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Shape of the Sampling Distribution of the Sample Proportion and the Central Limit Theorem for Beginning Statistics

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

Practice problems for these concepts can be found at:

When the sample size satisfies the inequalities np > 5 and nq > 5, the sampling distribution of the sample proportion is normally distributed with mean equal to p and standard error . This result is sometimes referred to as the central limit theorem for the sample proportion. This result is illustrated in Fig. 7-4.

EXAMPLE 7.20   Approximately 20% of the adults 25 and older have a bachelor's degree in the United States. If a large number of samples of adults 25 and older, each of size 100, were taken across the country, then the sample proportions having a bachelor's degree would vary from sample to sample. The distribution of sample proportions would be normally distributed with a mean equal to 20% and a standard error equal to = 4%. According to the empirical rule, approximately 68% of the sample proportions would fall between 16% and 24%, approximately 95% of the sample proportions would fall between 12% and 28%, and approximately 99.7% would fall between 8% and 32%. The sample proportion distribution may be assumed to be normally distributed since np = 20 and nq = 80 are both greater than 5.

Shape of the Sampling Distribution of the Sample Proportion and the Central Limit Theorem

Applications of the Sampling Distribution of the Sample Proportion

The theory underlying the sample proportion is utilized in numerous statistical applications. The margin of error, control chart limits, and many other useful statistical techniques make use of the sampling distribution theory connected with the sample proportion.

EXAMPLE 7.21   It is estimated that 42% of women and 36% of men ages 45 to 54 are overweight, that is, 20% over their desirable weight. The probability that one-half or more in a sample of 50 women ages 45 to 54 are overweight is expressed as P( ≥ .5). The distribution of the sample proportion is normal since np = 50 × .42 = 21 > 5 and nq = 50 × .58 = 29 > 5. The mean of the distribution is .42 and the standard error is = .07. The probability, P( ≥ .5), is shown as the shaded area in Fig. 7-5.

Applications of the Sampling Distribution of the Sample Proportion

To find the area under the normal curve shown in Fig. 7-5, it is necessary to transform the value .5 to a standard normal value. The value for z is z = = 1.14. The area to the right of = .5 is equal to the area to the right of z = 1.14 and is given as follows.

    P( ≥ .5) = P(z > 1.14) = .5 – .3729 = .1271

In Example 7.21, the value equal to .5 was transformed to a z value by subtracting the mean value of from .5 and then dividing the result by the standard error of the proportion. The equation for transforming a sample proportion value to a z value is given by

The solution to Example 7.21, using EXCEL, would look like the following.

The first three rows are EXCEL messages reiterating the Central Limit Theorem for the sample proportion. The fourth row contains a message, a computation, and a message explaining the computation. The function, =1-NORMDIST(0.5,0.42,0.07,1), asks EXCEL to find the area under the normal curve to the right of 0.5. The curve is centered at 0.42 and has standard deviation equal to 0.07. The 1 in the fourth position tells EXCEL to find the cumulative area or the area to the left of 0.5. The expression =1-NORMDIST(0.5,0.42,0.07,1) finds the area to the right of 0.5. See Figure 7-5. Note that the answers obtained from the table and those obtained from EXCEL differ a bit due to table interpolation error. The EXCEL answer is exact.

EXAMPLE 7.22   It is estimated that 1 out of 5 individuals over 65 have Parkinsonism, that is, signs of Parkinson's disease. The probability that 15% or less in a sample of 100 such individuals have Parkinsonism is represented as P( ≤ 15%). Since np = 100 × .20 = 20 > 5 and nq = 100 × .80 = 80 > 5, it may be assumed that has a normal distribution. The mean of is 20% and the standard error is 4%. Formula (7.10) is used to find an event involving z which is equivalent to the event ≤ 15. The event ≤ 15 is equivalent to the event z = , and therefore we have P( ≤ 15%) = P(z < – 1.25) = .5 – .3944 = .1056.

Practice problems for these concepts can be found at:

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