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Confidence Intervals

This page contains examples of the following:
  1. Z-confidence interval for the mean
  2. Z-confidence interval for the proportion
  3. t-confidence interval for the mean (with summary statistics given)
  4. t-confidence interval for the mean (with "raw" data given)

1.
Consider the following exercise: The inspection division of the Lee County Weights and Measures Department is interested in estimating the actual amount of soft drink that is placed in 2-liter bottles at the local bottling plant of a large nationally known soft-drink company.   The bottling plant has informed the inspection division that the standard deviation for 2-liter bottles is 0.05 liter.   A random sample of one hundred 2-liter bottles obtained from this bottling plant indicates a sample average of 1.99 liters.   Note that the standard deviation did not come from the sample.   Therefore, this will be a Z-interval, as opposed to a t-interval.   For this problem, we have $\sigma = 0.05$, n = 100, and $\overline{x} = 1.99$.
(a)
Set up a 95% confidence interval estimate of the true average amount of soft drink in each bottle.  
$\overline{x} \pm Z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}$
Here, $Z_{\alpha / 2} = Z_{.05 / 2} = Z_{.025} = 1.96$.   This value is found by drawing a normal curve (by hand) in order to illustrate exactly what we need for this problem.   Draw a curve and mark off the middle 95% of the area.   We need to know the Z-values that correspond to those marks, namely $\pm Z_{.025}$.   Use the invNorm function with .975 as the argument, since, as your hand-drawn curve should clearly show, the area from $-\infty$ to Z.025 is .95 + .025.

normal curve        TI-83 screen

$1.99 \pm 1.96 \frac{0.05}{\sqrt{100}}$
$1.99 \pm 0.0098$
(1.9802, 1.9998)
(b)
Does the population of soft-drink fill have to be normally distributed here?  
No; since $\sigma$ is known and n=100, we can use the central limit theorem and assume that the sampling distribution of $\overline{X}$ is approximately normal.  
(c)
Explain why an observed value of 2.02 liters is not unusual, even though it is outside the confidence interval you calculated.   An individual value of 2.02 liters is only 0.60 standard deviations above the sample mean of 1.99.   The confidence interval represents bounds on the average of a sample of 100, not an individual value.   A sample average of 2.02 would be unusual.  
(d)
Suppose that the sample average had been 1.97 liters.   What would be your answer to (a)?  
This shift of the sample average will result in a shift of the confidence interval by the same distance.   The width of the interval will not be changed, so we have:
$1.97 \pm 0.0098$
(1.9602, 1.9798)
(e)
Now let us re-do part (a) using the ZInterval function.   Under STAT TESTS, choose ZInterval.  

TI-83 screen

We do not have a list of data for this problem, instead we have summary statistics from the textbook.   Choose Stats.   Enter the information for this problem as you see below.   With the cursor on Calculate, press ENTER.  

TI-83 screen TI-83 screen

This gives the same results as before.

2.  
Consider the following Exercise: The dean of a graduate school of business wishes to estimate the proportion of MBA students enrolled who have access to a personal computer (PC) outside the school (either at home or at work).   A sample of 150 students reveals that 135 have access to a PC outside the school.   Since this is a confidence interval for the proportion, it will be a Z-interval.   For this problem, we have n = 150, and ps = 135/150 = .9.
(a)
Set up a 90% confidence interval estimate of the population proportion of students who have access to a PC outside the school.  
$p_{s} \pm Z_{\alpha / 2} \sqrt{\frac{p_{s}(1-p_{s})}{n}}$
Here, $Z_{\alpha / 2} = Z_{.10 / 2} = Z_{.05} = 1.645$.   Again, this value is found by drawing a normal curve (by hand) in order to illustrate exactly what we need for this problem.   Draw a curve and mark off the middle 90% of the area.   We need to know the Z-values that correspond to those marks, namely $\pm Z_{.05}$.   Use the invNorm function with .95 as the argument, since, as your hand-drawn curve should clearly show, the area from $-\infty$ to Z.05 is .90 + .05.  

normal curve        TI-83 screen

$.9 \pm 1.645 \sqrt{\frac{.9(.1)}{150}}$
$.9 \pm .0403$
(.8597, .9403)
(b)
How can the dean use the results in (a) to determine whether additional PCs should be purchased for the lab?  
We are 90% confident that the true proportion of students who have access to a PC outside the school falls between .8597 and .9403.   It follows that the proportion of students who do not have access to a PC outside the school falls between .0597 and .1403.   The proportion obtained by dividing the number of PCs in the lab by the total number of students should likewise fall in this interval.  
(c)
Now let us re-do part (a) using the 1-PropZInt function.   Under STAT TESTS, choose 1-PropZInt.  

TI-83 screen

Enter the information for this problem as you see below.   With the cursor on Calculate, press ENTER.  

TI-83 screen TI-83 screen

This gives the same results as before.  

3.
Consider the following exercise: A stationary store wants to estimate the average retail value of greeting cards that it has in its inventory.   A random sample of 20 cards indicates an average value of $1.67 and a standard deviation of $0.32.   Note that the standard deviation did come from the sample.   Therefore, this will be a t-interval, as opposed to a Z-interval.   For this problem, we have s = 0.32, n = 20, and $\overline{x} = 1.67$.
(a)
Assuming a normal distribution, set up a 95% confidence interval estimate of the average value of all greeting cards in the store.  
$\overline{x} \pm t^{(n-1)}_{\alpha / 2} \frac{s}{\sqrt{n}}$
Here, $t^{(n-1)}_{\alpha / 2} = t^{(19)}_{.05 / 2} =
t^{(19)}_{.025} = 2.093$.   This value is found by drawing a t curve (just draw a bell-shaped curve like usual) in order to illustrate exactly what we need for this problem.   Draw a curve and mark off the middle 95% of the area.  

t curve

We need to know the t-values that correspond to those marks, namely $\pm t^{(19)}_{.025}$.   For this, there is no invT function, so we use the following method to get the t critical value.   Under the MATH menu, choose Solver.  

