Week 2 Assignment-1 annie patel

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BST 322 Class #45728 Term 2308 (August 2023) Week 2 Assignment 1. (1 pt) What is the probability of rolling a four with a six-sided die? What is the probability that a person can roll four 3 times in a row? (Assume that rolling the die each time does not affect the outcome of the next roll) - Rolling a 4 is a 16.6% chance. - 1/6 = 0.166 * 100 = 16.6% - Rolling 4 three times in a row is 0.46% chance. - 1/6 * 1/6 = 1/36 * 1/6 = 1/216 = 0.00462963 * 100 = 0.46% 2. (1 pt) Population A and Population B both have a mean height of 70.0 inches with an SD of 6.0. A random sample of 50 people is picked from population A, and random sample of 20 people is selected from Population B. Which sample mean will probably yield a more accurate estimate of its population mean? Why? - the sample mean from population A will yield a more accurate estimate of its population means because the sample was from 50 people, and there is a random sample of population B of only 20 people. The central limit theorem states that with a larger sample, the distribution will be closer to the mean and the SEM will be the lower, which is better. 3. (1 pt) Suppose we obtained data on vein size after application of a nitroglycerin ointment in a sample of 50 patients. The mean vein size is found to be 4.7mm with an SD of 1.1. Using a t distribution table and/or StatCrunch, what are the confidence limits for a 95% confidence interval? For a 99% confidence interval? - One sample T summary confidence interval: μ: Mean of population 95% confidence interval results: Mean Sample Mean Std. Err. DF L. Limit U. Limit μ 4.7 0.15556349 49 4.3873835 5.0126165 - We can be confident that the mean vein size of the patients is between 4.38mm and 5.01mm. - 95% CL: 4.38 ≤ µ ≤ 5.01 - One sample T summary confidence interval: μ: Mean of population 99% confidence interval results: Mean Sample Mean Std. Err. DF L. Limit U. Limit μ 4.7 0.15556349 49 4.2830973 5.1169027 - We can be confident that the mean vein size of the patients is between about 4.28mm and 5.11mm. - 99% CL: 4.28 ≤ µ ≤ 5.11 4. (2 pts) In a pilot study evaluating the use of a new drug to lower resting heart rates (HR) of patients, the following data was recorded:

Subject # Resting HR 001 72 002 88 003 71 004 87 005 64 006 77 007 79 008 59 009 77 010 68 011 78 012 80 013 91 014 81 015 65 016 75 017 69 Given that the average resting HR of the general population for this study is 72, use StatCrunch to perform the appropriate t test. What is the value of t ? Part A: One sample T hypothesis test: μ: Mean of variable H 0 : μ = 72 H A : μ ≠ 72 Hypothesis test results: Variable Sample Mean Std. Err. DF T-Stat P-value Resting HR 75.352941 2.1419157 16 1.5653936 0.1371 - t -stat value = 1.57 Using an alpha of 0.05, is the t statistic significant? Why? - It is not significant because the p- value (0.1371) is greater than 0.05. What are the confidence limits for a 95% confidence interval here and what do they mean for this patient group? Copy and paste your work from StatCrunch into your Word document submission. Part B: - One sample T confidence interval: μ: Mean of variable 95% confidence interval results:

Variable Sample Mean Std. Err. DF L. Limit U. Limit Resting HR 75.352941 2.1419157 16 70.812283 79.8936 - The lower limit is 70.8 and the upper limit is 79.9 with 95% confidence interval, we can expect 95% of values are between the lower and upper limits. - The T-value is 1.5653936 and the critical value for .05 is 1.746 which means the t statistic is not significant because it is less than 1.746. The P value correlates that because the value is greater than .05. This shows that 95% of the heart rates should be between 70.812283 to 79.8936 HR’s. 5. (1 pt) Write a few sentences that could be used to report the results obtained for the t-test in Exercise 4. - We would retain the null hypothesis that this group of patients would report the same heart rates (µ = 72) because of the tabled values are greater than the absolute value of the t -stat at the 0.05 level of significance. The p- value is greater than 0.05 which makes it non-significant. We would therefore. Therefore, we would not be able to accept the alternative hypothesis that the drug alters the heart rates (µ ≠ 72). 6. (2 pts) A study by the Fellows Research Group evaluated the mental health status of 150 women in the inner city of Philadelphia. The results of the study are in an Excel file in the Week 2 assignment area. Load the data set into StatCrunch and perform a one sample t test on the mental health score given that the national average on this test is 45. Is the result significant? Copy and paste your StatCrunch output into your submission document for full credit. Part A: - One sample T hypothesis test: μ: Mean of variable H 0 : μ = 45 H A : μ ≠ 45 Hypothesis test results: Part B: - The findings are not significant because the calculated p-value of 0.3245 is greater than 0.05. 7. (1 pt) Write a few sentences that could be used to report the results obtained for the t -test in Exercise 6. - We would retain the null hypothesis that this group of subjects would report the same mental health scores (µ = 45) because the tabled value 1.97 is greater than the absolute value of the t-statistic (0.989) at the 0.05 level of significance. Also, the p-value is greater than .05 which also makes the result non- significant. We would therefore not be able to accept the alternative hypothesis that the subjects have a different average mental health score (µ ≠ 45). Variable Sample Mean Std. Err. DF T-Stat P-value Mental Health 45.88 0.89021125 149 0.98852941 0.3245

