Section 8.2 Estimating Population Means (Large Samples) HAWKES LEARNING SYSTEMS math courseware...

Section 8.2

Estimating Population Means (Large Samples)

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σ unknown, but a large sample (n ≥ 30) or population is normally distributed

Large sample:Use z, the normal distrib’n.

Small sample:Use the Student t-distribution.

z t

TODAY



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.



Find the critical value:



(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:



Critical Value, zc:



(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



Critical Value, zc:



(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

TI-84 confirmation of these values using invNorm(area to left)= z

c = area in the middleSo area in each tail = (1 – c) / 2(1 – 0.90) / 2 = .05(1 – 0.95) / 2 = .025(1 – 0.99) / 2 = .005

Take absolute value of –z.Round Responsibly.

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Margin of Error, E, for Large Samples:



(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-values = the sample standard deviationn = the sample size

This E is the value thatwe will + or – from the sample mean to build a Confidence Interval

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:



n = 100, s = 15.50, c = 0.99

z0.99 =

Solution:



(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:



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

z0.95 =

Solution:



(Large Samples)

1.96

$64.02 < < $69.98

($64.02, $69.98)

67 – 2.98 < < 67 + 2.98

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:





(Large Samples)

Large sample, so we choose z.STAT, TESTS, ZintervalUse “Data” if data is in a List.We have “Stats” already given.We don’t know σ so use s.Sample mean and sample size.Confidence Level as a decimal.Highlight Calculate.Press ENTER.

TI-84 easy way

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TI-84 Solution, continued

Inputs (as described on the previous slide.)

Outputs: “We are 95% confident that the mean monthly maintenance cost is $64.02 < μ < $69.98”

What’s the smallest sample size I need to use?

• Confidence Interval scenario:– I want to estimate the mean of a population– It’s going to take time and cost money to collect a

sample.– If sample is too small, my result is not accurate

enough.,– If sample is too big, I’ve worked too hard and

spent too much for precision I didn’t need.• There’s a way to find it out in advance, for z.

(added content by D.R.S.)



Finding the Minimum Sample Size for Means:



(Large Samples)

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

number.

zc = the critical z-values = the population standard deviationE = the margin of error

To find the minimum sample size necessary to estimate an average,

use the following formula:

This applies when you qualify to use z, not t.

always bump it upExample: 80.01 bumps up to n = 81 sample size needed

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:



c = 0.99, = 6.5, E = 2

z0.99 =

You will need a minimum sample size of 71.

Solution:



(Large Samples)

2.575

You can find zc from printed tables or by

using TI-84 invNorm(, as described earlier.

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:



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

You will need a minimum sample size of 1492 families.

Solution:



(Large Samples)

Section 8.2 Estimating Population Means (Large Samples) HAWKES LEARNING SYSTEMS math courseware...

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