1. In this chapter we cover sections 7-1 through 7-4.

2. In chapter 7, we will complete the shift from descriptive statistics to inferential statistics.
Remember, the whole point of statistics is this: it is often not feasible to study the whole population, so you study a sample. But samples rarely have exactly the same results as the whole population.
Inferential statistics takes the results from a study done on a sample, and decides how well those results translate to the larger population. This is a crucial part of statistics!

3. Estimating a Population Proportion
Sections 7-1 and 7-2
Estimating a Population Proportion
We will not be using the critical value method described in the book on p Pay attention to the calculator method in these notes.

4. Proportions
When estimating a population proportion, we are looking at fractions, percentages, or proportions. We look at the percentages of people who had a certain result in a study or survey, and decide how that translates to a percentage of the whole population who would have that same result.

5. Proportions
It is important to make a connection between proportions, probabilities, and percents. They are basically all the same, but can be written in different ways. For instance:
Proportions: 2/10 or 0.2
Probabilities: 0.20
Percents: 20%
All three of these are equivalent, but one is written as a fraction, one as a decimal, and one as a percent.

6. Notation p represents the population proportion
x represents the number of successes
(p-hat) represents the sample proportion of successes
(q-hat) represents the sample proportion of failures

7. Definition Point Estimate
A point estimate is a value used to approximate a population parameter.
This is our best guess for the population proportion, based on the sample we looked at.
The sample proportion is the best point estimate of the population proportion p.
This is saying that while we don’t know the proportion for the whole population, the sample proportion is a good starting estimate.

8. Definition Confidence Interval (CI)
A confidence interval is a range (or an interval) of values that (we are highly confident) contains the true value of a population parameter.
Example:
If someone’s spouse or partner asks them what time they are going to be home for dinner, they could say two types of things. They could say “around 6:30” or they could say “between 6 and 7.” “Around 6:30” would be similar to a point estimate. It is a single value estimating when they will be home. “Between 6 and 7” would be similar to a confidence interval. It is a range of time during which they are pretty confident they will be home.

9. Why use Confidence Intervals?
When a point estimate is used to estimate a parameter of interest, it is unlikely that the value of the point estimate will be equal to the value of the parameter. (Remember….samples are usually different from the population!)
Therefore, we will use the value of the point estimate to help construct an interval estimate for the parameter. We will be able to state, with some confidence, that the parameter lies within the interval, and because of this we refer to these as confidence intervals.

10. Confidence Intervals
We talked in Chp 6 about how you have many samples you could choose from one population. Each sample would produce a slightly different confidence interval, but most will overlap and most will contain the true population parameter somewhere within their range.

11. This is often 90%, 95%, or 99%, and will be given.
Definition
Confidence Level
A confidence level is the probability that out of all the confidence intervals you could produce from looking at all possible samples, this one actually contains the population parameter. In other words, it is how confident you are that the interval estimate contains the value of the parameter.
This is often 90%, 95%, or 99%, and will be given.
Typically, you do not see a confidence level below 90% (although sometimes you do). Remember you want to be CONFIDENT the interval contains the actual value.

12. Example
In a Gallup poll, 491 randomly selected adults were asked whether they are in favor of the death penalty for a person convicted of murder, and 65% of them said that they were in favor. A 95% confidence interval estimate of the percentage of all adults who are in favor of the death penalty is 60.8% < p < 69.2%. Interpret this information.
Point estimate = 65%, so = 0.65
Interpretation: We are 95% confident that the actual percentage of all adults that are in favor of the death penalty is between 60.8% and 69.2%.
The interpretation of the confidence interval is extremely important. I have highlighted in red the words that must appear in the interpretation to get full credit. We use the word “all” because we are relating this interpretation to the population, not the sample.

13. Confidence Interval Assumptions for p
Before estimating a population proportion, we must check a couple of things:
1. Find .
2. To be able to use these methods, we need np ≥ 5 and nq ≥ 5 to both be satisfied. (Check to see if there are at least 5 successes and 5 failures. Otherwise, this method is not accurate.) Because we do not actually know p or q, we can use as an estimate for p in this calculation.
Let’s try to do a couple of basic examples on checking our assumptions before we continue.

