PDF Stats Review Chapter 9

[Pages:22]Stats Review Chapter 9

Mary Stangler Center for Academic Success

Revised 8/16

Note:

This review is meant to highlight basic concepts from the course. It does not cover all concepts presented by your instructor. Refer back to your notes, unit objectives, handouts, etc. to further prepare for your exam. The questions are displayed on one slide followed by the answers are displayed in red on the next. This review is available in alternate formats upon request.

Mary Stangler Center for Academic Success

Confidence Interval for Population Proportion In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the 90% confidence interval for the population proportion, p?

Mary Stangler Center for Academic Success

9.1

Confidence Interval for Population Proportion

In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the 90% confidence interval for the population proportion, p?

? Condition 1: n(.05)N

? The sample size (10) is less than 5% of the population (millions of musicians), so the condition is met.

? Condition 2: np(1-p)10

? = 2 = .2

10

? 1 - = 10 .2 met.

1 - .2

= 1.6 . This is less than 10 so this condition is not

It would not be practical to construct the confidence interval.

Mary Stangler Center for Academic Success

9.1

Confidence Interval for Population Proportion

A poll conducted found that 944 of 1748 adults do not believe that people with tattoos are more rebellious. If appropriate construct a 90% confidence interval.

Mary Stangler Center for Academic Success

9.1

Confidence Interval for Population Proportion

A poll conducted found that 944 of 1748 adults do not believe that people with tattoos are more rebellious. If appropriate construct a 90% confidence interval.

Is it appropriate? Yes, it satisfies both conditions.

1) point estimate

=

944 1748

=

.54

2) Find /2.

First

take

1-.90 2

=

.05

The corresponding z is -1.645. We will ignore the negative and just use 1.645.

3) Find the Margin of Error

= /2

(1-)

=

1.645

.54(1-.54) 1748

=

.0196

4) Take the point estimate and add/subtract the margin of error

.54-.0196=.5204, .54+.0196=.5596 The confidence interval is (.5204,.5596)

Mary Stangler Center for Academic Success

9.1

Sample Size Needed of Population Proportion A researcher wants to estimate the proportion of Americans that have sleep deprivation. How large a sample is needed in order to be 95% confident and within 3% if

a) the researcher used a previous estimate of 60%? b) the researcher doesn't use a previous estimate?

Mary Stangler Center for Academic Success

Sample Size Needed of Population Proportion

A researcher wants to estimate the proportion of Americans that have sleep deprivation. How large a sample is needed in order to be 95% confident and within 3% if

a) the researcher used a previous estimate of 60%?

= 1 -

2

2

= .60 1 - .60

1.96 2 .03 1024.4

b) the researcher doesn't use a previous estimate?

= .25

2

2

= .25

1.96 2 .03 1067.1

Remember to always round up when finding sample sizes.

Mary Stangler Center for Academic Success

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