Download Grading of Oct 22 homework: If the student made some

Grading of Oct 22 homework: If the student made some effort to answer
#10, s/he received 25 points. For those who answered it correctly and
showed their work, 35 points. For the remaining 3 questions, 25 points each
IF the student provided the correct answer AND showed/explained how the
answer was obtained. If only the answer was provided, the student obtained
10 points (it’s too easy to get the answer from someone else and write it
down so I need you to explain exactly how you obtained the answer—it will
also help you to remember what you did).
Answer to #10 on page 235 (I provide this in more detail than I would
expect you to)
#10. For the population of a large southern city, mean family income is
$34,000, with a standard deviation (for the population) of $5,000.
a. Imagine that you are taking a subsample of 200 city residents.
What is the probability that your sample mean is between $33,000
and $34,000?
b. For this same sample size, what is the probability that the sample
mean exceeds $37,000?
If we go to page 227 the “learning check” asks the same exact question.
Answer: If we imagine a normal curve and draw a line at the $33,000 point
and another line at the $34,000 point (which happens to be the population
mean) we can then visually see the area where question 10a is referring.
What percentage of cases in our sample is expected to fall in this area
(between $33,000 and the mean of $34,000). That number will be the
probability that our sample mean would fall in this area (the answer to 10a).
So, since the mean is $34,000, all we have to do is determine the Z score
for $33,000, look at the table of Z scores and examine the percentage
score found under column B which is the area between Z and the mean.
What makes this problem more difficult is that we need to use a slightly
different formula in order to calculate the Z score since we have
information about the population that we can use. The slightly different
formula is the second formula provided in the learning check.
Substituting this formula for the formula above it (used when we have a
sample but don’t have information about the whole population) would be easy
to do except that you are only just learning the terminology. Consequently,
this slightly different formula uses terms that you are less familiar with
which makes it difficult to figure out how to plug numbers into the formula.
So, the learning check states “However, because here we are dealing with a
sampling distribution” (actually a sample of 200 that we will treat as if it
were a sampling distribution) “replace Y with the sample mean (or $33,000)
and replace ‘Y mean’ with the sampling distribution’s mean (actually the
population’s mean since we know that a sampling distribution’s mean is
equivalent to the population’s mean and we have the population mean
available), and the standard deviation with the standard error of the mean.”
So first we may want to calculate the standard error of the mean. This is
accomplished by taking the square root of the sample N (200) and then
dividing it into the population’s standard deviation (the formula is provided in
the learning check—it is found in the bottom formula, and within this
formula is the bottom half).
That is, the standard error of the mean = $5,000/ sqrt of 200 = $353.55
By fitting the SE ($353.55) into the whole formula we can then obtain the
Z score for a sample mean of $33,000.
Z = $33,000 - $34,000/ 353.55 = -2.83
The area between the Z score of 2.83 (converted from $33,000) and the
mean ($34,000) is about .4977 which is also the probability of a mean
between $33,000 and $34,000. That is, there is a 49.8% probability that
our sample mean would fall between $33,000 and $34,000.
To take this a step beyond the question being asked, what would be the
probability that our sample mean would fall between $35,000 and the mean
($34,000)? This is really easier than it looks. Since $33,000 is $1,000 less
than the mean and $35,000 is $1,000 more than the mean, their percentages
will be the same (.4977) with $33,000 drawn to the right of the mean
(34,000) and $35,000 drawn to the left of the mean. These could be viewed
as confidence intervals and we could say that we are 99 percent confident
that the sample mean will fall between $33,000 and $35,000 (the
confidence level for this confidence interval is .4977 + .4977).
The second part of the question (#10b) uses the same formula to determine
the Z score for $37,000 or:
Z = 37,000 – 34,000/353.55 = 8.48
When we look up this Z score, we find that the Z value is so large that it is
not represented in Appendix B. This means that, the probability of obtaining
a sample mean larger than $37,000, is essentially zero for all intents and
purposes.
Learning Check on Page 242: If you don’t understand the relationship
between the confidence level and Z, review the material in Chapter 6. What
would be the appropriate Z value for a 98% confidence interval?
A 98% confidence interval means that 98% of the cases fall within the
interval. So, if you drew a normal curve the lines going vertical through it
would be far to the left and far to the right since almost all the cases would
fall between the two (i.e., 98% of them). Since the Z table only provides one
side of the normal curve (with the mean dividing the two sides) and we want
to identify the Z value for a 98% confidence interval, we would divide 98 by
2. This would tell us that .4900 or 49 percent of the cases would fall on
each side of the mean. If we look up .4900 in the Z table in the column
“area between mean and Z”, we will find the Z score, or 2.33.
So, what does this tell us? Well, we know that if we go one standard
deviation from either side of the mean that 68% of the cases will fall within
this area. If we go, 2 standard deviations (1.96 to be exact) from either
side of the mean then 95% of the cases will fall within this wider area (since
we are now out two standard deviations instead of only one). If we go 2.33
standard deviations from either side of the mean (our answer above), then
98% of the cases will fall within this even wider area.
Learning Check on Page 244: What is the 90% confidence interval for the
mean commuting time? (Hint: First, find the Z value associated with a 90%
confidence level). How to calculated a confidence interval of a particular
percentage is explained in detail on pages 242-244 located just prior to the
learning check.
In order to determine a confidence interval, we must first calculate the
standard error (explained on page 243) which is equal to:
SE = standard deviation/sqrt of N or 1.5/sqrt of 500 = 1.5/22.36 = .07
Next, we multiply the standard error times the Z score representing the
confidence level of interest and then add and subtract this number from the
mean (7.5). The Z score for 90% is 1.65. (We obtain this by dividing 90 by
2, getting .4500 and then looking for the Z score associated with .4500
which is, 1.65.) When doing the math:
90% CI = 7.5 + (1.65 * .07) = 7.62
and 7.5 – (1.65 * .07) = 7.38
Thus, we have created a 90% confidence interval that ranges from
7.38 to 7.62 commuting hours per week . This tells us that 90 % of the
cases will fall between 7.38 and 7.62 hours per week.
If we are attempting to estimate the population mean (i.e, the average
commuting time for all persons in our population), then our analysis tells us
that we can be 90% confident that the population mean (average commuting
time) will be between 7.38 and 7.62 hours per week. Please see pages 242 –
244 if you’d like more clarification. Or, send me an email and we can get
together.