### Chapter 6. Parameter Estimation and Confidence Intervals: Estimation

### Confidence Interval for the Population Mean

A confidence interval for a population mean #\mu# is a range of values, based on sample data, which are highly plausible candidates for the true value of the population mean.

To construct a confidence interval for a population mean #\mu#, we will need to make use of the *sampling distribution of the sample mean*.

Remember that the sample mean #\bar{X}# (approximately) follows the #N\bigg(\mu, \cfrac{\sigma}{\sqrt{n}}\,\bigg)# distribution if at least one of the following conditions is satisfied:

- The population from which the sample is drawn is normally distributed
- The sample is large enough for the
*Central Limit Theorem*to apply: #n\geq 30#

The width of a confidence interval is determined by the *margin **of error*.

#\phantom{0}#

Margin of Error

The **margin of error **#(ME)# of a confidence interval for the population mean #\mu# is the distance from the center of the interval #\bar{X}# to either the lower bound #L# or the upper bound #U#.

To calculate the margin of error of a confidence interval for the population mean #\mu#, use the following formula:

\[\begin{array}{rcccl}ME &=& z^* \cdot \sigma_{\bar{X}} &=& z^* \cdot \cfrac{\sigma}{\sqrt{n}}\end{array}\]

Where* *#z^*# is the **critical value **of the *Standard Normal Distribution* such that:

\[\mathbb{P}(-z^* \leq Z \leq z^*) = \cfrac{C}{100}\]

Calculating z* with Statistical Software

Let #C# be the *confidence level *in #\%#.

To calculate the *critical value* #z^*# in Excel, make use of the function **NORM.INV()**:

\[=\text{NORM.INV}((100+C)/200, 0, 1)\]

To calculate the *critical value* #z^*# in R, make use of the function **qnorm()**:

\[\text{qnorm}(p=(100+C)/200, mean=0, sd=1,lower.tail = \text{TRUE})\]

Factors that Influence the Margin of Error

The margin of error of a confidence interval for the population mean #\mu# is dependent on #3# factors: the confidence level, the population standard deviation, and the sample size.

- As the confidence level increases, the margin of error increases and the confidence interval becomes wider.
- As the population standard deviation increases, the margin of error increases and the confidence interval becomes wider.
- As the sample size increases, the margin of error decreases and the confidence interval becomes narrower.

A researcher randomly selects #72# men from this age group and finds a mean BP of #\bar{X}=127.38#.

Calculate the

*margin of error*of the #93\%# confidence interval for the population mean #\mu#. Round your answer to #3# decimal places.

#ME=3.417#

There are a number of different ways we can calculate the *margin of error*. Click on one of the panels to toggle a specific solution.

The *margin of error *of a confidence interval for a population mean #\mu# is calculated with the following formula:

\[ME=z^* \cdot \sigma_{\bar{X}}\]

Since the population from which the sample is drawn is *normally *distributed, we know that the *sampling distribution of the sample mean* is the #N(\mu, \sigma / \sqrt{n})# distribution.

Determine the *standard error of the mean* #\sigma_{\bar{X}}#:

\[\sigma_{\bar{X}}=\cfrac{\sigma}{\sqrt{n}}=\cfrac{16}{\sqrt{72}}=1.88562\]

For a given *confidence level *#C#, the *critical value* #z^*# of the standard normal distribution is the value such that #\mathbb{P}(-z^* \leq Z \leq z^*)=\cfrac{C}{100}#.

To calculate this critical value #z^*# in Excel, make use of the following function:

NORM.INV(probability, mean, standard_dev)

probability: A probability corresponding to the normal distribution.mean: The mean of the distribution.standard_dev: The standard deviation of the distribution.

Here, we have #C=93#. Thus, to calculate #z^*# such that #\mathbb{P}(-z^* \leq Z \leq z^*)=0.93#, run the following command:

\[\begin{array}{c}

=\text{NORM.INV}((100+C)/200, 0, 1)\\

\downarrow\\

=\text{NORM.INV}(193/200, 0, 1)

\end{array}\]

This gives:

\[z^* = 1.81191\]

With this information, the *margin of error *can be calculated:

\[ME=z^* \cdot \sigma_{\bar{X}} = 1.81191 \cdot 1.88562 = 3.417\]

The *margin of error *of a confidence interval for the population mean #\mu# is calculated with the following formula:

\[ME=z^* \cdot \sigma_{\bar{X}}\]

Since the population from which the sample is drawn is *normally *distributed, we know that the *sampling distribution of the sample mean* is the #N(\mu, \sigma / \sqrt{n})# distribution.

Determine the *standard error of the mean* #\sigma_{\bar{X}}#:

\[\sigma_{\bar{X}}=\cfrac{\sigma}{\sqrt{n}}=\cfrac{16}{\sqrt{72}}=1.88562\]

For a given *confidence level *#C#, the *critical value* #z^*# of the standard normal distribution is the value such that #\mathbb{P}(-z^* \leq Z \leq z^*)=\cfrac{C}{100}#.