14. In each case, find and check the conditions that np ≥ 5 and nq ≥ 5
On a survey, 23 people said "yes" and 77 people said "no". You are interested in the people who said “no”.
n = ( )
p = unknown; use sample data instead
Check: There are 23 yes responses (np) and 77 no responses (nq), so there are more than 5 of each. We could then continue the problem.

15. In each case, find and check the condition that np ≥ 5 and nq ≥ 5
100 people were randomly chosen and tested for HIV. There were 2% of them who tested positive. You are interested in the ones that tested positive.
n = 100
p = unknown; use sample data instead
Check: np = (100)(0.02) = 2 positives Because there are less than five in this category, we would have to CONSULT A STATISTICIAN for more advanced methods. You cannot continue with the problem.

16. Calculating the Confidence Interval
We can use our calculator to find a confidence interval for a population proportion. You need to push STAT, arrow over to TESTS, down and choose
A: 1-PropZInt. You will see a screen like this:
You need to enter x, n, and your confidence level (written as a decimal). Then go to Calculate and hit ENTER. Because we are working with proportions, it is good to note that x and n will always be whole numbers.

17. Round-Off Rule for Confidence Interval Estimates of p
These are connected to probabilities, so round to
three significant digits
They can be written like this:
0.123 < p < 0.345
or like this: (0.123, 0.345)

18. CONFIDENCE INTERVAL FLOWCHART
What parameter do you
want a C.I. for?
Is the population normal?
Are there at least 5 items in
each category? No
n > 30? No
Yes
Yes
Yes No
Use ZInterval (s known)
Use TInterval (s unknown)
Consult a Statistician
1 – Prop ZInterval
We will be using this flowchart for all of chapter 7. It is a flowchart to help you decide which confidence interval you should choose.

19. Example
In a September 1999 poll conducted by the Gallup Organization, 230 of 1000 randomly selected adults aged 18 years or older stated that they knew someone capable of committing an act of violence in their workplace.
a.) Obtain a point estimate for the proportion of adults 18 years or older who know someone capable of committing an act of violence in their workplace.
b.) Verify that the requirements for constructing a
confidence interval about p are satisfied, using the sample proportion as an estimate.
To find the point estimate, we need to find x and n: x = 230 and n = 1000
p = 230/1000 = 0.23 ˆ ˆ
np = 1000(0.23) = 230 and nq = 1000(1-0.23) or = 770
Over 5 in each category! (Since np = x, we could have figured this out by looking instead of calculating.)

20. Example
In a September 1999 poll conducted by the Gallup Organization, 230 of 1000 randomly selected adults aged 18 years or older stated that they knew someone capable of committing an act of violence in their workplace.
c.) Construct a 98% confidence interval for the proportion of adults 18 years old or older who know someone capable of committing an act of violence in their workplace.
d.) Interpret the confidence interval.
x = 230, n = 1000, C-Level = 0.98
Calculate: (0.199, 0.261)
We are 98% confident that the proportion of ALL adults 18 years of age or older who know someone capable of committing an act of violence in their workplace is between and
Remember the conclusion relates to the population. It is very important you can write a correct conclusion.

21. Example
In a 1995 national survey conducted by the Centers for Disease Control, 55.2% of 10,847 women who had a child between 1990 and 1993 indicated having breast-fed the baby.
a.) Verify that the requirements for constructing a
confidence interval about p are satisfied.
The point estimate is 55.2% = 0.552
We aren’t given x, so we need to find it using the percent of successes: 10,847(0.552) = = 5988 breast-fed (round to nearest whole number, up or down as appropriate)
np = 10847(0.552) = and nq = 10847( ) = 4859
Over 5 in each category!

