Figure 1
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Problem Statement: A population has a known mean=486.0 and
standard deviation=24.15. We are going to take repeated samples,
with replacement, of size 34 from this population.
The standard deviation of the sampling distribution is `sigma_(barx)`.
Find the value, rounded to 3 decimal places, of `sigma_(barx)`?
(Correct answer :4.142.)
We really do not need to know the mean value for this. The
standard deviation of the sample means is `sigma/sqrt(n)` where
`sigma` is the standard deviation of the population and `n` is the sample size.
Thus, in Figure 1, we compute `24.15/sqrt(34)` to get the answer
`4.141690831` which we round, to 3 decimal place, to `4.142`.
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Figure 2
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Problem Statement: A population has a known mean=129.0 and standard deviation=15.44.
We are going to take repeated samples, with replacement,
of size 20 from this population.
Find the standard error, rounded to 3 decimal places,
of the sample means?
(Correct answer: 3.452.)
The standard error is just the standard deviation of the sample means.
Thus, we compute the same value `sigma_(barx) = sigma/sqrt(n)` which in this case is
`15.44/sqrt(20)`. This produces `3.452488957` which we round to `3.452`.
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Figure 3
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Problem Statement:
A population has a known mean=–179.0 and
standard deviation=28.50. We are going to take repeated
samples, with replacement, of size 34 from this population.
What is the probability that the sample mean, `barx`, will be
less than –185.989? (Express answer
rounded to 4 decimal places.)
(Correct answer: 0.0764.)
We will look at two ways to do this problem. The first
method immitates the process that we would go through if we had
the normal distribution table. To use that table we need to
convert the –185.989 of the problem into a z-score
We do that by dividing the difference between that value and the given mean by the
standard deviation of the sample means. We will do this in separate steps. First we
compute the standard deviation of the sample means as `28.5/sqrt(34)` and we store that in `X`.
Next we find the difference between our point value, `-185.989` and the given mean `-179`.
Then we divide that answer by `X` to get the z-score.
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Figure 4
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Now that we have the z-score we want to find the area under the
standard normal curve (with mean=0, standard deviation=1), to the left of the z-score.
The command normalcdf(–9999,Ans) will do that.
We round the answer to 4 decimal places to get `0.0764`.
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Figure 5
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An alternative approach is to use the functionality of the calculator
to process the data without making an external change to a z-score.
To do this we use the extended form of the
normalcdf( command, the one that allows us to enter the mean
and standard deviation of the population. We just have to remember that for us, the "population"
is now the population of the means of 34-item samples. Therefore, the mean of the poplation,
`mu_(barx)=mu` and `sigma_(barx)=sigma/sqrt(n)`. Note that in Figure 5, the extended
normalcdf( command uses a very low "left end", namely `-99999`, and the value `28.5/sqrt(34)`
as the appropriate standard deviation.
By using this extended form we are able to obtain the answer in one step.
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Figure 6
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Problem Statement:
A population has a known mean=371.0 and standard deviation=13.90.
We are going to take repeated samples, with replacement, of
size 18 from this population. What is the probability that the
sample mean, `barx`, will be greater than 377.651?
(Express answer rounded to 4 decimal places.)
(Correct answer: 0.0212.)
As was the case starting in Figure 3,
we will look at two ways to do this problem. The first
method immitates the process that we would go through if we had
the normal distribution table.
In that case we would be solving the problem by recognizing
that `P( barx ) > 377.651 = 1 - P(barx < 377.651)`.
We can evaluate the expression on the right by using
standard normal distribution table if we convert
`377.651` to a z-score.
We do that by dividing the difference between that value and the given mean by the
standard deviation of the sample means. We will do this in separate steps.
First we
compute the standard deviation of the sample means as `13.9/sqrt(18)`
and we store that in `X`.
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Figure 7
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Next we find the difference between our point value, `377.651` and
the given mean `371`.
Then we divide that answer by `X` to get the z-score.
At that point we can evaluate `1-P(barx < Ans)` which we express on the calculator as
normalcdf(`-999`,Ans).
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Figure 8
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We could do the same problem without calculating the z-score
by using the extended form of the normalcdf( command. We do this in Figure 8
in a single command where we specify the mean and standard deviation of the population as the
third and fourth arguments in the extended
normalcdf( command.
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Figure 9
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Given the flexibility of the normalcdf( command, we have a third
alternative wehere we use that command to look at the area under the normal curve ro the
right of the value `377.651`.
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Figure 10
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Problem Statement:
A population has a known mean=334.0 and standard
deviation=15.44. We are going to take repeated samples,
with replacement, of size 33 from this population.
What is the probability that the sample mean, `barx`,
will be greater than 334.968 but less than 335.962?
(Express answer rounded to 4 decimal places.)
(Correct answer: 0.1267.)
Again we can look at this in the same way that we would have proceed
if we were using a table. That is we would convert
`P( 334.968 < barx < 335.962 )` to
`P( barx < 335.962) - P( barx < 334.968)`.
To actually use such a table we need to convert the two values into z-scores..
We start by calculating the standard deviation of the sample means,
`15.44/sqrt(33)` and storing that in X.
Then we can compute the two differences from the mean, storing those
results in A and B.
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Figure 11
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Now change A and B by dividing each by the value stored in X.
We conclude by translating the probability expreession given in Figure 10
into TI-84 commands normalcdf(`-9`,B) - normalcdf(`-9`,A).
This gives our answer.
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Figure 12
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As before, we really did not need to tanslate things into N(0,1)
values. The extended normalcdf( command allows us to specify teh appropraite mean
and standard deviation. This makes for a long command, as shown in Figure 12.
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Figure 13
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Figure 13 uses the flexibility of the normalcdf(
to replace the difference of the two values by just a straight
command to get the area between two points given
the mean of the population of
sample means and the associated standard deviation of such means.
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