TI-83 screen TI-83 screen

We want the t values such that there is 95% of the area under the curve between them, symmetrically.   Therefore, we use the equation solver to obtain these values from the tcdf function, which is under DISTR.  

TI-83 screen TI-83 screen

The tcdf function takes three arguments, i.e. (lower bound, upper bound, df), where df stands for degrees of freedom.   We want to know the lower bound and the upper bound, but we know that one is just the negative of the other, due to the symmetry of the curve.   So we can just pick one of them to solve for.   For the arguments, enter L for the lower bound, U for the upper bound, and D for the degrees of freedom.   Also, set the cdf equal to A, representing the area between L and U.   Once these variables are entered, press ENTER.   Now we need to enter actual values for the arguments.   There are a few different ways we could do this.   Consider the left tail area of the curve, which is .025.   Let's have the calculator solve for the t critical value on the left side.   So L will be negative infinity, U is what we want to solve for, D = df = n-1 = 20-1 = 19 for this problem, and A = .025.   Enter in these values for L (use -1E99), D and A.  

TI-83 screen TI-83 screen

The solver expects a "guess" for U.   We can simply enter "0" for this, since all t-curves are centered about zero.   With the cursor on the U=0 line, press SOLVE.   This is done by pressing ALPHA ENTER.   The calculation will take about 20 seconds; be patient!
We now see that $\pm t^{(19)}_{.025} = 2.093.$
$1.67 \pm 2.093 \frac{0.32}{\sqrt{20}}$
$1.67 \pm 0.15$
(1.52, 1.82)

TI-83 screen

(b)
How might the results obtained in (a) be useful in assisting the store owner to estimate the total value of her inventory?  
We are 95% confident that the true mean value of the cards falls between $1.52 and $1.82.   Therefore, we could multiply these bounds by the total number of cards in the store to get an interval for the total value of the stock.  
(c)
Now let us re-do part (a) using the TInterval function.   Under STAT TESTS, choose TInterval.  

TI-83 screen

We do not have a list of data for this problem, instead we have summary statistics from the textbook.   Choose Stats.   Enter the information for this problem as you see below.   With the cursor on Calculate, press ENTER.  

TI-83 screen TI-83 screen


This gives the same results as before.  

4.
Consider the following exercise: The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan.   A random sample of 10 employees reveals the following family dental expenses (in dollars) for the preceding year:
     110    362    246   85    510    208    173    425    316    179
We were not given a mean or standard deviation; we'll have to get them ourselves from the data.   Of course, the standard deviation we get will be a sample standard deviation, which makes this a t-interval, as opposed to a Z-interval.   Enter the data into your calculator, into L1, say.  

TI-83 screen

Obtain the summary statistics from STAT CALC 1-Var Stats.  

TI-83 screen TI-83 screen

TI-83 screen

So for this data, we have $\overline{x} = 261.4$, s = 138.8, and n = 10.  
(a)
Set up a 90% interval estimate of the average family dental expenses for all employees of this corporation.  
$\overline{x} \pm t^{(n-1)}_{\alpha / 2} \frac{s}{\sqrt{n}}$
Here, $t^{(n-1)}_{\alpha / 2} = t^{(9)}_{.10 / 2} =
t^{(9)}_{.05} = 1.8331$.   This value is found by the same method illustrated above.   Again let's use the left tail area, which is .05, and solve for the critical value on the left side.   Now df = 9.  

t curve   TI-83 screen

TI-83 screen TI-83 screen

$261.4 \pm 1.8331 \frac{138.8}{\sqrt{10}}$
$261.4 \pm 80.459$
(180.941, 341.859)
(b)
What assumption about the population distribution must be made in (a)?  
The population of dental expenses must be approximately normally distributed.  
(c)
Give an example of a family dental expense that is outside the confidence interval but is not unusual for an individual family, and explain why this is not a contradiction.  
A dental expense of $350 would not be unusual for a single family, since it is less than one standard deviation from the mean.   The confidence interval is for the true mean, and the sampling distribution of the mean has a much smaller standard deviation; remember that we divide the standard deviation by $\sqrt{n}$.  
(d)
Suppose you used a 95% confidence interval in (a).   What would be your answer to (a)?  
Draw another curve to see that now we need $ \pm t^{(9)}_{.025}$, which is 2.2622.  

t curve   TI-83 screen

TI-83 screen TI-83 screen

$261.4 \pm 2.2622 \frac{138.8}{\sqrt{10}}$
$261.4 \pm 99.293$
(162.107, 360.693)
(e)
Suppose the fourth value was $585 instead of $85.   What would be your answer to (a)?   What effect does this change have on the confidence interval?  
Go to STAT Edit and change the 4th value to 585.  

TI-83 screen

Obtain the summary statistics again.   The mean and standard deviation will be different.  

TI-83 screen

$311.4 \pm 1.8331 \frac{157.1}{\sqrt{10}}$
$311.4 \pm 91.067$
(220.333, 402.467)
The confidence interval is shifted to the right, and is a bit wider.  
(f)
Now let's re-do part (a) using the TInterval function.   Under STAT TESTS, choose TInterval.   We do have a list of data for this problem, so choose Data.   Enter the information for this problem as you see below.   With the cursor on Calculate, press ENTER.  

TI-83 screen TI-83 screen

This gives the same results as before.  



 
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Updated 21 August 2003