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Splash Screen

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Splash Screen Lesson Menu 5-Minute Check 1 5-Minute Check 2 5-Minute Check 3 5-Minute Check 4 5-Minute Check 5 CCSS Then/Now Vocabulary Concept Example 1 Example 1 ... – PowerPoint PPT presentation

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Title: Splash Screen


1
Splash Screen
2
Lesson Menu
Five-Minute Check (over Lesson 115) CCSS Then/Now
New Vocabulary Key Concept Maximum Error of
Estimate Example 1 Maximum Error of Estimate Key
Concept Confidence Interval for the Population
Mean Example 2 Real-World Example Confidence
Interval Example 3 Claims and Hypotheses Concept
Summary Tests of Significance Key Concept
Steps for Hypothesis Testing Example
4 Real-World Example One-Sided Hypothesis
Test Example 5 Two-Sided Hypothesis Test
3
5-Minute Check 1
A normal distribution has a mean of 79 and a
standard deviation of 13. Find the range of
values that represent the middle 68 of the
distribution.
A. 53 lt X lt 105 B. 66 lt X C. 66 lt X lt 92 D. X
lt 66 or X gt 92
4
5-Minute Check 2
A normal distribution has a mean of 79 and a
standard deviation of 13. Find the range of
values that represent the outside 5 of the
distribution.
A. X lt 40 or X gt 118 B. X lt 53 or X gt 105 C. 53
lt X lt 105 D. 40 lt X lt 118
5
5-Minute Check 3
A normal distribution has a mean of 79 and a
standard deviation of 13. What percent of the
data will be greater than 85?
A. 18 B. 32 C. 68 D. 82
6
5-Minute Check 4
JOBS For a particular company, the number of
applicants per position in a sample of 90
positions is normally distributed with a mean of
46.2 and a standard deviation of 8.3. How many
positions had at least 50 applicants?
A. about 54 B. about 49 C. about 45 D. about
29
7
5-Minute Check 5
LIGHT BULBS The life of a particular light bulb
is normally distributed with a mean of 96.3 hours
and a standard deviation of 7.3 hours. About what
percent of these light bulbs will last between 90
and 100 hours?
A. 19.4 B. 30.6 C. 50.0 D. 80.6
8
CCSS
Content Standards S.IC.1 Understand statistics as
a process for making inferences about population
parameters based on a random sample from that
population. S.IC.4 Use data from a sample survey
to estimate a population mean or proportion
develop a margin of error through the use of
simulation models for random sampling.
Mathematical Practices 7 Look for and make use
of structure.
9
Then/Now
You applied the standard normal distribution and
z-values.
  • Find confidence intervals for normally
    distributed data.
  • Perform hypothesis tests on normally distributed
    data.

10
Vocabulary
  • inferential statistics
  • alternative hypothesis
  • critical region
  • left-tailed test
  • two-tailed test
  • right-tailed test
  • statistical inference
  • confidence interval
  • maximum error of estimate
  • hypothesis test
  • null hypothesis

