Download presentation
Presentation is loading. Please wait.
2
Practice For an SAT test = 500 = 100
What is the probability that a single person will have a SAT score of 550 or higher?
3
Step 1: Sketch out question
4
Step 2: Calculate the Z score
( ) / 100 = .50
5
Step 3: Look up Z score in Table
Z = .50; Column C =.3085 .3085
6
Practice There is a probability that any single person will have an SAT score of 550 or higher.
7
Practice For an SAT test = 500 = 100
What is the probability that a group of 10 people have a mean SAT score of 550 or higher?
8
Step 1: Sketch out question
9
Step 2: Calculate the standard error
100 / √10 = 31.62
10
Step 3: Calculate the Z score
( ) / = 1.58
11
Step 4: Look up Z score in Table
Z = 1.58; Column C =.0571 .0571
12
The next step Based on a sample of participants how accurate is a given sample mean at estimating the population mean? Example: Opinion polls
13
Confidence Intervals Using a sample of US citizens you ask if they approve of the president. You find 43% approve “The findings of the survey are accurate within 3 points, using a 95% level of confidence”
14
Confidence Intervals Using a sample of US citizens you ask if they approve of the president. You find 43% approve Thus you are 95% confident the population values falls between 40% to 46%
15
Confidence Intervals A recent survey found that on average a person each 10 pieces of fruit a week. The findings of the survey are accurate within 5 pieces of fruit, using a 95% level of confidence What does this mean?
16
Confidence Intervals A recent survey found that on average a person each 10 pieces of fruit a week. The findings of the survey are accurate within 5 pieces of fruit, using a 95% level of confidence 95% confident that on average a person eats between 5 and 15 pieces of fruit a week.
17
Note You can not use the Z formula to answer this type of question
You usually don’t know the or the If you did you wouldn’t need a sample! Better to use a “t distribution”
18
t distribution Different sample sizes have different t distributions
When using t distributions you must know the degrees of freedom (df) df = N - 1 Thus, the df can range from 1 to
19
t distribution
20
t distribution There are an infinite number of t distributions; however, once df is greater than 120 it doesn’t change much Table D page 392 (new book page 402) First Column are df For this chapter you will look at the top row
21
t distribution 95% 2.5% 2.5% -1.96 1.96 df =
22
t distribution 95% 2.5% 2.5% -1.96 1.96 df =
NOTICE: Similarities to the z table
23
t distribution 95% 2.5% 2.5% -2.26 2.26 df = 9
24
t distribution 95% 2.5% 2.5% -2.26 2.26 df = 9
25
t distribution 95% 2.5% 2.5% -4.30 4.30 df = 2
26
t distribution 95% 2.5% 2.5% -4.30 4.30 df = 2
27
Confidence Intervals Around a Mean
Inferential statistic Based on a sample you calculate a mean With a certain degree of confidence (usually 95%) you can state that the population mean is in the interval
28
Confidence Intervals To find the lower and upper limits:
LL = X - t (sx) UL = X + t(sx) Where: X is the mean of the sample t is a value from the t distribution table sx is the standard error of the mean, calculated from a sample
29
Confidence Intervals Note: Similar to the backwards z LL = X - t (sx)
UL = X + t(sx) X = + (z)()
30
Confidence Intervals To find the lower and upper limits:
LL = X - t (sx) UL = X + t(sx) Where: X is the mean of the sample t is a value from the t distribution table sx is the standard error of the mean, calculated from a sample
31
Confidence Intervals Sx = S / N
32
Remember S = -1
33
Example You are curious about how many hours a week students study. You sample 5 students: 20, 22, 19, 23, 17 What is the mean score? What are its 95% confidence intervals?
34
Step 1: Calculate S
35
Step 1: Calculate S NOTE: In other examples you may be given X and X2. Make sure you understand where they came from!
36
101 2.39 = 2063 5 5 - 1 N = 5 X = 101 X2 = 2063
37
Step 2: Calculate Sx Sx = S / N 1.07 = 2.39 / 5
38
Step 3: Look up t value df = N - 1 4 = 5 - 1 Want 95% confidence
39
Step 4: Calculate LL and UL
LL = X - t (sx) UL = X + t(sx) X = 20.2 Sx = 1.07 t = 2.776
40
Step 4: Calculate LL and UL
X = 20.2 Sx = 1.07 t = 2.776
41
Example Thus, you are 95% confident that the average student studies between and hours a week
42
Example Using the same data -- what if you want to be 99.9% confident.
43
Step 3: Look up t value df = N - 1 4 = 5 - 1 Want 99.9% confidence
44
Step 4: Calculate LL and UL
LL = X - t (sx) UL = X + t(sx) X = 20.2 Sx = 1.07 t = 8.610
45
Step 4: Calculate LL and UL
X = 20.2 Sx = 1.07 t = 2.776
46
Example Thus, you are 95% confident that the average student studies between and hours a week You are 99.9% confident that the average student studies between and hours a week. As you get more confident your intervals will get larger
47
Practice Old book Page 168 8.24 New book Page 172 8.23
48
Practice M = 1.20 S hat = .10 SE = .0577 t(2) = 31.598 LL = -.62
UL = 3.02 You can be 99.9% confident that the average burn rate of pots in this box is below 4 inches per minute.
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.