22. Example
In a 1995 national survey conducted by the Centers for Disease Control, 55.2% of 10,847 women who had a child between 1990 and 1993 indicated having breast-fed the baby.
b.) Construct a 90% confidence interval for the percentage of women who had a child between 1990 and 1993 who breast-fed their baby. Interpret this interval.
x = 5988, n = 10847, C-Level = 0.90
Calculate: (0.544, 0.560)
We are 90% confident that the percentage of ALL women who had a child between 1990 – 1993 who breast-fed their baby falls between 54.4% and 56.0%.
(Percents are used here instead of decimals because of the phrasing of the question.)

23. Example
In a 1995 national survey conducted by the Centers for Disease Control, 55.2% of 10,847 women who had a child between 1990 and 1993 indicated having breast-fed the baby.
c.) Construct a 99% confidence interval for the proportion of women who had a child between 1990 and 1993 who breast-fed their baby. Interpret this interval.
d.) What is the effect of increasing the level of confidence on the width of the interval?
x = 5988, n = 10847, C-Level = 0.99
Calculate: (0.540, 0.564)
We are 99% confident that the percentage of ALL women who had a child between 1990 – 1993 who breast-fed their baby is between 54.0% and 56.4%.
The confidence interval got bigger when we did a 99% confident interval. The interval must be bigger to be more sure that the real value is in it.

24. Margin of Error
The margin of error, E, is the maximum likely difference between the point estimate and the parameter. Since the point estimate is the middle of the confidence interval, the margin of error E is the distance between the point estimate and either end. This gives us another way to write confidence intervals:

25. Example
Given the confidence interval (0.931, 0.949), find the margin of error and rewrite the confidence interval in form.
= ( )/2 = (middle of interval)
E = – 0.94 = = 0.9% margin of error
CI:

26. Sample Size
What if you know the margin of error you are willing to accept, and want to know how many people to interview to stay within that margin?

27. Sample Size for Estimating Proportions
(Note that invNorm finds the critical value referred to in the book’s formula. If your calculator has the pop-up screen for invNorm and you have to enter three values, use area=α/2, μ=0, σ=1.)
α = 1 – confidence level
is a preliminary estimate of the proportion. If you don’t know , then use 0.5.
E should be in decimal form, not a percent.
n is the number to have in your sample, and should always be rounded up to the nearest whole number. (Rounding down means you interview less people, causing a less accurate study with a larger margin of error.)

28. Example
Spalding Corporation wants to estimate the proportion of golfers who are left-handed.
How many golfers must be surveyed if we want 98% confidence that the sample proportion has a margin of error of 2.5%?
b. A USA Today study found that 15% of golfers in their sample were left-handed. Using this information, how many golfers must be surveyed?

29. Solution
Spalding Corporation wants to estimate the proportion of golfers who are left-handed.
a. How many golfers must be surveyed if we want 98% confidence that the sample proportion has a margin of error of 2.5%?
We don’t have any guess about the proportion of left-handed golfers, so we use 0.5.

30. Solution
Spalding Corporation wants to estimate the proportion of golfers who are left-handed.
b. A USA Today study found that 15% of golfers in their sample were left-handed. Using this information, how many golfers must be surveyed?
This time, we are given a previous study to use as a guess for the proportion of left-handed golfers.

31. Estimating µ when σ is Known
Section 7-3
Estimating µ when σ is Known
We will not be using the z-score critical value method described in the book. Pay attention to the calculator method in these notes.

32. Means
When you look at the mean value of a sample, it may or may not be the same as the mean of the whole population. In this section, we will use a sample mean to estimate a population mean.

33. Point Estimate
This is our best guess for the population mean, based on the sample we looked at.
The sample mean, , is the best point estimate of the population mean .
This is saying that while we don’t know the mean for the whole population, the sample mean is a good starting estimate.