To calculate this critical value #z^*# in R, make use of the following function:

qnorm(p, mean, sd, lower.tail)

p: A probability corresponding to the normal distribution.mean: The mean of the distribution.sd: The standard deviation of the distribution.lower.tail: If TRUE (default), probabilities are #\mathbb{P}(X \leq x)#, otherwise, #\mathbb{P}(X \gt x)#.

Here, we have #C=93#. Thus, to calculate #z^*#such that #\mathbb{P}(-z^* \leq Z \leq z^*)=0.93#, run the following command:

\[\begin{array}{c}

\text{qnorm}(p = (100+C)/200, mean = 0, sd = 1, lower.tail = \text{TRUE})\\

\downarrow\\

\text{qnorm}(p =193/200, mean = 0, sd = 1, lower.tail = \text{TRUE})

\end{array}\]

This gives:

\[z^* = 1.81191\]

With this information, the *margin of error *can be calculated:

\[ME=z^* \cdot \sigma_{\bar{X}} = 1.81191 \cdot 1.88562 = 3.417\]

#\phantom{0}#

General Formula for a Confidence Interval for a Population Mean

Assuming the *sampling distribution of the sample mean *is (approximately) normal, the general formula for computing a #C\%# CI for the population mean #\mu#, based on a random sample of size #n#, is:

\[CI_{\mu}=\bigg(\bar{X} - z^*\cdot \cfrac{\sigma}{\sqrt{n}},\,\,\,\, \bar{X} + z^*\cdot \cfrac{\sigma}{\sqrt{n}} \bigg)\]

*total*weight to be #2312#g when placed together on a scale.

Assume the standard deviation in weight for a single one-euro coin is #\sigma=0.0507#.

Construct a #98\%# confidence interval for the population mean #\mu#. Round your answers to #3# decimal places.

There are a number of different ways we can compute the *confidence interval*. Click on one of the panels to toggle a specific solution.

A sample size of #n=1000# is considered large enough for the *Central Limit Theorem *to apply.

This means that, although the sample in question comes from a population having an *unknown *distribution, the *sampling distribution* *of the sample* *mean* is approximately normal.

Assuming the *sampling distribution of the sample mean *is (approximately) normal, the general formula for computing a #C\%\, CI# for a population mean #\mu#, based on a random sample of size #n#, is:

\[CI_{\mu}=\bigg(\bar{X} - z^*\cdot \cfrac{\sigma}{\sqrt{n}},\,\,\,\, \bar{X} + z^*\cdot \cfrac{\sigma}{\sqrt{n}} \bigg)\]

For a given *confidence level *#C#, the *critical value* #z^*# of the standard normal distribution is the value such that #\mathbb{P}(-z^* \leq Z \leq z^*)=\cfrac{C}{100}#.

To calculate this critical value #z^*# in Excel, make use of the following function:

NORM.INV(probability, mean, standard_dev)

probability: A probability corresponding to the normal distribution.mean: The mean of the distribution.standard_dev: The standard deviation of the distribution.

Here, we have #C=98#. Thus, to calculate #z^*# such that #\mathbb{P}(-z^* \leq Z \leq z^*)=0.98#, run the following command:

\[\begin{array}{c}

=\text{NORM.INV}((100+C)/200, 0, 1)\\

\downarrow\\

=\text{NORM.INV}(198/200, 0, 1)

\end{array}\]

This gives:

\[z^* = 2.32635\]

Calculate the lower bound #L# of the confidence interval:

\[L = \bar{X} - z^* \cdot \cfrac{\sigma}{\sqrt{n}} = 2.312 - 2.32635 \cdot \cfrac{0.0507}{\sqrt{1000}}=2.308\]

Calculate the lower bound #U# of the confidence interval:

\[U = \bar{X} + z^* \cdot \cfrac{\sigma}{\sqrt{n}} = 2.312 + 2.32635 \cdot \cfrac{0.0507}{\sqrt{1000}}=2.316\]

Thus, the #98\%# confidence interval for the population mean #\mu# is:

\[CI_{\mu,\,98\%}=(2.308,\,\,\, 2.316)\]

A sample size of #n=1000# is considered large enough for the *Central Limit Theorem *to apply.

This means that, although the sample in question comes from a population having an *unknown *distribution, the *sampling distribution* *of the sample* *mean* is approximately normal.

Assuming the *sampling distribution of the sample mean *is (approximately) normal, the general formula for computing a #C\%\, CI# for a population mean #\mu#, based on a random sample of size #n#, is:

\[CI_{\mu}=\bigg(\bar{X} - z^*\cdot \cfrac{\sigma}{\sqrt{n}},\,\,\,\, \bar{X} + z^*\cdot \cfrac{\sigma}{\sqrt{n}} \bigg)\]

For a given *confidence level *#C#, the *critical value* #z^*# of the standard normal distribution is the value such that #\mathbb{P}(-z^* \leq Z \leq z^*)=\cfrac{C}{100}#.

