1. Open the file COTININE using menu option Datasets and then Elementary Stats, 9th Edition.
This file contains some information about a exposure to second-hand smoke.
How many observations are there in this file?
In this file, there are three variables, labeled Smokers, ETS, and No ETS.
The dataset was collected for a study of second-hand smoke.
The sample data consists of the measured serum cotinine levels in three different groups of people.
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.
2. What results do you expect to find in this data?
Part II. Descriptive Statistics
3. -6Generate descriptive statistics for all three groups of people and complete the following table. Round all results to 2
decimal
places.
Variable
Sample
Mean
Sample
Standard Deviation
Sample Size
Smokers
ETS
No ETS
4. Did you get the results you expected here? Explain why.
5 .In which of the three groups did we experience the MOST variation (highest deviation from the mean)?
Part III. Confidence Intervals
6. Generate a 95% interval for the mean of the SMOKERS group. Paste your results here.
7. Generate a 95% interval for the mean of the ETS group. Paste your results here.
8. Generate a 95% interval for the mean of the No ETS group. Paste your results here
9. Create a graph below by illustrating all three confidence intervals on one graph using the tools in your word processor
(example below). Statdisk cannot do this for you. Create your graph and turn the font red. For this process, I just use 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 14---------------------42
Case 2 35-------------------------70
________________________________________
0 20 40 60
Your Solution:
10.. Based on the confidence intervals shown above, does there appear to be some evidence to indicate that exposure to
tobacco smoke corresponds to higher levels of cotinine?
Part IV. Hypothesis Testing
11. Let's say that you are trying to show that the level of cotinine is equal to zero (0) for people who do not smoke and are not
exposed to Environmental Smoke.
Compose and test a hypothesis (use Significance of 0.01) that the cotinine levels are significantly different from 0 for the No
ETS group. Show all steps.
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 did this in an above step in this assignment.]
Step 3. Determine the likelihood that the population mean is actually equal to 0 by completing a Hypothesis Test: One Mean
in STATDISK. Use significance of 0.01. Paste results here.
Step 4. State your conclusion. Your conclusion should relate the p-value and level of significance to whether you reject the null
hypothesis.
12. Now test the claim of a tobacco company spokesmen that people in the ETS population have "positive" levels of cotinine
(significantly greater than zero), suggesting that this measure is not a good indicator of exposure to tobacco smoke.
Use a significance level of 0.05.
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 did this in an above step in this assignment.]
Step 3. Determine the likelihood that the population mean is actually equal to 0 by completing a Hypothesis Test:
One Mean in STATDISK. Use significance of 0.05. Paste results here.
Step 4. State your conclusion in terms of the claim by the tobacco company.
13. Answer the following questions based on the above hypothesis test.
a. What is the p-value in this example and what does it represent?
b. Given that your data and p-value do not change, what would need to be different in order for us to FAIL TO REJECT the null
hypothesis here?