What is the best point estimate of the population mean

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Chapter 8: Confidence Intervals

Module Objectives:

Estimate Population Means with known Standard Deviation
Learn about the Student's t-distribution
Estimate Population Means with unknown Standard Deviation
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Chapter 8

Confidence Intervals

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Point Estimate – a single number estimate of a population parameter. The best point estimate of a population mean is the sample mean.
Interval Estimate – a range of possible values for the population parameter.
Level of Confidence, c – the degree of certainty that the interval estimate contains the population parameter.
Confidence Interval – an interval estimate associated with a certain level of confidence.
Margin of Error, E – the largest possible distance from the point estimate that a confidence interval will cover.
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Definitions:

Confidence Intervals

8.1 Introduction to Estimating Population Means

Find the best point estimate for the population mean of test scores on a standardized biology final exam. The following is a simple random sample taken from these test scores.

Find the best point estimate:

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The best point estimate of a population mean is the sample mean.

Solution:

Confidence Intervals

8.1 Introduction to Estimating Population Means

45 68 72 91 100 71
69 83 86 55 89 97
76 68 92 75 84 70
81 90 85 74 88 99
76 91 93 85 96 100
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Confidence Interval for Population Means:

Confidence Intervals

8.1 Introduction to Estimating Population Means

A college student researching study habits collects data from a random sample of 250 college students on campus and calculates that the sample mean is 15.7 hours per week. If the margin of error for the data using a 95% level of confidence is 2.2 hours, construct a 95% confidence interval for the data.

Construct the confidence interval:

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Lower endpoint:

15.7 - 2.2 = 13.5 hours

Upper endpoint:

15.7 + 2.2 = 17.9 hours

Solution:

13.5 <  < 17.9

Confidence Intervals

8.1 Introduction to Estimating Population Means

How to choose z or t:

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Confidence Intervals

8.1 Introduction to Estimating Population Means

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Criteria for estimating the population mean for large samples:

Confidence Intervals

8.2 Estimating Population Means (Large Samples)

All possible samples of a given size have an equal probability of being chosen.
The size of the sample is at least 30 (n ≥ 30).
The population’s standard deviation is unknown.
When all of the above conditions are met, then the distribution used to calculate the margin of error for the population mean is the Student t-distribution.

However, when n ≥ 30, the critical values for the t-distribution are almost identical to the critical values for the normal distribution at corresponding levels of confidence.

Therefore, we can use the normal distribution to approximate the t-distribution.

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Find the critical value:

Confidence Intervals

8.2 Estimating Population Means (Large Samples)

Find the critical value for a 95% confidence interval.

To find the critical value, we first need to find the values for –z0.95 and z0.95.

Since 0.95 is the area between –z0.95 and z0.95, there will be 0.05 in the tails, or 0.025 in one tail.

Solution:

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Critical Value, zc:

Confidence Intervals

8.2 Estimating Population Means (Large Samples)

Critical z-Values for Confidence Intervals
Level of Confidence, c zc
0.80 1.28
0.85 1.44
0.90 1.645
0.95 1.96
0.98 2.33
0.99 2.575
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Margin of Error, E, for Large Samples:

Confidence Intervals

8.2 Estimating Population Means (Large Samples)

When calculating the margin of error, round to one more decimal place than the original data, or the same number of places as the standard deviation.

zc = the critical z-value

s = the sample standard deviation

n = the sample size

Find the margin of error for a 99% confidence interval, given a sample of size 100 with a sample standard deviation of 15.50.

Find the margin of error:

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n = 100, s = 15.50, c = 0.99

z0.99 =

Solution:

Confidence Intervals

8.2 Estimating Population Means (Large Samples)

2.575

A survey of 85 homeowners finds that they spend on average $67 a month on home maintenance with a standard deviation of $14. Find the 95% confidence interval for the mean amount spent on home maintenance by all homeowners.

Construct a confidence interval:

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c = 0.95, n = 85, s = 14, = 67

z0.95 =

Solution:

Confidence Intervals

8.2 Estimating Population Means (Large Samples)

1.96

$64.02 <  < $69.98

($64.02, $69.98)

67 – 2.98 <  < 67 + 2.98

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Finding the Minimum Sample Size for Means:

Confidence Intervals

8.2 Estimating Population Means (Large Samples)

When calculating the sample size, round to up to the next whole number.

zc = the critical z-value

= the population standard deviation

E = the margin of error

To find the minimum sample size necessary to estimate an average, use the following formula:

Determine the minimum sample size needed if you wish to be 99% confident that the sample mean is within two units of the population mean given, that  = 6.5. Assume that the population is normally distributed.

