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    In summary, as a result of the central limit theorem:

  • X is normally distributed, that is, X ~ N ( μ X , σ n ).
  • When the population standard deviation σ is known, we use a Normal distribution to calculate the margin of error.

Calculating the confidence interval:

To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. The steps to construct and interpret the confidence interval are:

  • Check the assumptions and conditions
  • Calculate the sample mean x from the sample data. Remember, in this section, we already know the population standard deviation σ .
  • Find the Z-score that corresponds to the confidence level.
  • Calculate the margin of error ME
  • Construct the confidence interval
  • Write a sentence that interprets the estimate in the context of the situation in the problem. (Explain what the confidence interval means, in the words of the problem.)

We will first examine each step in more detail, and then illustrate the process with some examples.

Finding z for the stated confidence level

When we know the population standard deviation σ, we use a standard normal distribution to calculate the margin of error ME and construct the confidence interval. We need to find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution Z~N(0,1).

This value can be found by using technology or the confidence levels found at the bottom of the t Table in the appendix of the text. Below is the part of the t Table needed to find the z value corresponding to the level of confidence asked for in a problem.

If asked to calculate a 95% confidence interval, replace the z in the confidence formula with 1.960. x + z σ n and x - z σ n

The confidence level, CL , is the area in the middle of the standard normal distribution. CL = 1 - α . So α is the area that is split equally between the two tails. Each of the tails contains an area equal to α 2 .

The z-score that has an area to the right of α 2 is denoted by z α 2

For example, when CL = 0.95 then α = 0.05 and α 2 = 0.025 ; we write z 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 α 2 = z 0.025 = 1.96 , using a calculator, computer or a Standard Normal probability table.

Me: margin of error

The margin of error formula for an unknown population mean μ when the population standard deviation σ is known is

  • ME = z α 2 σ n

    Constructing the confidence interval

  • The confidence interval estimate has the format ( x ME , x + ME ) .

The graph gives a picture of the entire situation.

CL + α 2 + α 2 = CL + α = 1 .

Normal distribution curve displaying the confidence interval formulas and corresponding area formulas.

Writing the interpretation

The interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated (here, a population mean ), the 'who' and the 'what' of the 5 W's, and should state the confidence interval (both endpoints). "We estimate with ___% confidence that the true population mean (include context of the problem) is between ___ and ___ (include appropriate units)."

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 mean score) of 68. Find a confidence interval estimate for the population mean exam score (the mean score on all exams).

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

  • Check the assumptions and conditions
  • The solution is shown step-by-step (Solution A).

Checking assumptions and conditions: Start by discussing the assumptions and conditions that support your model. 1) Randomization - Is the sample selected from a randomized

Solution a

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

  • x = 68
  • ME = z α 2 ( σ 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

α 2 = 0.05 z 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 α 2 = z .05 = 1.645

ME = 1.645 ( 3 36 ) = 0.8225

x - ME = 68 - 0.8225 = 67.1775

x + ME = 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 mean 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 ME.

  • x = 68
  • ME = z α 2 ( σ 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

α 2 = 0.025 z 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 α 2 = z .025 = 1.96

ME = 1.96 ( 3 36 ) = 0.98

x - ME = 68 - 0.98 = 67.02

x + ME = 68 + 0.98 = 68.98

Interpretation

We estimate with 95 % confidence that the true population mean 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 meanstatistics 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 margin of error, making the confidence interval wider.
  • Decreasing the confidence level decreases the margin or error, making the confidence interval narrower.

Changing the sample size:

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

Leave everything the same except the sample size. Use the original 90% confidence level. What happens to the margin of error 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
  • ME = z α 2 ( σ n )
  • σ = 3 ; The confidence level is 90% (CL=0.90) ; z α 2 = z .05 = 1.645

If we decrease the sample size n to 25, we increase the margin of error.

When n = 25 : ME = z α 2 ( σ n ) = 1.645 ( 3 25 ) = 0.987

    Summary: effect of changing the sample size

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

Practice Key Terms 3

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Source:  OpenStax, Collaborative statistics using spreadsheets. OpenStax CNX. Jan 05, 2016 Download for free at http://legacy.cnx.org/content/col11521/1.23
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