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Chapter 7: Problem 43

According to a 2017 Gallup poll, \(80 \%\) of Americans report being afflicted by stress. Suppose a random sample of 1000 Americans is selected. a. What percentage of the sample would we expect to report being afflicted by stress? b. Verify that the conditions for the Central Limit Theorem are met. c. What is the standard error for this sample proportion? d. According to the Empirical Rule, there is a \(95 \%\) probability that the sample proportion will fall between what two values?

Short Answer

Expert verified
a) We would expect \(80\%\) of the sample to report being afflicted by stress. b) The conditions for the Central Limit Theorem are met because the sample size is more than 30 and the sampling is random. c) The standard error for this sample proportion is \( \sqrt{\frac{0.8(1-0.8)}{1000}} \). d) According to the Empirical Rule, there is a \(95\%\) probability that the sample proportion will fall within two standard deviations from the mean, between the two values calculated via \(\bar{x}\pm z*SE\).

Step by step solution

01

Calculating Expected Percentage

As per the Gallup poll, \(80 \%\) Americans report being afflicted by stress. This would be expected to carry over into a sample of 1000 as well, so you would expect \(80 \%\) of the 1000 people, or 800 people, to report being stressed.
02

Verifying Central Limit Theorem

The Central Limit Theorem can be applied if the sample size is large enough (usually at least 30) and the sampling is random. Both conditions apply in this case, where the sample size is 1000 and it's mentioned that a random sample was selected.
03

Calculating Standard Error

The standard error of a proportion is computed using the formula: \[ SE = \sqrt{ \frac{ p(1-p)}{n}} \] where 'p' indicates the proportion (in this case, 0.8) and 'n' indicates the sample size (in this case, 1000). Substituting these values in the formula, you would get the standard error as \(SE =\sqrt{\frac{0.8(1-0.8)}{1000}} \)
04

Applying the Empirical Rule

The Empirical Rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean. Here, we're asked for where \(95\%\) of the sample proportion will fall, which is within two standard deviations. To find these values, use the formula \(\bar{x}\pm z*SE\), where \(\bar{x}\) = sample mean, SE = standard error, and z = z score (1.96 for \(95\%\)).

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