34. Confidence Interval Assumptions for m
1. The sample is a simple random sample.
2. The value of the population standard deviation  is known.
3. Either or both of these conditions is satisfied:
The population is normally distributed or n > 30.
Remember if your sample size is bigger than 30 we can assume normality. If the population is not normal and the sample size is not over 30, then we cannot continue with the methods we have in this class. Consult a statistician.
If all these requirements are satisfied, we use a ZInterval to find the confidence interval
Note: In real world situations,  is rarely known, but we discuss how to test it in this section. In Section 7-4 we will be discussing intervals when  is unknown

35. CONFIDENCE INTERVAL FLOWCHART
What parameter do you
want a C.I. for?
Is the population normal?
Are there at least 5 items in
each category? No
n > 30? No
Yes
Yes
Yes No
Use ZInterval (s known)
Use TInterval (s unknown)
Consult a Statistician
1 – Prop ZInterval

36. Calculating the Confidence Interval
To find confidence intervals in this section, we will be using ZInterval: push STAT, go over to TESTS, scroll down and choose 7: ZInterval
Once you are in the ZInterval screen, you can choose “Data” or “Stats”. We will be using “Stats” when the statistics are given to us in the problem. (When you arrow over to Stats, be sure to hit ENTER to choose it.) Then enter s, , and n, and the confidence level. Go to Calculate and press ENTER.

37. Round-Off Rule for Confidence Interval Estimates of m
Because these intervals are related to the mean, round one decimal place farther than the original data, or the same as the sample mean.
They can be written like this:
3.4 < µ < 4.2
Or like this: (3.4, 4.2)

38. Example
A manager at a paper company wants to estimate the mean time required for a new machine to produce all reams of paper. A random sample of 36 reams required a mean machine time of 1.5 minutes for each ream. Assume the population standard deviation is 0.3 minutes.
a. What is the parameter m ?
b. What is the point estimate for the population mean?
c. Construct an interval estimate for m with a confidence level of 95%.
d. Interpret the interval.
n = =
s =

39. Solution n = 36 = 1.5 s = 0.3 a. What is the parameter m ?
A manager at a paper company wants to estimate the mean time required for a new machine to produce all reams of paper. A random sample of 36 reams required a mean machine time of 1.5 minutes for each ream. Assume the population standard deviation is 0.3 minutes.
n = 36
= 1.5
s = 0.3
a. What is the parameter m ?
We do not know the actual value, but we do know m represents the mean amount of time to produce ALL reams
b. What is the point estimate for the population mean?
The point estimate for m is 1.5 minutes

40. Solution
A manager at a paper company wants to estimate the mean time required for a new machine to produce all reams of paper. A random sample of 36 reams required a mean machine time of 1.5 minutes for each ream. Assume the population standard deviation is 0.3 minutes.
c. Construct an interval estimate for m with a confidence level of 95%.
(1.4, 1.6) is a 95% confidence interval for m
You can find this Zinterval with s = 0.3, = 1.5, n = 36, C-Level = 0.95.
d. Interpret the interval.
We are 95% confident that the mean time to produce all reams of paper is between 1.4 and 1.6 minutes.

41. Example n = What is the point estimate for the population mean? = s =
The Ledd Pipe Company has received a shipment of lengths of pipe, and a quality control inspector wants to estimate the average diameter of the pipes to see if they meet minimum standards. She takes a random sample of 20 pipes, and the sample produces an average diameter of 2.55 in and a standard deviation of 0.08 in. Assume the population of diameters is normally distributed with a standard deviation of 0.07 inches.
n = =
s =
What is the point estimate for the population mean?
b. Construct an interval estimate for m with a confidence level of 99%.
c. Interpret the interval.

42. Solution
The Ledd Pipe Company has received a shipment of lengths of pipe, and a quality control inspector wants to estimate the average diameter of the pipes to see if they meet minimum standards. She takes a random sample of 20 pipes, and the sample produces an average diameter of 2.55 in and a standard deviation of 0.08 in. Assume the population of diameters is normally distributed with a standard deviation of 0.07 inches.
n = 20
= 2.55
s = 0.07
We were given two standard deviations in this problem. Always use the standard deviation for the whole population if you have it.
a. What is the point estimate for the population mean?
The point estimate for m is 2.55 in