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8. (2 pts) Looking again at the Fellows Research Group data used in exercise 6 above, perform a two- sample t test on the mental health score of the married and divorced subjects. Is the result significant? What hypothesis did you test? Copy and paste your StatCrunch output into your submission document for full credit. - Part A: Two sample T hypothesis test: μ 1 : Mean of Married Mental Health score μ 2 : Mean of Divorced Mental Health score μ 1 - μ 2 : Difference between two means H 0 : μ 1 - μ 2 = 0 H A : μ 1 - μ 2 ≠ 0 (with pooled variances) Hypothesis test results: Difference Sample Diff. Std. Err. DF T-Stat P-value μ 1 - μ 2 5.3409091 2.3386091 86 2.2837973 0.0248 PART B: - Two groups from our subject pool were compared with a two-sample t-test based on marriage status. We have evidence to reject the null hypothesis that this group of subjects would report the same mental health scores (H 0 : μ 1 = μ 2 ) because the tabled value 1.664 is less than the absolute value of the t-statistic 2.2837973 at the 0.05 level of significance. Also, the p-value is 0.0249, less than .05 which also makes the result significant. We would therefore be able to accept the alternative hypothesis that the subjects have a different average mental health score (H A : μ 1 ≠ μ 2 ). 9. (2 pt) For which of the following situations is the independent groups t-test appropriate (if inappropriate, why?): a. The independent variable is infant birth weight at one week (normal vs high); the dependent variable is resting heart rate. - Appropriate b. The independent variable is radiation treatment on throat cancer patients (one group getting a low dose and the other a high dose treatment); the dependent variable is white blood cell count. – Appropriate c. The IV is infant birth weight (grouped as low vs high); the DV is number of days absent from school in first grade. – Appropriate d. The IV is marital status (single vs divorced vs married); the DV is a happiness score, based on a 50- point scale. – Inappropriate, because there are more than two groups for the independent variable. 10. ( 2 pt) For which of the following situations is the dependent groups t-test appropriate (if not appropriate, why?) a. The IV is presence or absence of conversation directed to comatose patients (same patients with and without conversation); the DV is the patients’ intracranial pressure.

- Appropriate b. The IV is birth type (home vs hospital); the DV is perceived functional ability of the patient on a 20- point scale 48 hours after surgery. – Inappropriate, because there are more two groups that are unrelated and independent of each other. c. The IV is time since incarceration for a single group of ex-cons (measured at 1 months and 6 months); the DV is body weight. – Appropriate d. The IV is menopausal state (pre vs during vs post) in the same women over time; the DV is attitudes toward menopause (measured in a survey score 1-20). – Inappropriate, because there are more than three groups for the independent variable. 11. (1 pt) Suppose we wanted to test the idea that a control group of cancer patients (Group 1) would report higher mean pain ratings than an experimental group receiving special massage treatments (Group 2). Use the following information. Compute a t -statistic for independent groups: mean group 1 = 78.1 SD 2 1 (variance) = 49.1 n 1 = 25 mean group 2 = 74.1 SD 2 2 (variance) = 59.7 n 2 = 25 What are the degrees of freedom and the value of t ? Using α=0.05 for a two-tailed test, is this t statistic significant? Show your calculations (or StatCrunch output) for full credit. Part A: Two sample T summary hypothesis test: μ 1 : Mean of Population 1 μ 2 : Mean of Population 2 μ 1 - μ 2 : Difference between two means H 0 : μ 1 - μ 2 = 0 H A : μ 1 - μ 2 ≠ 0 (with pooled variances) Hypothesis test results: Difference Sample Diff. Std. Err. DF T-Stat P-value μ 1 - μ 2 4 2.087 48 1.9165935 0.0613 - t- stat = 1.92 Part B: - For df = 48, the tabled critical value is 2.011 and is greater than the calculated t statistic 1.9165935. The result is statistically non-significant at an alpha of 0.05. 12. (1 pt) For each of the following t values (a-d), indicate whether the t statistic is statistically significant for a two-tailed test, at the specified alphas: a. t = 2.90, df = 25, α = 0.01 - significant b. t = 2.20, df = 25, α = 0.05 - significant c. t = 5.52, df = 10, α = 0.01 - significant d. t = 2.02, df = 20, α = 0.05 - not significant

13. (3 pts) Looking again at the Fellows Research Group data used in exercise 6 above, perform another two-sample t test on the General Practitioner Assessment of Cognition (GPCOG) score of the married and never married subjects. Is the result significant? What hypothesis did you test? Did you try a “pooled” variances test here in StatCrunch? Why or why not? Copy and paste your StatCrunch output into your submission document for full credit. Part A: Summary statistics: Column n Mean Variance Divorced GPCOG 44 8.9545455 15.300211 Married GPCOG 44 10.113636 19.033298 Never married GPCOG 62 10.032258 12.687467 Part B: Two sample T hypothesis test: μ 1 : Mean of Married GPCOG μ 2 : Mean of Never married GPCOG μ 1 - μ 2 : Difference between two means H 0 : μ 1 - μ 2 = 0 H A : μ 1 - μ 2 ≠ 0 (without pooled variances) Hypothesis test results: Difference Sample Diff. Std. Err. DF T-Stat P-value μ 1 - μ 2 0.081378299 0.7982553 80.592827 0.1019452 0.9191 Part C: Two sample T hypothesis test: μ 1 : Mean of Married GPCOG μ 2 : Mean of Never married GPCOG μ 1 - μ 2 : Difference between two means H 0 : μ 1 - μ 2 = 0 H A : μ 1 - μ 2 ≠ 0 (with pooled variances) Hypothesis test results: Difference Sample Diff. Std. Err. DF T-Stat P-value μ 1 - μ 2 0.081378299 0.77132198 104 0.10550497 0.9162 - The result here is not significant. You cannot do a pooled test without checking the variances – and the groups are not equal in size. Either way, the result is not significant – marriage has no impact on cognitive score here.

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Week 2 Assignment-1 annie patel

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