- 1.
- Z-confidence interval for the mean
- By hand (although we'll still use the TI-83 to get the Z "critical" value)
- With the TI-83
*ZInterval*function

- 2.
- Z-confidence interval for the proportion
- By hand, but using the TI-83 to get the Z critical value
- With the TI-83
*1-PropZInt*function

- 3.
- t-confidence interval for the mean (with summary
statistics given)
- By hand (and using the TI-83 to get a t-critical value - we need to use the EQUATION SOLVER for this)
- With the TI-83
*TInterval*function

- 4.
- t-confidence interval for the mean (with "raw"
data given)
- By hand (using the TI-83 to get the t-critical value - and the summary statistics)
- With the TI-83
*TInterval*function

- 1.
- Consider Exercise 7.9, on page 423 of your textbook.

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 ,*n*= 100, and .- (a)
- Set up a 95% confidence interval estimate of the true average
amount of soft drink in each bottle.
*invNorm*function with .975 as the argument, since, as your hand-drawn curve should clearly show, the area from to*Z*_{.025}is .95 + .025.(1.9802, 1.9998) - (b)
- Does the population of soft-drink fill have to be normally
distributed here?

No; since is known and*n*=100, we can use the central limit theorem and assume that the sampling distribution of 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:(1.9602, 1.9798) - (e)
- Now let's re-do part (a) using the
*ZInterval*function. Under STAT TESTS, choose ZInterval. 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.

- 2.
- Consider Exercise 7.29, on page 439 of your textbook.

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*p*_{s}= 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.
*invNorm*function with .95 as the argument, since, as your hand-drawn curve should clearly show, the area from to*Z*_{.05}is .90 + .05.(.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's re-do part (a) using the
*1-PropZInt*function. Under STAT TESTS, choose 1-PropZInt. Enter the information for this problem as you see below. With the cursor on Calculate, press ENTER.

- 3.
- Consider Exercise 7.17, on page 433 of your textbook.

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 .- (a)
- Assuming a normal distribution, set up a 95% confidence
interval estimate of the average value of all greeting
cards in the store.
*invT*function, so we use the following method to get the t critical value. Under the MATH menu, choose Solver.*tcdf*function, which is under DISTR.*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.

The calculator will solve for "X". Note that we are setting the cdf equal to .95, the amount of area between the two t critical values. Note also that we specify the degrees of freedom,*df*=*n*-1 = 20-1 = 19 for this problem. Once this information is entered, press ENTER.

We now see that(1.52, 1.82) - (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's re-do part (a) using the
*TInterval*function. Under STAT TESTS, choose TInterval. 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.

This gives the same results as before.

- 4.
- Consider Exercise 7.18, on page 433 of your textbook.

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*L*_{1}, say. Obtain the summary statistics from STAT CALC 1-Var Stats.*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.
*df*= 9 and the appropriate area is .90.(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 . - (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 , which is 2.2622.(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 4^{th}value to 585. Obtain the summary statistics again. The mean and standard deviation will be different.(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.