11
Concept
12
Example 1
Maximum Error of Estimate
ORAL HYGIENE A poll of 422 randomly selected
adults showed that they brushed their teeth an
average of 11.4 times per week with a standard
deviation of 1.6. Use a 99 confidence interval
to find the maximum error of estimate for the
number of times per week adults brush their teeth.
In a 99 confidence interval, 0.5 of the area
lies in each tail. The corresponding z-value is
2.576.
13
Example 1
Maximum Error of Estimate
This means that you can be 99 confident that the
number of times per week that adults brush their
teeth will be within 0.20 of the sample mean of
11.4 times.
Answer 0.20
14
Example 1
READING A poll of 385 randomly selected adults
showed that they read for recreation an average
of 39.3 minutes per week with a standard
deviation of 9.6 minutes. Use a 95 confidence
interval to find the maximum error of estimate
for the number of minutes per week adults read
for recreation.
A. 0.49 B. 0.80 C. 0.96 D. 1.26
15
Concept
16
Example 2
Confidence Interval
PACKAGING A poll of 156 randomly selected
members of a golf course showed that they play an
average of 4.6 times every summer with a standard
deviation of 1.1. Determine a 95 confidence
interval for the population mean.
17
Example 2
Confidence Interval
The 95 confidence interval is 4.43 µ 4.77.
Therefore, we are 95 confident that the average
number of times played per summer is between 4.43
and 4.77 times.
Answer 4.43 µ 4.77
18
Example 2
CAREERS A poll of 114 randomly selected members
of a large corporation showed that they stay an
average of 6.9 years with the company with a
standard deviation of 1.8 years. Determine a 90
confidence interval for the population mean.
A. 6.73 µ 7.07 B. 6.62 µ 7.18 C. 6.57 µ
7.23 D. 6.47 µ 7.33
19
Example 3
Claims and Hypotheses
Identify the null and alternative hypotheses for
the statement. Then identify which statement
represents the claim. A. Cindy thinks her
employees spend over 60 minutes per day surfing
the Internet at work.
over 60 minutes µ gt 60 not over 60 minutes µ
60 The claim is µ gt 60, and it is the alternative
hypothesis because it does not include equality.
The null hypothesis is µ 60, which is the
complement of µ gt 60.
Answer H0 µ 60 Ha µ gt 60 (claim)
20
Example 3
Claims and Hypotheses
Identify the null and alternative hypotheses for
the statement. Then identify which statement
represents the claim. B. Mr. Bacon thinks that he
tells at least five jokes every class period.
at least 5 jokes µ 5 not at least 5 jokes µ lt
5 The claim is µ 5, and it is the null
hypothesis because it includes equality. The
alternative hypothesis is µ lt 5, which is the
complement of µ 5.
Answer H0 µ 5 (claim) Ha µ lt 5
21
Example 3
Angelene thinks that the manuscript can be
produced in no more than 14 days. Identify the
null and alternative hypotheses. Then identify
which statement represents the claim.
A. H0 µ 14 (claim) Ha µ lt 14 B. H0 µ gt 14
(claim) Ha µ 14 C. H0 µ lt 14 (claim) Ha µ
14 D. H0 µ 14 (claim) Ha µ gt 14
22
Concept
23
Concept
24
Example 4
One-Sided Hypothesis Test
MAGAZINES Jenny thinks there are more than 10
pages of advertisements in every issue of a
particular magazine. Using a sample of 42
magazines, she calculated a mean of 10.5 pages
with a standard deviation of 1.6. Test the
hypothesis at 1 significance.
Step 1 State the claim and hypotheses. H0 µ
10 and Ha µ gt 10 (claim) Step 2 Determine the
critical region. The alternative hypothesis is µ
gt 10, so this is a right-tailed test. We are
testing at 1 significance, so we need to
identify the z-value that corresponds with the
upper 1 of the distribution.
25
Example 4
One-Sided Hypothesis Test
Step 3 Calculate the z-statistic for the sample
data.
26
Example 4
One-Sided Hypothesis Test
27
Example 4
One-Sided Hypothesis Test
Step 4 Decide whether to reject the null
hypothesis. H0 is not rejected because the
z-statistic for the sample does not fall within
the critical region.
Step 5 Make a conclusion about the claim. There
is not enough evidence to support Jennys claim
that there are more than 10 pages of
advertisements in every issue.
Answer H0 is not rejected. There is not enough
evidence to support Jennys claim that there are
more than 10 pages of advertisements in every
issue.
28
Example 4
TRAFFIC Kate thinks a certain traffic light is
usually red for at least 150 seconds. Using a
sample of 106 times at the intersection, she
calculated a mean of 153.9 seconds with a
standard deviation of 4.3 seconds. Test the
hypothesis at 5 significance.
A. There is not enough evidence to support Kates
claim. B. There is enough evidence to support
Kates claim.
29
Example 5
Two-Sided Hypothesis Test
CALORIES A label on a package of food states
that there are 120 Calories per serving. Using a
sample of 38 servings, Libby calculated a mean of
120.6 Calories with a standard deviation of 2.2.
Test the hypothesis at 5 significance.
Step 1 State the claim and hypotheses. H0 µ
120 (claim) and Ha µ ? 120 Step 2 Determine the
critical region. The alternative hypothesis is µ
? 100, so this is a two-tailed test. We are
testing at 5 significance, so we need to
identify the z-value that corresponds with upper
and lower 2.5 of the distribution.
30
Example 5
Two-Sided Hypothesis Test
Step 3 Calculate the z-statistic for the sample
data.
31
Example 5
Two-Sided Hypothesis Test
32
Example 5
Two-Sided Hypothesis Test
Step 4 Reject or fail to reject the null
hypothesis. H0 is not rejected because the
z-statistic for the sample does not fall within
the critical region.
Step 5 Make a conclusion about the claim.
Therefore, there is not enough evidence to
reject the claim that there are 12 Calories per
servicing.
Answer H0 is not rejected. Therefore, there is
not enough evidence to reject the claim that
there are 12 Calories per servicing.
33
Example 5
HAMBURGER A restaurant advertisement claims
there is exactly 0.25 pound of hamburger in every
cheeseburger. Using a sample of 67 cheeseburgers,
Julie calculated a mean of 0.246 pound with a
standard deviation of 0.03 pound. Test the
hypothesis at 10 significance.
A. There is not enough evidence to reject the
claim. B. There is enough evidence to reject the
claim.
34
End of the Lesson
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