8.1 Confidence intervals for a population mean, population standard (Page 3/7)

Collaborative statistics: custom Page 3 / 7

Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of 3 points. A random sampleof 36 scores is taken and gives a sample mean (sample average score) of 68. Find a confidence interval estimate for the population mean exam score (the average score on all exams).

Find a 90% confidence interval for the true (population) mean of statistics exam scores.

The solution is shown step-by-step.

To find the confidence interval, you need the sample mean, $x$ , and the EBM.

$x = 68$
$EBM = z_{\frac{α}{2}} \cdot (\frac{σ}{\sqrt{n}})$
$σ = 3$ ; $n = 36$ ; The confidence level is 90% (CL=0.90)

$CL = 0.90$ so $α = 1 - CL = 1 - 0.90 = 0.10$

$\frac{α}{2} = 0.05 z_{\frac{α}{2}} = z_{.05}$

The area to the right of $z_{.05}$ is 0.05 and the area to the left of $z_{.05}$ is 1−0.05=0.95

$z_{\frac{α}{2}} = z_{.05} = 1.645$

using invnorm(.95,0,1) on the TI-83,83+,84+ calculators. This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the Standard Normal distribution.

$EBM = 1.645 \cdot (\frac{3}{\sqrt{36}}) = 0.8225$

$x - EBM = 68 - 0.8225 = 67.1775$

$x + EBM = 68 + 0.8225 = 68.8225$

The 90% confidence interval is (67.1775, 68.8225).

Interpretation

We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82.

Explanation of 90% confidence level

90% of all confidence intervals constructed in this way contain the true average statistics exam score. For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to contain the true population mean exam score.

Changing the confidence level or sample size

Changing the confidence level

Suppose we change the original problem by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score.

To find the confidence interval, you need the sample mean, $x$ , and the EBM.

$x = 68$
$EBM = z_{\frac{α}{2}} \cdot (\frac{σ}{\sqrt{n}})$
$σ = 3$ ; $n = 36$ ; The confidence level is 95% (CL=0.95)

$CL = 0.95$ so $α = 1 - CL = 1 - 0.95 = 0.05$

$\frac{α}{2} = 0.025 z_{\frac{α}{2}} = z_{.025}$

The area to the right of $z_{.025}$ is 0.025 and the area to the left of $z_{.025}$ is 1−0.025=0.975

$z_{\frac{α}{2}} = z_{.025} = 1.96$

using invnorm(.975,0,1) on the TI-83,83+,84+ calculators. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the Standard Normal distribution.)

$EBM = 1.96 \cdot (\frac{3}{\sqrt{36}}) = 0.98$

$x - EBM = 68 - 0.98 = 67.02$

$x + EBM = 68 + 0.98 = 68.98$

Interpretation

We estimate with 95 % confidence that the true population average for all statistics exam scores is between 67.02 and 68.98.

Explanation of 95% confidence level

95% of all confidence intervals constructed in this way contain the true value of the population averagestatistics exam score.

Comparing the results

The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider.

Normal distribution curve with 0.90 confidence interval area blocked off and corresponding residual areas.

Normal distribution curve with 0.95 confidence interval area blocked off and corresponding residual areas.

Summary: effect of changing the confidence level

Increasing the confidence level increases the error bound, making the confidence interval wider.
Decreasing the confidence level decreases the error bound, making the confidence interval narrower.

Changing the sample size:

Suppose we change the original problem to see what happens to the error bound if the sample size is changed.

Leave everything the same except the sample size. Use the original 90% confidence level. What happens to the error bound and the confidence interval if we increase the sample size and use n=100 instead of n=36? What happens if we decrease the sample size to n=25 instead of n=36?

$x = 68$
$EBM = z_{\frac{α}{2}} \cdot (\frac{σ}{\sqrt{n}})$
$σ = 3$ ; The confidence level is 90% (CL=0.90) ; $z_{\frac{α}{2}} = z_{.05} = 1.645$

If we decrease the sample size $n$ to 25, we increase the error bound.

When $n = 25$ : $EBM = z_{\frac{α}{2}} \cdot (\frac{σ}{\sqrt{n}}) = 1.645 \cdot (\frac{3}{\sqrt{25}}) = 0.987$

Summary: effect of changing the sample size

Increasing the sample size causes the error bound to decrease, making the confidence interval narrower.
Decreasing the sample size causes the error bound to increase, making the confidence interval wider.

Working backwards to find the error bound or sample mean

Working backwards to find the error bound or the sample mean

When we calculate a confidence interval, we find the sample mean and calculate the error bound and use them to calculate the confidence interval. But sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backwards to find both the error bound and the sample mean.

Finding the error bound

From the upper value for the interval, subtract the sample mean
OR, From the upper value for the interval, subtract the lower value. Then divide the difference by 2.

Finding the sample mean

Subtract the error bound from the upper value of the confidence interval
OR, Average the upper and lower endpoints of the confidence interval

Notice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know.

Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. We may know that the sample mean is 68. Or perhaps our source only gave the confidence interval and did not tell us the value of the the sample mean.

Calculate the error bound:

If we know that the sample mean is 68: $EBM = 68.82 - 68 = 0.82$
If we don't know the sample mean: $EBM = \frac{(68.82 - 67.18)}{2} = 0.82$

Calculate the sample mean:

If we know the error bound: $x = 68.82 - 0.82 = 68$
If we don't know the error bound: $x = \frac{(67.18 + 68.82)}{2} = 68$

Calculating the sample size n

If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.

The error bound formula for a population mean when the population standard deviation is known is $EBM = z_{\frac{α}{2}} \cdot (\frac{σ}{\sqrt{n}})$

The formula for sample size is $n z 2 σ 2 EBM 2$ , found by solving the error bound formula for $n$

In this formula, $z$ is $z_{\frac{α}{2}}$ , corresponding to the desired confidence level. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study.

The population standard deviation for the age of Foothill College students is 15 years. If we want to be 95% confident that the sample mean age is within 2 years of the true population mean age of Foothill College students , how many randomly selected Foothill College students must be surveyed?

From the problem, we know that $σ = 15$ and EBM=2
$z = z_{.025} = 1.96$ , because the confidence level is 95%.

$n z 2 σ 2 EBM 2$ = $1.96 2 152 22$ =216.09 using the sample size equation.
Use $n$ = 217: Always round the answer UP to the next higher integer to ensure that the sample size is large enough.

Therefore, 217 Foothill College students should be surveyed in order to be 95% confident that we are within 2 years of the true population age of Foothill College students.

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Source: OpenStax, Collaborative statistics: custom version modified by r. bloom. OpenStax CNX. Nov 15, 2010 Download for free at http://legacy.cnx.org/content/col10617/1.4

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