Find the minimum sample size:

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c = 0.99,  = 6.5, E = 2

z0.99 =

You will need a minimum sample size of 71.

Solution:

Confidence Intervals

8.2 Estimating Population Means (Large Samples)

2.575

The electric cooperative wishes to know the average household usage of electricity by its non-commercial customers. They believe that the mean is 15.7 kWh per day for each family with a variance of 3.24 kWh.

How large of a sample would be required in order to estimate the average number of kWh of electricity used daily per family at the 99% confidence level with an error of at most 0.12 kWh?

Find the minimum sample size:

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c = 0.99,  = 1.8, E = 0.12, z0.99 = 2.575

You will need a minimum sample size of 1492 families.

Solution:

Confidence Intervals

8.2 Estimating Population Means (Large Samples)

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Criteria for estimating the population mean for small samples:

All possible samples of a given size have an equal probability of being chosen.
The size of the sample is less than 30 (n < 30).
The distribution of the population is approximately normal.
The population’s standard deviation is unknown.
Confidence Intervals

8.3 Estimating Population Means (Small Samples)

When all of the above conditions are met, then the distribution used to calculate the margin of error for the population mean is the Student t-distribution.

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Margin of Error, E, for Small Samples:

Confidence Intervals

8.3 Estimating Population Means (Small Samples)

When calculating the margin of error, round to one more decimal place than the original data, or the same number of places as the standard deviation.

ta/2 = the critical t-value

a = 1 – c

s = the sample standard deviation

n = the sample size

Assuming the population is approximately normal and all samples have an equal probability of being chosen, find the margin of error for a sample size of 10 given that s = 15.5 and the level of confidence is 95%.

Find the margin of error:

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n = 10, s = 15.5, c = 0.95, a = 1 – 0.95 = 0.05

t0.05/2 = t0.025 =

Solution:

Confidence Intervals

8.3 Estimating Population Means (Small Samples)

2.262

A student records the repair cost for 20 randomly selected computers. A sample mean of $216.53 and a standard deviation of $15.86 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for all computers. Assume the criteria for this section are met.

Construct a confidence interval:

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n = 20, = 216.53, s = 15.86, c = 0.98, a = 1 – 0.98 = 0.02

t0.02/2 = t0.01 =

Solution:

Confidence Intervals

8.3 Estimating Population Means (Small Samples)

2.539

$207.53 <  < $225.53

($207.53, $225.53)

216.53 – 9.00 <  < 216.53+ 9.00

Population proportion (p) – the percentage of a population that has a certain characteristic.
Sample proportion ( ) – the percentage of a sample that has a certain characteristic.
Point Estimate – the best point estimate of a population proportion is the sample proportion.
Margin of Error, E, – the largest possible distance from the point estimate that a confidence interval will cover.
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Definitions:

Confidence Intervals

8.4 Estimating Population Proportions

When calculating a proportion, round to three decimal places.

A graduate student wishes to know the proportion of American adults who speak two or more languages. He surveys 565 American adults and finds that 226 speak two or more languages. Estimate the proportion of all American adults that speak two or more languages.

Find the best point estimate:

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The best point estimate of a population proportion is the sample proportion.

Solution:

Confidence Intervals

8.4 Estimating Population Proportions

= 0.400

We estimate the population proportion to be 40.0%.

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Margin of Error, E, for Proportions:

zc = the critical z-value

= the sample proportion

n = the sample size

When calculating the margin of error for proportions, round to three decimal places.

Confidence Intervals

8.4 Estimating Population Proportions

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Critical Value, zc:

Confidence Intervals

8.4 Estimating Population Proportions

Critical z-Values for Confidence Intervals
Level of Confidence, c zc
0.80 1.28
0.85 1.44
0.90 1.645
0.95 1.96
0.98 2.33
0.99 2.575
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Confidence Interval for Population Means:

Confidence Intervals

8.4 Estimating Population Proportions

A survey of 200 computer chips is obtained and 192 are found to not be defective. Find the 99% confidence interval for the percentage of all computer chips that are defective.

