1. Section 6.1 Confidence Intervals for the Mean(Large Samples)
Objective: SWBAT find a point estimate and a margin of error
And construct and interpret confidence intervals for the population mean
And determine the minimum sample size required when estimating p

2. In this section you will learn how to use sample statistics to estimate the value of an unknown population parameter ц when the sample size is 30 or when the population is normally distributed. And the standard deviation σ is known.

3. Definition
A point estimate Is a single value estimate for a population parameter The most unbiased point estimate of the population mean ц is the sample mean ҳ

4. Example 1 Finding a Point Estimate:
Market researchers use the number of sentences per advertisement as a measure of reliability for magazine advertisements. The following represents a random sample of the number of sentences found in 54 advertisements. Find a point estimate of the population mean ц. 20

5. Solution The sample mean of the data is
So your point estimate for the mean length of all magazine advertisements is
12.4 sentences.

6. Try it yourself
Another random sample of the number of sentences found in 30 magazine advertisements is listed on page 280 use this sample to find the point estimate for ц.
Find the sample mean
b. Estimate the mean sentence length of the population.
Sum= n= /30 =

7. Definition
An interval estimate is an interval or range of values used to estimate a population
Parameter.

8. To form an interval estimate use the point estimate as the center of the interval then add and subtract a margin of error. For instance if the point estimate is 12.4 and the margin of error is 2.1 then an interval estimate would be given by ± 2.1 or < ц < 14.5

9. Before finding an Interval estimate you first determine how confident you need to be that your interval estimate contains the population mean ц. Definition The level of confidence c is the probability that the interval estimate contains the population parameter.

10. The level of confidence c is the area under the standard normal curve between the critical values –zc and zc the area remaining is 1- c and the area in each tail is ½ (1-c)

11. Margin of Error Definition
Given a level of confidence c , the margin of error sometimes also called the maximum error of estimate or error tolerance) E is the greatest possible distance between the point estimate and the value of the parameter it is estimating.
When n ≥ 30 the sample standard deviation s can be used in place of σ.

12. Example 2
Use the data in example 1 and a 95% confidence level to find the margin of error for the mean number of sentences in all magazine advertisements. Solution : The z score that corresponds to a 95% confidence level is 1.96 This implies that 95% of the area under the standard normal curve falls within 1.96 standard deviations of the mean. n=54 ≥ 30 We don’t know the population standard deviation. But because n ≥ 30 you can use s in place of σ.

13. Solution (cont.) Using the values
Interpretation: You are 95% confident that the margin of error for the population mean is about 1.3 sentences.
Using the values

14. Try it Yourself
Use the data in try it yourself 1 and a 95% confidence level to find the margin of error for the mean number of sentences in a magazine advertisement. a.) Identify z, n, and s. b.) Find E using zc, σ≈ s , and n c.) State the margin of error.
Z = n = 30 s = (b) ( c ) You are 95% confident that the maximum error is about sentences per magazine advertisement.

15. Confidence Intervals for the Population Mean
Guidelines
Find the sample statistic n and x
2. Specify σ if known . Otherwise if n≥30
Find the sample standard deviation s and
use it as an estimate for σ.
3. Find the critical value Zc that corresponds
To the given level of confidence.
4. Find the margin of error E
5. Find the left and right endpoints Left endpoint x - E
And form the confidence intervals Right endpoint x + E
Interval x – E < ц < x + E

16. Example 3:
Constructing a Confidence Interval Construct a 95% confidence interval for the mean number of sentences in all magazine advertisements. Solution in Examples 1 and 2 you found that the mean = 12.4 and E = 1.3 The confidence interval is as follows X – E = 12.4 – 1.3 = 11.1 x+ E = = < ц < 13.7

17. Example 5
Constructing a Confidence Interval, σ known A college admissions director wishes to estimate the mean age of all students enrolled. In a random sample of 20 students, the mean age is found to be 22.9 years. From the past studies the standard deviation is known to be 1.5 years and the population tis normally distributed. Construct a 90% confidence interval of the population mean age.

18. Example 5 Cont.
Solution: Using n=20,mean =22.9,σ=1.5 and Zc = the margin of error at the 90% confidence interval is The 90% confidence interval can be written as x ± E = 22.9 ± .6 or as follows = 22.3 or = < ц < 23.5 Interpretation: With 90% confidence you can say that the mean age of all students lies between 22.3 and 23.5 years.

19. Sample Size
For the sample statistics as the level of confidence increases, the confidence interval widens as the interval widens the precision of the estimate decreases. One way to improve the precision of an estimate without decreasing the level of confidence is to increase the sample size. But how large a sample size is needed to guarantee a certain level of confidence for a given margin of error.