43. Solution
The Ledd Pipe Company has received a shipment of lengths of pipe, and a quality control inspector wants to estimate the average diameter of the pipes to see if they meet minimum standards. She takes a random sample of 20 pipes, and the sample produces an average diameter of 2.55 in and a standard deviation of 0.08 in. Assume the population of diameters is normally distributed with a standard deviation of 0.07 inches.
c. Construct an interval estimate for m with a confidence level of 99%.
(2.51, 2.59) is a 99% confidence interval for m
You can find this Zinterval with s = 0.07, = 2.55, n = 20, C-Level = 0.99.
d. Interpret the interval.
We are 99% confident that the MEAN diameter of ALL the pipes in the shipment is between 2.51 inches and 2.59 inches.

44. ZInterval with Data
The next slide is an example when you are not given the statistics (just given a set of data). What you need to do if given a set of data is to plug it into a list. (STAT, 1: Edit, enter data under L1).
Once you have that done, go to ZInterval. Instead of choosing Stats, it needs to be Data. Hit enter to choose Data. Enter the population standard deviation, which should be given. For List: make sure it says the list you entered the data into (usually L1). For Freq: it should always be 1. Then enter your confidence level and Calculate. Your interval will appear.

45. Example
A random sample of quiz scores was obtained. They were: 14, 23, 26, 27, 28, 31, 33, 35, 37. Assume that the population of quiz scores has a normal distribution with σ = 6.7.
a. Find the point estimate for the population mean.
b. Find a 90% confidence interval for the population mean. Interpret.

46. Solution
a. Use 1-Var Stats or ZInterval to find the sample mean, which is our point estimate: 28.2
b. The confidence interval is (24.5, 31.9)
We are 90% confident that the mean quiz score for the population is between 24.5 and 31.9 points.

47. Margin of Error
Remember that the margin of error E is the distance between the point estimate and either end. This gives us another way to write confidence intervals:

48. Example
On a previous page, we found a 99% confidence interval for the mean diameter of pipes in a Ledd Pipe Company shipment, which was (2.51, 2.59). Use it to find the margin of error, and write the confidence interval in form.
The point estimate was 2.55, which you can find again by finding the middle of the interval.
E = 2.59 – 2.55 = 0.04 in, so the confidence interval is 2.55 ± 0.04
We are 99% confident that are sample mean is within 0.04 in of the population mean.

49. Sample Size
What if you know the margin of error you are willing to accept, and want to know how many people to interview to stay within that margin?

50. Sample Size for Estimating Means
α = 1 – confidence level
E is the margin of error, in the same units as the data
n is the number to have in your sample, and should always be rounded up to the nearest whole number
(Note that invNorm finds the critical value referred to in the book’s formula. If your calculator has the pop-up screen for invNorm and you have to enter three values, use area=α/2, μ=0, σ=1.)

51. Example
An automobile manufacturer wants to estimate the mean amount of money now being spent on the purchase of new cars in the United States. Estimate the sample size necessary to be 95% confident that you are within $100 margin of error. A pilot study showed a standard deviation of $10,000.

52. Solution
An automobile manufacturer wants to estimate the mean amount of money now being spent on the purchase of new cars in the United States. Estimate the sample size necessary to be 95% confident that you are within $100 margin of error. A pilot study showed a standard deviation of $10,000.
I find it easiest to type everything in the fraction first, hit ENTER, then square the whole thing after that with the x2 button. (See next slide)

53. Calculator

54. Estimating µ When s is Unknown
Section 7-4
Estimating µ When s is Unknown
We will not be using the critical t-value method described in the book. Pay attention to the calculator method in these notes.

55. The information in this section will be very similar to the last.
TInterval
In this section, sigma will be UNKNOWN. In this case we use a t Distribution, and TInterval on the calculator. We have not discussed this distribution before, and will briefly discuss it in the slides. What is important to know is when your sample size gets larger and larger, the t distribution gets closer and closer to a normal distribution. This is good because we know how to work with normal distributions!
The information in this section will be very similar to the last.

