Download week7quizhelp_oct2007_v03

Week 7 Quiz Help
Things to remember for this week’s quiz. There’s some good info here, pay close
attention.
With the standard normal distribution
Remember that the total area under the curve is equal to 1 or 100% (half on each side).
The mean of the standard normal is 0 and the standard deviation and variance are 1. For
any normal distribution regardless of the mean and standard deviation the area will
always be 1 or 100%. We see the normal distribution all around us as noted in one of our
discussion topics this week. Also remember that if you ever told that you have a normal
distribution and then asked what the probability x = ?, it is zero. We can give the
probability that it is less than a value, greater than a value, or between two values – but
not that it is exactly one value because it is a “continuous distribution”. (For example
given that mu = 9, sigma = 2.1, what is the probability that x = 7? It’s 0 because with a
continuous distribution we are “slicing jello” as I like to say.
On the Central Limit Theorem, remember that it states that given a distribution with a
mean μ and variance σ², the sampling distribution of the mean approaches a normal
distribution with a mean (μ) and a variance σ²/N as N, the sample size, increases. The
amazing and counter-intuitive thing about the central limit theorem is that no matter what
the shape of the original distribution, the sampling distribution of the mean approaches a
normal distribution. Also another important fact is that any sample size is big enough
when we know the population is normal. Using the Central Limit Theorem answer the
following question.
Assuming you have a normal distribution for your population and you take 64 samples of
size 25 each. Calculate the standard deviation of the sample means if the population’s
variance is 16.
Since the population is normally distributed with a variance of 16, then the sample means
have a variance equal to 16/25 according to the Central Limit Theorem. Hence their
standard deviation will be SQRT(16/25) = 4/5 = .800
Be able to use the normal distribution to solve problems. Examples
Bob scored a 190 on his entrance exam, where
the average was 165 and the standard deviation
was 12. Where does he stand in relation to the
rest of his class?
He scored in the “top 2 %”, see excel
attachment.
In a normal distribution with mu = 35 and sigma
= 6 what is the z score for a value of 41?
Z= +1 see excel attachment
In a normal distribution with mu = 35 and sigma
= 6 what number corresponds to z = -2?
23, see excel attachment
We have an area of .4840. What z-score
corresponds to this area?
Using the Standard Normal Table in your book,
or the one on the attached excel you can see it is
a z score of -0.04
Find P(80 < x < 86) when mu = 82 and sigma =
4. Write your steps in probability notation.
I did this in excel (see attachment – fourth tab),
but I still need to show my work:
The z-score corresponding to x = 86 is z = (8682)/4 = 4/4 = 1.0.
The area corresponding to z = 1.0 is .8413
The z-score corresponding to x = 80 is (80 82)/4 = -2/4 = -0.5.
The area corresponding to z = -0.5 is .3086.
Thus, P(80 < x < 86)
= P(-0.5 < z < 1.0)
= P(z < 1.0) - P(z < -0.5)
= 0.8413 – 0.3086
= 0.5327 (my excel calculated it to be 0.5328 so
I feel good about it)
My excel file will be in a separate post.
Interpret a 90% confidence interval of (63.3,
83.4).
You should note that there here is a 90%
probability that the interval (63.3 to 83.4)
contains µ, the true population mean.
What is the critical value that corresponds to a
confidence level of 96%
100 – 96 = 4
Divide 4 by 2 (tails) and get 2
Add 2 to the original 96% and get 98% and find
the critical value (z-score that corresponds to
.9800 which is 2.05 (closest to it)
Compute the population mean margin of error
for a 90% confidence interval when sigma is 7
and the sample size is 81.
E = z * sigma / sqrt(n) = 1.645 * 7 / sqrt(81) =
1.279 (remember +/-)
A Military entrance exam has a mean of 120 and
a standard deviation of 9. We want to be 95%
certain that we are within 6 points of the true
mean. Determine the sample size.
n = ( z * sigma / error ) ^ 2 = (1.96*9/6)^2 =
2.94^2 = 8.6436. Round up to 9.
A researcher wants to get an estimate of the true
mean performance measure of its product. It
randomly samples 180 of its machines. The
mean performance measure was 900 with a
standard deviation of 60. Find a 95%
confidence interval for the true mean
performance measure of the machines.
The population standard deviation is unknown
and the sample size is 180. Thus, since the
sample size is greater than 30, this confidence
interval will use a z-value. For a 95%
confidence interval, the z-value = 1.96. Sample
mean = 900 and sample standard deviation = 60.
Population mean = 900 +/- 1.96 * 60/sqrt(180) =
900 +/- 8.765. 891.235 and 908.765
A researcher wants to get an estimate of the true
mean performance measure of its product. The
researcher needs to be within 15 of the true
mean. The researcher estimates the true
population standard deviation is around 30. If
the confidence level is 95%, find the required
sample size in order to meet the desired
accuracy.
For a 95% confidence level, the z-value = 1.96.
The formula for sample size is n = ( z-value *
standard deviation / error ) ^ 2 = ( 1.96 * 30/ 15)
^ 2 = ( 3.92 ) ^ 2 = 15.3664. Thus, the
researcher must sample at least 16 to obtain the
desired accuracy.
A researcher wants to estimate the mean cost to
develop a product. The researcher tests 18 cases
and finds the mean cost to be $3000 with a
standard deviation of $400. Find a 95%
confidence interval for the true mean cost to
develop this product.
The population standard deviation is unknown
and the sample size is 18. Thus, since the
population standard deviation is unknown AND
the sample is less than 30, we must use the tvalue for this confidence interval. For a 95%
confidence interval and degrees of freedom = 17
(from 18-1), the t-value = 2.110. Sample mean
= 3000 and sample standard deviation = 400.
Population mean = 3000 +/- 2.110 *
400/sqrt(18) = 3000 +/- 198.93. So our bounds
are 2801.07 and 3198.93
A researcher wants to estimate what proportion
of failures that are due to poor workmanship.
The researcher randomly samples 50 failures
and finds 18 are due to poor workmanship.
Using a 95% confidence interval, estimate the
true proportion of poor workmanship for all
failures.
For a 95% confidence level, the z-value is 1.96.
The sample proportion = 18/50 = 0.36, thus p
hat = 0.36 and 1-p hat = 0.64 . The sample size
= 50. The population proportion is between 0.36
+/- 1.96 * sqrt ( 0.36 * 0.64 / 50 ) = 0.36 +/- .13
So the bounds are .23 and .49
Here is the link to a lecture on preparing for the final exam, I did
about a year ago.
https://sas.elluminate.com/mr.jnlp?suid=M.27531350C5A252320BBBBE48B8CF0F
Excel Spreadsheets Follow (Double
Click to see all tabs)
Normal Distribution
Mean
165
Stdev
12
P(X<x)
x
x
190
P(X>x)
0.0186
x
P(>x)
x
P(x <X<x )
x
Inverse Calculations
P(<x)
x
x
Symmetric Intervals
P(x <X<x )
x
Remember the green areas are
red lettering is the resulting ca
It is important to note that the c
to show you what part of the g