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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

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

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-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

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

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

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.

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.

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

Concept

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.

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

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

Concept

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.

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

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

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)

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

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

Concept

Concept

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.

Example 4

One-Sided Hypothesis Test

Step 3 Calculate the z-statistic for the sample

data.

Example 4

One-Sided Hypothesis Test

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.

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.

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.

Example 5

Two-Sided Hypothesis Test

Step 3 Calculate the z-statistic for the sample

data.

Example 5

Two-Sided Hypothesis Test

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.

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.

End of the Lesson