1 - Statistics Department
Stat 112 – Spring 2004
Homework 3 Solutions
1. (a) From JMP:
Height - Moments
Mean (μ) = 68.096774
Std Dev (sample) = s = 4.2845057
N = 31
Compute population std dev (σ): [pic] = 4.215
(b)
Distributions Height
Moments
|Mean |69.60000 |
|Std Dev |5.68331 |
|Std Err Mean |2.54165 |
|upper 95% Mean |76.65666 |
|lower 95% Mean |62.54334 |
|N |5.00000 |
Std Err (sampling without replacement) = [pic]2.365
(c)
Distributions
Height
|Mean |68.09677 |
|Std Dev |4.28451 |
|Std Err Mean |0.76952 |
|upper 95% Mean |69.66833 |
|lower 95% Mean |66.52522 |
|N |31.00000 |
[pic]
Mean Sample Height
|Mean |67.74000 |
|Std Dev |1.50569 |
|Std Err Mean |0.47614 |
|upper 95% Mean |68.81712 |
|lower 95% Mean |66.66288 |
|N |10.00000 |
[pic]
As expected by the law of the large numbers, the distribution of the sample mean is less varied than the distribution for the population of heights.
2. The result is not trustworthy. The confidence interval’s guarantee that it has a 95% chance of containing the true population mean (where the population is the citizens of the city) is based on the data being collected through a simple random sample. The numbers here are collected by voluntary response rather than a random sample and the sample is likely to be biased.
3. (a). There are many useful graphs that one can draw. Two particularly useful ones are a histogram of the variable difference (after - before) to give you a basic idea about the general characteristics of a data set and a box plot which provides information about the skewness and spread of the data set as well as identifying outliers. There are not that many data points but the histogram shows that the distribution of the differences does not have strong skewness and does not deviate dramatically from the normal distribution. Other useful plots include a stem-and-leaf diagram, a plot of before on the x-axis vs. difference on the y-axis to check for any patterns between the two variables and separate box plots and histograms for the before and after variables.
Distributions:
Before After
[pic][pic]
Distributions: Difference
[pic]
(b). First, note that this is a matched pairs problem and consequently, the one sample t procedure should be used. The before and after samples are not independent and consequently, the two sample t-test is not appropriate. We are interested in the population mean of the variable difference; in effect, we have one sample of the variable difference. Let [pic] denote the mean of the variable difference. Several different choices for the formulation of the null and alternative hypotheses are possible. One reasonable choice is:[pic]:[pic]=0 vs.[pic]: [pic][pic]0. Another reasonable choice would be [pic]:[pic]=0 vs.[pic]: [pic]|t|=0.0016. A 95% confidence interval for the population mean of difference ([pic], where difference equals after minus before) is (4.91, 15.63). The test provides strong evidence that the mean of platelet aggregation after smoking is not the same as the mean of platelet aggregation before smoking. The confidence interval suggests that a range of plausible values for the mean increase in platelet aggregation after smoking is (4.91, 15.63).
(c). The lettuce leaf cigarettes were controls to ensure that the effects of the experiment were due to tobacco specifically, not just due to smoking a lit cigarette. The unlit cigarettes were controls to ensure that the effects were due to lit tobacco, not just unlit tobacco. In designing an experiment, it is important to exercise control in design to ensure that influences other than the treatment of interest (the tobacco) operate equally on the groups.
4. (a). Use JMPIN to open the data bumpusweight.JMP, choose “Fit Y by X”, then choose “Weight” as “Y” and “Status” as “X”. In the pop-up window, click the red triangle before “One Way Analysis” and choose “Quantiles”. You will get the following “Comparing Box-plots” and “Quantiles”:
[pic]
(b). Choose “Mean/Anova/t Test” and you can get the following result. The difference between perished and survived average weights is 0.812g, with a standard error of 0.358g. Student’s t=2.27 with p-value=0.0269. There is strong evidence of a difference.
[pic]
(c). Choose “Unequal Variances” and you will get the following output by JMPIN. By Levene F-test, p-value=0.9606, so we accept[pic], which means that there is no strong statistical evidence that the variances of the weights of the birds in two groups are different.
[pic]
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