56.  Not Known Assumptions
1) The sample is a simple random sample.
2) The value of the population standard deviation  is
unknown. (Although you might know s, the SAMPLE standard deviation)
3) Either the sample is from a normally distributed
population, or n > 30.

57. Random Trivia
The t distribution was developed by William Sealy Gosset, master brewer at the Guinness Brewery in Dublin at the time. He used it to do quality testing on small samples of beer in which the population standard deviation was not known. (So these things DO have real-life applications!) The brewery prohibited publications by their staff, so he published under the pseudonym “Student”. Thus this is often called the Student t distribution.

58. Statistics Geek
In Dublin for a conference, I had a statistics teacher geek-out moment on a Guinness brewery tour:

59. Important Properties of the Student t Distribution
The t distribution is different for different sample sizes
The t distribution has the same general symmetric bell shape as the normal distribution but it reflects the greater variability (with wider distributions) that is expected with small samples.
3. This distribution also has wider intervals to compensate for the added uncertainty you get when you don’t know .

60. CONFIDENCE INTERVAL FLOWCHART
What parameter do you
want a C.I. for?
Is the population normal?
Are there at least 5 items in
each category? No
n > 30? No
Yes
Yes
Yes No
Use ZInterval (s known)
Use TInterval (s unknown)
Consult a Statistician
1 – Prop ZInterval
Here is our flowchart again. We are still working with the population mean. Notice you use TInterval when sigma is unknown.

61. ZInterval or TInterval?
If for a sample of size 20 the sample mean and standard deviation are 85.2 and 7.5 respectively, calculate an interval estimate for m with a 90% confidence level.
A sample of 20 measurements is taken from a population that has a normal distribution. The sample mean and standard deviation are 85.2 and 7.5 respectively, calculate an interval estimate for m with a 90% confidence level .
A sample of 20 measurements is taken from a population that has a normal distribution. The sample mean and standard deviation are 85.2 and 7.5 respectively, and has a population standard deviation of 7.3, calculate an interval estimate for m with a 90% confidence level .

62. ZInterval or TInterval?
If a sample of 45 measurements has a sample mean and standard deviation of 85.2 and 7.5 respectively, calculate an interval estimate for m with a 90% confidence level.
5. If a sample of 45 measurements has a sample mean and standard deviation of 85.2 and 7.5 respectively, and a population standard deviation of 7.3, calculate an interval estimate for m with a 90% confidence level.

63. Solutions
If for a sample of size 20 the sample mean and standard deviation are 85.2 and 7.5 respectively, calculate an interval estimate for m with a 90% confidence level.
A sample of 20 measurements is taken from a population that has a normal distribution. The sample mean and standard deviation are 85.2 and 7.5 respectively, calculate an interval estimate for m with a 90% confidence level .
A sample of 20 measurements is taken from a population that has a normal distribution. The sample mean and standard deviation are 85.2 and 7.5 respectively, and has a population standard deviation of 7.3, calculate an interval estimate for m with a 90% confidence level .
Check if normal? No, so check sample size – 20. Because the sample size is not bigger than 30 we will have to CONSULT A STATISTICIAN (Make sure you always check!)
Check if normal? Yes, says in problem. Your sample standard deviation is known, but your population standard deviation is not. Hence, you would use TInterval.
Check if normal? Yes, says in problem. Your sample standard deviation is known, and your population standard deviation is known. Hence, you would use ZInterval.

64. Solutions
If a sample of 45 measurements has a sample mean and standard deviation of 85.2 and 7.5 respectively, calculate an interval estimate for m with a 90% confidence level.
5. If a sample of 45 measurements has a sample mean and standard deviation of 85.2 and 7.5 respectively, and a population standard deviation of 7.3, calculate an interval estimate for m with a 90% confidence level.
Check if normal? No, check sample size We can continue since our sample is bigger than 30. Your sample standard deviation is known, but your population standard deviation is not. Hence, you would use TInterval.
Check if normal? No, check sample size We can continue since our sample is bigger than 30. Your sample standard deviation is known, and your population standard deviation is known. Hence, you would use ZInterval.

