Part IV. Hypothesis Testing 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. The parameter of interest is the population mean continine level for the ETS group. As we want to test whether it is not equal to zero or not so the hypotheses are, H 0 : =0H a : 0 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. Sample Size, n: 40 Mean: 60.575 St. Dev., s: 138.0839 Alternative Hypothesis: not equal to (hyp) t Test Test Statistic, t: 2.7745 Critical t: 2.0227 P-Value: 0.0084 95% Confidence interval: 16.41367 < < 104.7363 The p-value is smaller than the significance level of 0.05 thus we are rejecting the null hypothesis at 0.05 significance level. The conclusion is that the population mean is significantly different than 0 at 0.05 significance level. 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. The parameter of interest is the population mean continine level for the NO ETS group. As we want to test whether it is greater than zero or not so the hypotheses are, H 0 : 0H a : >0 Step 2. Determine the sample mean, sample standard deviation, and sample size. [Hint: You recorded these previously in Part II, #3-6] Sample Size, n: 40 Mean: 16.35 St. Dev., s: 62.5335 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. Alternative Hypothesis: > (hyp) 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 not smaller than the significance level of 0.01 thus we are not rejecting the null hypothesis at 0.01 significance level. The conclusion is that the population mean is not significantly larger than 0 at 0.01 significance level. t Test Test Statistic, t: 1.6536 Critical t: 2.4258 P-Value: 0.0531 98% Confidence interval: -7.635266 < < 40.33527 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. It represents that, if the null hypotheses is true then the probability of obtaining the test statistic which we obtained i.e. 1.6536 or larger is 0.0531. The significance level should be higher than the p-value of 0.0531. Week 4 Project - STAT 3001 Student Name: Date: 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 samples with 3 different observations made on each sample for a total of 120 different observations. 2. What would you expect to find relative to the continine level in the groups? The NOETS group lists the cotinine levels for subjects who are nonsmokers and have no exposure to environmental tobacco smoke at home or work. The ETS group lists cotinine levels for subjects who are nonsmokers exposed to tobacco smoke at home or work. The SMOKERS group lists cotinine levels for subjects who report tobacco use. Serum cotinine is a metabolite of nicotine, meaning that cotinine is produced when nicotine is absorbed by the body. Higher levels of cotinine correspond to higher levels of exposure to smoke that contains nicotine. Part II. Descriptive Statistics 3-6 Generate descriptive statistics for No ETS, Smokers and ETS groups and Variable Sample Mean 1 Sample Standard Deviation Sample Size 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? Part III. Confidence Intervals 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 16.35 62.534 40 172.475 119.498 40 60.575 138.084 40 ETS - the variable in stat disk is the highest at 19067.17 Margin of error, E = 16.65918 90% Confident the population mean is within the range: -0.3091758 < mean <33.00918 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: No ETS Smoker ETS -0.3---------------------33.0 140.6-------------------204.3 23.8-----------------97.4 ________________________________________________________ 0 50 100 150 200 250 13. Based on the confidence The smoking is definitely leads to higher continine levels intervals shown above, what due to having confidence interval containing larger values conclusion can you draw than the two groups. As there is no common region between 2 about whether the exposure to smoke leads to higher continine levels? Why? the confidence interval for Smoker and \"ETS and NO ETS\" so clearly smoking leads to higher continine levels. But there is a common region between the confidence interval for ETS and NO ETS thus there is no significant different between the continine levels for them indicating the exposure to smoke does not lead to higher continine levels. Part IV. Hypothesis Testing 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. The parameter of interest is the population mean continine level for the ETS group. The null and alternative hypotheses which we can form from the given information are, H 0 : =0H 1 : 0 Sample mean = 60.575 Sample Standard 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: 16.41367 < < 104.7363 We reject the null hypothesis if the p-value is smaller than the significance level. Based on the result it can be observed that the p-value is smaller than the significance level therefore the null hypothesis should be rejected in conclusion that the continine level for the ETS group is not equal to zero. 3 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. Step 4. State your conclusion. Your conclusion statement should include the p-value and level of significance to phrase your conclusion. The parameter of interest is the population mean continine level for the NO ETS group. The null and alternative hypotheses which we can form from the given information are, H 0 : 0H 1 :> 0 Sample mean = 16.350 Sample Standard deviation = 62.534 Sample Size = 40 Alternative Hypothesis: > (hyp) t Test Test Statistic, t: 1.6536 Critical t: 2.4258 P-Value: 0.0531 98% Confidence interval: -7.635266 < < 40.33527 We reject the null hypothesis if the p-value is smaller than the significance level. Based on the result it can be observed that the p-value is larger than the significance level therefore the null hypothesis should be retained in conclusion that continine level in the NO ETS group is not greater than zero. 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 probability of obtaining the test statistic as obtained or extreme under the null hypothesis is the p-value which is 0.0531 in this case. We reject the null hypothesis if the p-value is smaller than the significance level. Therefore to REJECT the null hypothesis we need to set up the significance level larger than 0.0531. A 0.10 significance level would do the job. 4