To calculate this critical value #z^*# in R, make use of the following function:

qnorm(p, mean, sd, lower.tail)

p: A probability corresponding to the normal distribution.mean: The mean of the distribution.sd: The standard deviation of the distribution.lower.tail: If TRUE (default), probabilities are #\mathbb{P}(X \leq x)#, otherwise, #\mathbb{P}(X \gt x)#.

Here, we have #C=98#. Thus, to calculate #z^*#such that #\mathbb{P}(-z^* \leq Z \leq z^*)=0.98#, run the following command:

\[\begin{array}{c}

\text{qnorm}(p = (100+C)/200, mean = 0, sd = 1, lower.tail = \text{TRUE})\\

\downarrow\\

\text{qnorm}(p =198/200, mean = 0, sd = 1, lower.tail = \text{TRUE})

\end{array}\]

This gives:

\[z^* = 2.32635\]

Calculate the lower bound #L# of the confidence interval:

\[L = \bar{X} - z^* \cdot \cfrac{\sigma}{\sqrt{n}} = 2.312 - 2.32635 \cdot \cfrac{0.0507}{\sqrt{1000}}=2.308\]

Calculate the lower bound #U# of the confidence interval:

\[U = \bar{X} + z^* \cdot \cfrac{\sigma}{\sqrt{n}} = 2.312 + 2.32635 \cdot \cfrac{0.0507}{\sqrt{1000}}=2.316\]

Thus, the #98\%# confidence interval for the population mean #\mu# is:

\[CI_{\mu,\,98\%}=(2.308,\,\,\, 2.316)\]

#\phantom{0}#

Controlling the Margin of Error

Suppose you would like the *margin of error *for a #C\%# confidence interval for the population mean #\mu# to be no larger than #k#.

Then the minimum sample size required is

\[n=\Big(\cfrac{z^* \cdot \sigma}{k}\Big)^2,\]

rounded up to the next whole number.

If the researcher wants the

*margin of error*of the #99\%# confidence interval for the population mean #\mu# to be no larger than #1#, what is the minimum sample size he needs?

#n \geq 1122#

There are a number of different ways we can calculate the *minimum* *sample size*. Click on one of the panels to toggle a specific solution.

For a given *confidence level *#C#, the *critical value* #z^*# of the standard normal distribution is the value such that #\mathbb{P}(-z^* \leq Z \leq z^*)=\cfrac{C}{100}#.

To calculate this critical value #z^*# in Excel, make use of the following function:

NORM.INV(probability, mean, standard_dev)

probability: A probability corresponding to the normal distribution.mean: The mean of the distribution.standard_dev: The standard deviation of the distribution.

Here, we have #C=99#. Thus, to calculate #z^*# such that #\mathbb{P}(-z^* \leq Z \leq z^*)=0.99#, run the following command:

\[\begin{array}{c}

=\text{NORM.INV}((100+C)/200, 0, 1)\\

\downarrow\\

=\text{NORM.INV}(199/200, 0, 1)

\end{array}\]

This gives:

\[z^* = 2.57583\]

With this information, the *minimum sample size *can be calculated:

\[n=\Big(\cfrac{z^* \cdot \sigma}{k}\Big)^2=\Big(\cfrac{2.57583 \cdot 13}{1}\Big)^2=1121.2975\]

Rounding this value up gives #n=1122#.

Thus, to obtain a *margin of error *no larger than #1#, you need a sample size of at least #1122#.

For a given *confidence level *#C#, the *critical value* #z^*# of the standard normal distribution is the value such that #\mathbb{P}(-z^* \leq Z \leq z^*)=\cfrac{C}{100}#.

To calculate this critical value #z^*# in R, make use of the following function:

qnorm(p, mean, sd, lower.tail)

p: A probability corresponding to the normal distribution.mean: The mean of the distribution.sd: The standard deviation of the distribution.lower.tail: If TRUE (default), probabilities are #\mathbb{P}(X \leq x)#, otherwise, #\mathbb{P}(X \gt x)#.

Here, we have #C=99#. Thus, to calculate #z^*#such that #\mathbb{P}(-z^* \leq Z \leq z^*)=0.99#, run the following command:

\[\begin{array}{c}

\text{qnorm}(p = (100+C)/200, mean = 0, sd = 1, lower.tail = \text{TRUE})\\

\downarrow\\

\text{qnorm}(p =199/200, mean = 0, sd = 1, lower.tail = \text{TRUE})

\end{array}\]

This gives:

\[z^* = 2.57583\]

With this information, the *minimum sample size *can be calculated:

\[n=\Big(\cfrac{z^* \cdot \sigma}{k}\Big)^2=\Big(\cfrac{2.57583 \cdot 13}{1}\Big)^2=1121.2975\]

Rounding this value up gives #n=1122#.

Thus, to obtain a *margin of error *no larger than #1#, you need a sample size of at least #1122#.

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