65. Calculating the Confidence Interval
TInterval is in the same area of your calculator as ZInterval. Go to STAT, over to TESTS, scroll down to 8: TInterval. If you know the mean and standard deviation, then choose Stats and fill in , s, n, C-Level, and then Calculate.
The interval will appear on the screen once you hit enter. Once you have your interval, round and label correctly.

66. Round-Off Rule for Confidence Interval Estimates of m
Because these intervals are related to the mean, round one decimal place farther than the original data, or the same as the sample mean.
They can be written like this:
3.4 < µ < 4.2
Or like this: (3.4, 4.2)

67. Example
A study found the body temperatures of 106 healthy adults. The sample mean was degrees and the sample standard deviation was 0.62 degrees.
a) What is the point estimate for the population mean?
b) Construct an interval estimate for m with a confidence level of 95%.
c) Interpret the interval.
Try to work this problem on your own, use the flowchart. Check next slides for solutions.

68. Solution
a) What is the point estimate for the population mean? b) Construct an interval estimate for µ with a confidence level of 95%. c) Interpret the interval.
98.20 degrees
Because our sample is bigger than 30 and sigma is unknown: TInterval
(98.08, 98.32)
We are 95% confident the mean body temperature for all healthy adults is between and degrees.

69. Example
A purchasing agent at the Kelly Bread Company wants to estimate the mean daily usage of rye flour. She takes a sample of 14 days and finds a mean of lb. with a standard deviation of 45.0 lb. Assume the daily rye flour usage is normally distributed.
a) What is the point estimate for the population mean?
b) Construct an interval estimate for m with a confidence level of 95%.
c) Interpret the interval.
Try to work this problem on your own, use the flowchart. Check next slides for solutions.

70. Solution
a) What is the point estimate for the population mean? b) Construct an interval estimate for µ with a confidence level of 95%. c) Interpret the interval.
173.0 pounds
Because the problem states normal and sigma is unknown: TInterval
(147.0, 199.0)
We are 95% confident that the mean daily usage of rye flour for all days is between and pounds.

71. TInterval with Data
The next slide is an example when you are not given the statistics (just given a set of data). What you need to do if given a set of data is to plug it into a list. (STAT, 1: Edit, enter data under L1).
Once you have that done, go to TInterval (assuming the population standard deviation is not given—the calculator will use the data to figure the sample standard deviation instead) Instead of choosing Stats, it needs to be Data. Hit enter to choose Data. For List: make sure it says the list you entered the data into (usually L1). For Freq: it should always be 1. Then enter your confidence level and Calculate. Your interval will appear.

72. Example
A random sample of quiz scores was obtained. They were: 14, 23, 26, 27, 28, 31, 33, 35, 37. Assume that the population of quiz scores has a normal distribution.
Find a 90% confidence interval for the population mean. Interpret.

73. Solution
No population standard deviation is given, so we use TInterval. The confidence interval is (23.9, 32.5).
We are 90% confident that the mean quiz score for the population is between 23.9 and 32.5 points.
Compare this to the similar problem from section With TInterval, the confidence interval is wider. This is because without a population standard deviation, we are less sure of what the mean will be.

74. Comparing Confidence Intervals
When comparing two data sets, if their confidence intervals overlap then it is possible that their means could be the same. If the confidence intervals do not overlap, then most likely one mean is higher than the other.

75. Example
Maximum breadth of samples of male Egyptian skulls are studied from different time periods to measure the amount of interbreeding with people from other regions. Here are two samples:
4000 B.C.:
150 A.D.:
Use 95% confidence intervals to determine whether the head sizes appear to have changed significantly between these two years. Assume normality.

76. Solution
No population standard deviation for either, so we’ll use Tinterval.
For 4000 BC, our confidence interval is (125.7, 131.6)
For 150 AD, our confidence interval is (130.2, 136.5)
4000 BC
150 AD
The confidence intervals overlap, so there is no significant difference in the mean head sizes.
125.7
131.6
130.2
136.5