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Confidence Intervals

8.4 Estimating Population Proportions

Construct a confidence interval:

c = 0.99, n = 200,

z0.99 =

Solution:

2.575

0.004 < p < 0.076

(0.4%, 7.6%)

0.04 – 0.036 < p < 0.04 + 0.036

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Finding the Minimum Sample Size for Means:

When calculating the sample size, round to up to the next whole number.

zc = the critical z-value

= the population proportion

E = the margin of error

To find the minimum sample size necessary to estimate an average, use the following formula:

Confidence Intervals

8.4 Estimating Population Proportions

The FBI wants to determine the effectiveness of their 10 Most Wanted List. To do so, they need to find out the fraction of people who appear on the list that are actually caught. They have estimated the fraction to be about 0.31.

How large of a sample would be required in order to estimate the fraction of people who are captured after appearing on the list at the 85% confidence level with an error of at most 0.04?

Find the minimum sample size:

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= 0.31, c = 0.85, E = 0.04, z0.99 = 2.575

You will need a minimum sample size of 278 people.

Solution:

Confidence Intervals

8.4 Estimating Population Proportions

Point Estimate – a single number estimate of a population parameter. The best point estimate of a population variance is the sample variance. Likewise, the best point estimate of a population standard deviation is the sample standard deviation .
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Definitions:

Confidence Intervals

8.5 Estimating Population Variance

When calculating the variance and standard deviation, round to one more decimal place than given in the original data.

General Auto is testing the variance in the length of its windshield wiper blades. A sample of 12 windshield wiper blades is randomly selected, and the following lengths are measured:

Find the best point estimate:

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Give the best point estimate for the population variance.

0.05

Give the best point estimate for the population standard deviation.

0.22

Confidence Intervals

8.5 Estimating Population Variance

22.1 22.0 22.1 22.4 22.3 22.5
22.3 22.1 22.2 22.6 22.5 22.7
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Confidence Interval for Population Variance:

Confidence Intervals

8.5 Estimating Population Variance

Confidence Interval for Population Standard Deviation:

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The steps for creating a confidence interval for a population variance (or standard deviation) are as follows:

Confidence Intervals

8.5 Estimating Population Variance

From the sample data, calculate s2.

Based on the level of confidence given, calculate and 1 – .

Use the 2 distribution table to find the critical values for and 1 – with n – 1 degrees of freedom.

Substitute the necessary values into the confidence interval formula above.

A commercial bakery is testing the variance in the weights of the cookies it produces. A sample of 15 cookies is randomly chosen and found out to have a variance of 3.4 grams. Build a 95% confidence interval for the variance of the weights of all the cookies produced by the company.

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Confidence Intervals

8.5 Estimating Population Variance

Construct a confidence interval:

s2 = 3.4, c = 0.95, a = 1 – 0.95 = 0.05,

d.f. = 14 so

Solution:

0.025 and

1.8 < s 2 < 8.5

0.975

26.119 and

5.629

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Finding the Minimum Sample Size:

Confidence Intervals

8.5 Estimating Population Variance

Minimum Sample Sizes at 95% Confidence
s2 is within the percentage of the value s 2 Minimum value of n needed s is within the percentage of the value s Minimum value of n needed
1% 77,209 1% 19,206
5% 3150 5% 769
10% 807 10% 193
20% 212 20% 49
30% 99 30% 22
40% 58 40% 13
50% 39 50% 9
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Finding the Minimum Sample Size:

Confidence Intervals

8.5 Estimating Population Variance

Minimum Sample Sizes at 99% Confidence
s2 is within the percentage of the value s 2 Minimum value of n needed s is within the percentage of the value s Minimum value of n needed
1% 133,450 1% 33,220
5% 5459 5% 1337
10% 1403 10% 337
20% 370 20% 86
30% 173 30% 39
40% 102 40% 23
50% 69 50% 15
A market research team wants to estimate the standard deviation of home prices in a metropolitan area in the Northeast. They need to be 99% confident that their results are within 5% of the true standard deviation. Assuming that the housing prices in that area are normally distributed, what is the minimum number of housing prices they must acquire?

Find the minimum sample size:

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According to the table, we see that for a 99% confidence of being within 5% of the true standard deviation, the minimum sample size is 1337.

Solution:

Confidence Intervals