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Confidence Interval for the Population Mean: Large Samples for Beginning Statistics

By — McGraw-Hill Professional
Updated on Aug 12, 2011

Practice problems for these concepts can be found at:

Estimation and Sample Size Determination Solved Problems for Beginning Statistics

According to the central limit theorem, the sample mean, x , has a normal distribution provided the sample size is 30 or more. Furthermore, the mean of the sample mean equals the mean of the population and the standard error of the mean equals the population standard deviation divided by the square root of the sample size. The variable given in formula below was shown to have a standard normal distribution provided n ≥ 30.

Since 95% of the area under the standard normal curve is between z = –1.96 and z = 1.96, and since the variable in formula (7.6) has a standard normal distribution, we have the result shown in formula (8.1).

The inequality, , is solved for μ and the result is given in formula (8.2).

The interval given in formula (8.2) is called a 95% confidence interval for the population mean, μ. The general form for the interval is shown in formula (8.3), where z represents the proper value from the standard normal distribution table as determined by the desired confidence level.

EXAMPLE 8.2 The mean age of policyholders at Mutual Insurance Company is estimated by sampling the records of 75 policyholders. The standard deviation of ages is known to equal 5.5 years and has not changed over the years. However, it is unknown if the mean age has remained constant. The mean age for the sample of 75 policyholders is 30.5 years. The standard error of the ages is years. In order to find a 90% confidence interval for μ, it is necessary to find the value of z in formula (8.3) for confidence level equal to 90%. If we let c be the correct value for z, then we are looking for that value of c which satisfies the equation P(–c < z < c) = .90. Or, because of the symmetry of the z curve, we are looking for that value of c which satisfies the equation P(0 < z < c) = .45. From the standard normal distribution table, we find P(0 < z < 1.64) = .4495 and P(0 < z < 1.65) = .4505. The interpolated value for c is 1.645, which we round to 1.65. Figure 8-1 illustrates the confidence level and the corresponding values of z. Now the 90% confidence interval is computed as follows. The lower limit of the interval is years, and the upper limit is . We are 90% confident that the mean age of all 250,000 policyholders is between 29.5 and 31.5 years. It is important to note that μ either is or is not between 29.5 and 31.5 years. To say we are 90% confident that the mean age of all policyholders is between 29.5 and 31.5 years means that if this study were conducted a large number of times and a confidence interval were computed each time, then 90% of all the possible confid ence intervals would contain the true value of μ.

Confidence Interval for the Population Mean: Large Samples

Since it is time consuming to determine the correct value of z in formula (8.3), the values for the most often used confidence levels are given in Table 8.1. They are found in the same manner as illustrated in Example 8.2.

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