20. Find a Minimum sample size to estimate ц
Given a confidence level c and a margin of error E find the minimum sample size the population mean needed to estimate the population mean ц. If σ is unknown you can estimate it using s provided you have a preliminary sample with at least 30 members.

21. Example 6
Determining a Minimum sample size You want to estimate the mean number of sentences in a magazine advertisement. How many magazine advertisements must be in the sample to be 95% confident that the sample mean is within one sentence of the population mean?

22. Solution
Using c= 0.95 zc = 1.96 , σ≈ s ≈ 5.0 (from example 2) and E =1. You can solve for the Minimum sample size is
When necessary round up to the nearest whole number. So you could include at 97 magazine advertisements in your sample.

23. Try it Yourself
How many magazine advertisements must be included in the sample if you want to be 95% confident that the sample mean is within 2 sentences of the population mean
zc = 1.96, E=2, and s ≈ 5.0

24. Point Estimate DEFINITION:
A point estimate is a single value estimate for a population parameter. The best point estimate of the population mean
is the sample mean
An unbiased estimator is just as likely to overestimate as to underestimate the parameter it is estimating. The sample mean is an unbiased estimator of the population mean. From chapter 5 students know that the mean of the sample means is the population mean.

25. Example: Point Estimate
A random sample of 35 airfare prices (in dollars) for a one-way ticket from Atlanta to Chicago. Find a point estimate for the population mean, . 99
101
107
102
109 98
105
103
101
105 98
107
104 96
105 95 98 94
100
104
111
114 87
104
108
101 87
103
106
117 94
103
101
105 90
The sample mean is
Sometimes students will work from raw data and other times summary statistics. The raw data are presented here but summary statistics are given in the interest of time. After raw data is given, ask students what single number could be used to estimate the population mean.
The point estimate for the price of all one way tickets from Atlanta to Chicago is $

26. Interval Estimates • • Point estimate
101.77
An interval estimate is an interval or range of values used to estimate a population parameter. ( ) •
101.77
The chances the the archer hits the exact point center are virtually 0. The probability he hits within the center ring is his level of confidence for that ring. Discuss as the rings widen, the level of confidence goes up but accuracy is sacrificed.
The level of confidence, x, is the probability that the interval estimate contains the population parameter.

27. Distribution of Sample Means
When the sample size is at least 30, the sampling distribution for is normal.
Sampling distribution of
For c = 0.95
0.95
0.025
0.025
Have students calculate other critical numbers for other values of c.
For c = 90%, z.90= 1.645
For c = 99%, z.99 =1.575 z
-1.96
1.96
95% of all sample means will have standard scores between z = and z = 1.96

28. Maximum Error of Estimate
The maximum error of estimate E is the greatest possible distance between the point estimate and the value of the parameter it is, estimating for a given
level of confidence, c.
When n  30, the sample standard deviation, s, can be used for
Find E, the maximum error of estimate for the one-way plane fare from Atlanta to Chicago for a 95% level of confidence given s = 6.69.
Calculate the error of estimate for c = .90 and c= .95. Have students compare the widths of the intervals at various levels of confidence.
Using zc = 1.96, s = 6.69, and n = 35,
You are 95% confident that the maximum error of estimate is $2.22.

29. Confidence Intervals for
Definition: A c-confidence interval for the population mean is
Find the 95% confidence interval for the one-way plane fare from Atlanta to Chicago.
You found = and E = 2.22
Left endpoint
Right endpoint
Encourage students to use a number line to form confidence intervals. •
101.77 ( )
103.99
99.55
With 95% confidence, you can say the mean one-way fare from Atlanta to Chicago is between $99.55 and $

30. Sample Size
Given a c-confidence level and an maximum error of estimate, E, the minimum sample size n, needed to estimate , the population mean is
You want to estimate the mean one-way fare from Atlanta to Chicago. How many fares must be included in your sample if you want to be 95% confident that the sample mean is within $2 of the population mean?
Rework the problem with different values of E. Compare. What effect does increasing E, the maximum tolerance have on the minimum required sample?
You should include at least 43 fares in your sample. Since you already have 35, you need 8 more.

31. Homework 1-22 Page 287 Day2: 23-30 all, 31-51 odd pgs 288-290
Answers: 1.) – interval estimate, 3.) (d) 99% , 5.) c=.80 , z=1.28 , (7.) c= .75 z=1.15 , (9.) – 0.47, (11.) – 1.76, (13.) , (15.) – 0.685,
(17.) – 0.197, (19.) – c
(20.) - d