same mean age. Use Statistics>2-Sample Inference> t, choose the option Both samples are in one column", select column BMI into samples field, and select column group into sample IDs field. Click OK. 6. Based on the p-value and 0.05 level of significance, what decision is reached? (a) The populations represented by samples that received experimental versus conventional treatments have the same mean BMI; (b) The populations represented by samples that received experimental versus conventional treatments do not have the same mean BMI; For the categorical variable sex, we will conduct a test of Ho: two populations have the same distribution of a For the categorical variable sex, we will conduct a test of Ho: two populations have the same distribution of a categorical variable versus Ha: not so. Use Statistics>Tables>Cross Tabulation and Chi-Square. Select group as categorical variable for rows, select sex as categorical variable for columns. Click on the Display tab, and check the boxes Percent of row total and Chi-square test for association. Click OK. When submitting your answer to LON-CAPA, enter the 7. Females make up what percent of group 1? number between 0 and 100 without % symbol. 8. Females make up what percent of group 2? number between 0 and 100 without % symbol. When submitting your answer to LON-CAPA, enter the 9. For the chi-square test comparing the distribution of sex in two groups, what is the value of the chi-square test statistic (found under Pearson chi-square on Minitab's output)? 10. What is the p-value? 11. Is the distribution of sex the same in two populations, represented by the samples that received experimental versus conventional treatment? While randomization appears to be successful in this case, often the researchers choose to deliberately control for important factors when designing randomization, so that not everything is left to chance. To choose these important factors, the researchers need to know whether these factors affect the outcome, ICU stay in this case. Let's see if age, BMI and sex affect the duration of stay in the ICU. Use Statistics>2-Sample Inference>t, choose the option Both samples are in one column, select ICU into samples field, and select sex into sample IDs field. Click on the Options tab and check that the alternative is not equal: we are conducting a two-tailed test to see if there are differences in the duration of stay in the ICU between males and females: Ho: Ufemale - Umale = 0 versus Ha: Ufemale - Umale 70. Do not check assume equal variances. The p-value for the test of Ho: Ufemale - Umale = 0) versus Ha: Ufemale - Umale #0 is_ 12. Based on the p-value for the test of Ho: Ufemale - Umale = 0 versus Ha: Ufemale - Umale 70, what decision do we make at .05 level of significance, retain the null hypothesis or reject it