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University of Delaware *
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Course
200
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Statistics
Date
Apr 3, 2024
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Pages
5
Uploaded by BaronJaguar4308
1 STAT 200 Guided Exercise 5
For On-Line Students, be sure to:
•
Submit your answers in a Word file at the same place you downloaded the file •
Remember you can paste any Excel or JMP output into a Word File (use Paste Special for best results). •
Put your name and the Assignment # on the file name: e.g. Ilvento Guided5.doc Key Topics •
Sampling Distributions
•
Confidence Intervals
•
Answer as completely as you can and try to show your work. You will learn best when you have more complete answers. Then upload the file to get credit. 1. Alpha is the Greek term α. In Inferential Statistics alpha represents the probability of making an error when we make an inference from a sample to the population. In most cases α is the probability of finding a sample with our z-score value or beyond - i.e., the area in the tail of the distribution, either positive or negative. In a confidence Interval, we divide α by 2 to spread this error in both tails.
We need to calculate the values of α for different Confidence Intervals. Fill in the following table. It will require you to find the z value that is associated with an α/2 level of probability in the Standard Normal Table. Hint: the standard normal table shows the probability from the mean up to a value. •
If we subtract that probability from .5, we get the probability after the value (in the tail). •
We start this exercise by dividing α/2. We do this because half of α represents the area in the upper and lower tails. •
We will find inside the table the probability of .5 - α/2
. •
Then we read from inside the table to the outside to get the z-value that corresponds to it. •
For example, for the first example: o
α = .25
o
α/2 = .125, o
Find inside the table the probability for .5 - .125 = .375 (or something very close to it) o
We then read the z-value of 1.15 that corresponds to this probability. Confidence Level
α
α/2
Z α/2
100(1 - α)
75% .25 .125 1.15 80% .20 .100 1.28 90% .10 .050 1.645 95% .05 .025 1.96 99% .01 .005 2.575 These values you calculate will become the z-values you will use for various Confidence Intervals for proportions.
2 2. An experiment was conducted at MIT on the effect of melatonin on inducing sleep. Young male volunteers were either given melatonin or a placebo. They were then placed in a dark room at midday and told to close their eyes for 30 minutes. The length of time it took them to fall asleep was recorded. With the placebo, the researchers found it took on average 15 minutes to fall asleep. We will assume this is the population parameter for young males. That is, μ = 15.
= 5
Here is the stem and leaf plot and the JMP descriptive statistics for a random sample of 40 young men who were given melatonin. Note: the values in the stem and leaf are 1.5 minutes, 1.6 minutes and so forth. Two values are much higher than the others. a.
Summarize the descriptive statistics of the data and note any outliers in the data (there will be more on this in part e). -
The mean is 5.55 minutes. The median is slightly less at 5.05 minutes.
-
The standard deviation is 2.94 minutes which reflects a fair amount of spread in the data the Coefficient of
-
Variation is 53%.
-
There are two values that are considerably larger than the rest at 15.6 and 16.2 minutes, both of which are
more than 3.4 standard deviations away from the mean.
b. Calculate the Standard Error
of the data using the sample statistics. The formula for the Standard Error of the mean is: S.E. = SQRT(s
2
/n) = s/SQRT(n) = 2.941/SQRT(40) = 2.941/6.3246 = .4650
3 c. Construct a
95% Confidence Interval
for this data using the Standard Error you just calculated. Use the Z-value from the table you calculated from the previous problem. Z-value Approach 5.553 ± 1.96*(.465) = 5.553 ± .911 4.642 to 6.464
d. Calculate the Z-score for this sample mean as if it were part of a sampling distribution with
= 15 and
= 5. So the SE = 5/(40)
.5
that is, as if the sample was from a population that was similar to the placebo group. We are asking, if the population values are true, how far away is our sample from those population values? I will help you here with the formula below. This z-value will be a large value –
many standard deviations away from the mean of 15 –
so don’t worry about it being large
. z = (5.553 –
15)/[5/SQRT(40)] = - 11.950 This result is based on using population values of μ = 15 and s = 5.
