Question: Week 4 Project - STAT 3001 Student Name: Date: November 5, 2015 Instructions: To complete this project, you will need the following materials: STAT DISK

Week 4 Project - STAT 3001 Student Name: Date: November 5, 2015 Instructions: To complete this project, you will need the following materials: STAT DISK User Manual (found in the classroom in Doc Sharing) Access to the internet to download the Stat Disk program. Part I. Analyze Data Instructions Answers 1. Open the file Passive and Active Smoke using menu option Datasets and then Elementary Stats, 11th Edition. This file contains some information on the continine levels in smokers, nonsmokers exposed to smoke (ETS), and nonsmokers not exposed to smoke (No ETS). How many observations are there in this file? There are 40 observations per group in this Dataset. 2. What would you expect to find relative to the continine level in the groups? I expect smokers to have higher levels of continine, followed by the ETS and finally the non exposed to smoke group. Part II. Descriptive Statistics 3-6 Generate descriptive Variable statistics for No ETS, Smokers and ETS groups and complete the following table. No ETS Round mean and standard deviation to 3 decimal places. Smokers ETS 7. Did you get the results you expected here? Explain why. 8. Which of the three groups experienced the MOST variation? How do you know? Sample Mean Sample Size 16.35 Sample Standard Deviation 62.534 172.475 119.498 40 60.575 138.084 40 40 I did get the expected results from the means. The (No ETS) group had the lowest mean and standard deviation. The (ETS) group experienced the most variation. Part III. Confidence Intervals 1 9. Generate a 90% interval for the mean of the No ETS group. Paste your results here. 10. Generate a 90% interval for the mean of the Smoker group. Paste your results here. 11. Generate a 90% interval for the mean of the ETS group. Paste your results here Margin of error, E = 16.65918 90% Confident the population mean is within the range: -0.2591758 < mean <33.05918 Margin of error, E = 31.83449 90% Confident the population mean is within the range: 140.6405 < mean <204.3095 Margin of error, E = 36.78584 90% Confident the population mean is within the range: 23.78916 < mean <97.36084 12. Create a graph below by illustrating all three confidence intervals on one graph using the tools in your word processor (example below). Stat Disk cannot do this for you. Create your graph and turn the font red. For this process, I just used the dashes and wrote a scale below the axes. Here is an example, but it is not based on the data you are analyzing: Case 1 Case 2 14---------------------42 35-------------------------70 ________________________________________ 0 20 40 60 Your Solution: Case 1 -.2592 --------------------33.060 Case 2 140.641--------------------------204.310 Case 3 23.789-----------------------------97.361 ________________________________________________________ 0 25 50 75 100 125 150 175 200 13. Based on the confidence Based on the confidence intervals shown above, I can conclude that intervals shown above, what exposure to smoke leads to higher continine levels, however, the conclusion can you draw about difference between first and second hand smoke is not well defined. whether the exposure to smoke leads to higher continine levels? Why? Part IV. Hypothesis Testing 2 14. Your researcher's claim is that the continine level for the ETS group is not equal to zero. Compose a hypothesis test with a level of significance of .05 to test your claim. Step 1. Determine parameter of interest and compose null and alternative hypotheses. Step 2. Determine the sample mean, sample standard deviation, and sample size. [Hint: You recorded these previously in Part II, #3-6] Step 3. Determine the likelihood that the population mean is actually not equal to 0 by completing a Hypothesis Test: One Mean in STAT DISK. Use significance of 0.05. Remember to change the pulldown option in Stat Disk to agree with the alternate hypothesis. Paste your results here. Step 4. State your conclusion. Your conclusion statement should include the p-value and level of significance to phrase your conclusion. Null: ETS have a continine level =0 Alternative: ETS have continine levels 0 Sample Mean- 60.575 Sample std. deviation- 138.084 Sample size-40 Alternative Hypothesis: not equal to (hyp) t Test Test Statistic, t: 2.7745 Critical t: 2.0227 P-Value: 0.0084 95% Confidence interval: 16.41364 < < 104.7364 The p value is less than .05 therefore we reject the null hypothesis. 15. For your second hypothesis test, your researcher's claim is that the continine level in the NO ETS group is greater than zero. Perform a hypothesis test using a significance level of .01 to test your claim. Step 1. Determine parameter of interest and compose null and alternative hypotheses. Step 2. Determine the sample mean, sample standard deviation, and sample size. [Hint: You recorded these previously in Part II, #3-6] Step 3. Determine the likelihood that the population mean is actually greater than 0 by completing a Hypothesis Test: One Mean in STAT DISK. Use significance of 0.01. Remember to change the pulldown option in Stat Disk to agree with the alternate hypothesis. Paste results here. Null: NOETS groups have continine levels =0 Alternative: NOETS groups have continine 0 Sample mean-16.35 Sample std deviation-62.534 Sample size-40 Alternative Hypothesis: > (hyp) t Test Test Statistic, t: 1.6536 Critical t: 1.6849 P-Value: 0.0531 90% Confidence interval: -0.3091758 < < 33.00918 3 Step 4. State your conclusion. Your conclusion statement should include the p-value and level of significance to phrase your conclusion. The p value is greater than 0.01 therefore the data is likely with a true null. 16. Answer the following questions based on the above hypothesis test. a. What is the p-value for the hypothesis test in #15 and what does it represent? (Look at page 383 in the text) b. Given that your data, hypotheses, and p-value do not change, what would need to be different in order for you to REJECT the null hypothesis? (What do you compare to make your rejection decision?) The p-value is 0.0531 and it tells us how likely it is to get a result like this if the null hypothesis is true. Submit your final draft of your Word file by going to Week 4, Project, and follow the directions under Week 4 Assignment 2. Please use the naming convention "WK4Assgn2+first initial+last name" as the Submission Title. 4

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