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It is important that the "standard deviation" used must be appropriate for the parameter we are estimating, so in this section we need to use the standard deviation that applies to sample means, which is σ n . The fraction σ n , is commonly called the "standard error of the mean" in order to distinguish clearly the standard deviation for a mean from the population standard deviation σ .

    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 error bound.

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:

  • 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 error bound EBM .
  • 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 the z -score for the stated confidence level

When we know the population standard deviation σ , we use a standard normal distribution to calculate the error bound EBM 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).

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, α = 0.05 and α 2 = 0.025; we write z α 2 = z 0.025 .

The area to the right of z 0.025 is 0.025 and the area to the left of z 0.025 is 1 – 0.025 = 0.975.

z α 2  =  z 0. 025  = 1 .96 , using a calculator, computer or a standard normal probability table.

invNorm (0.975, 0, 1) = 1.96

Note

Remember to use the area to the LEFT of z α 2 ; in this chapter the last two inputs in the invNorm command are 0, 1, because you are using a standard normal distribution Z ~ N (0, 1).

Calculating the error bound ( EBM )

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

  • EBM = ( z α 2 ) ( σ n )

Constructing the confidence interval

  • The confidence interval estimate has the format ( x ¯ E B M , x ¯ + E B M ) .

The graph gives a picture of the entire situation.

CL + α 2 + α 2 = CL + α = 1.

This is a normal distribution curve. The peak of the curve coincides with the point x-bar on the horizontal axis. The points x-bar - EBM and x-bar + EBM are labeled on the axis. Vertical lines are drawn from these points to the curve, and the region between the lines is shaded. The shaded region has area equal to 1 - a and represents the confidence level. Each unshaded tail has area a/2.

Writing the interpretation

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

Questions & Answers

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the Reply
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Lokesh Reply
7.The following data give thenumber of car thefts that occurred in a city in the past 12 days. 63711438726915 Calculate therange, variance, and standard deviation.
Mitu Reply
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Xx Reply
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Shazain Reply
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sharmin Reply
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Ashfat Reply
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Ashfat
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Ashfat
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Ashfat
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nazrul Reply
3. The following are the number of mails received in different days by different organizations: Days (x) : 23, 35, 38, 50, 34, 60, 41, 32, 53, 67. Number of mails (y) : 18, 40, 52, 45, 32, 55, 50, 48, 26, 25. i) Fit a regression line of y on x and test the significance of regression. ii) Estimate y
Atowar Reply
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Tamim Reply
Is the severity of the drug problem in high school the same for boys and girls? 85 boys and 70 girls were questioned and 34 of the boys and 14 of the girls admitted to having tried some sort of drug. What can be concluded at the 0.05 level?
Ashfat Reply
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Pratik
a quality control specialist took a random sample of n=10 pieces of gum and measured their thickness and found the mean 7.6 and standered deviation 0.10. Do you think that the mean thickness of the spearmint gum it produces is 7.5?
Shanto Reply
99. A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct? a
Niaz Reply
A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct?
Niaz
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Niaz
what is null Hypothesis
Niaz
when median is greater than mode?
Hafiza Reply
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Amaano
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Worthy
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Worthy
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Jungjoon
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Worthy
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Yoliswa
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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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