What is the probability of finding a sample mean equal to or less than the value you calculated if the population parameter for the mean is really 15? Hint: Use the z-score you calculated above, then look up this z-value in the table to find a probability associated with being less than the z-value. If the z-value is off the chart on the standard normal table, you simply say the probability is < .001, the farthest reaches of the table. Based on this result, did the melatonin seem to work? As noted above, the probability of our z-score is very, very small (way off the table we have been using). It is less than
.001.
It sure looks like the melatonin worked!!!! But here is how we will express it:
It was a rare event to get a random sample of 40 young men with a mean of 5.55 if it really came from a
population with a mean of 15. In fact, it would be extremely rare.
We have evidence to support that our sample is different from the placebo group.
With a low probability of being wrong, we have evidence that melatonin does lead to a reduction in time to get to
sleep.
This example is a Hypothesis Test where the Null Value is 15. Hypothesis Tests are the next topic after
Confidence Intervals.
e. There are two extreme values in the data. What do you think about these? How might we express them as outliers or not? Are they unusual? For the observation 15.6 Z = (15.6 –
5.55)/2.94 = 3.42
For the observation 16.2 Z = (16.2 –
5.55)/2.94 = 3.62
Both observations are more than 3 standard deviations from the mean. This is very rare. Perhaps something
different happened with these two subjects –
it is worth investigating.
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4 3. Pond's Age-Defying Complex is a cream with alpha-hydroxy acid, a product that is advertised to improve the skin.
In a study, 130 women over age 40 used a cream with alpha-hydroxy acid, for 22 weeks. At the end of the study period the women were examined by dermatologists and 72 were determined to exhibit skin improvement. This is a proportion problem.
a.
Calculate a 95% Confidence Interval for the proportion of women who exhibited improvement. Remember, the formula for the confidence interval for a proportion is: p
±
Z
a
/2
*
(
p
*(1
-
p
))
n
= p = 72/130 = .554 q = .446
Standard Error = [(.554)(.446)/130].5 = .0439 rounded to .044
.554 ± 1.96(.044)
.554 ± .0855
.4684 to .6393
b.
Does there seem to be support from this study that the cream improved more than half of the women who use it? No, we do not have sufficient evidence that more than half of the women using the cream would show
improvement. Even though nearly 55% of the sample showed improvement based on the sample estimate,
based on our confidence interval, .5 is within the confidence interval.
Therefore, it is possible that less than 50% showed improvement
5 4. A study was done that looked at the affect of an oil spill on plant growth. Random plots were selected within the oil spill and counts were made of the number of plants growing in the plot. Later this will be compared to random plots not affected by the spill. For now we will focus on the oil spill plots, 40 in total. For now we will treat this as a large sample and you can use a z-value for the confidence interval. The information from JMP is given below. All the information you need for a confidence interval is given below. a.
Briefly summarize the data using measures of central tendency and variability. The mean number of plants per plot was 26.925, or nearly 27 plants per plot. The median was slightly lower at
26 plants per plot. Based on the histogram and the stem and leaf plot, the distribution appears symmetrical,
and mound shaped. An assumption of normality of the distribution is reasonable. The range of the data is
from a low of 5 plants to a high of 52. The standard deviation is 9.882. The Coefficient of Determination is 36.7
indicating a moderate amount of spread in this variable.
b.
Construct a 99% C.I. for the number of plants in the oil spill plots. Remember, the formula for a confidence interval is: X ±
Z
a
/2
*
s
n
The standard error is actually given in the JMP output as 1.563. This can also be calculated by us as
9.882/SQRT(40) = 9.882/6.3246 = 1.5625 (some slight rounding error). We will use a z-value of 2.575
for the 99% C.I.
26.925 ± 2.575*1.563
26.925 ± 4.025
22.900 to 30.950
c. Say what this confidence interval means in words. We are 99% confident that the true mean lies between 22.900 and 30.950 plants per plot. By this we
mean if we took many samples and constructed a confidence interval for each sample, 99% of them
would contain the true population mean. It is highly likely that the true population mean value is within
our confidence interval.
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