300 CHAPTER 6 Confidence Intervals The difference between the point estimate and the actual parameter value is called the sampling error. When m is estimated, the sampling error is the difference x - m. In most cases, of course, m is unknown, and x varies from sample to sample. However, you can calculate a maximum value for the error when you know the level of confidence and the sampling distribution. Given a level of confidence c, the margin of error E (sometimes also called the maximum error of estimate or error tolerance) is the greatest possible distance between the point estimate and the value of the parameter it is estimating. For a population mean m where s is known, the margin of error is E = zcsx = zc s2 n Margin of error for m (s known) when these conditions are met. 1. The sample is random. 2. At least one of the following is true: The population is normally distributed or n Ú 30. (Recall from the Central Limit Theorem that when n Ú 30, the sampling distribution of sample means approximates a normal distribution.) DEFINITION Finding the Margin of Error Use the data in Example 1 and a 95% confidence level to find the margin of error for the mean number of hours spent per week on required athletic activities by all student-athletes in the conference. Assume the population standard deviation is 2.4 hours. SOLUTION Because s is known 1s = 2.42, the sample is random (see Example 1), and n = 40 Ú 30, use the formula for E given above. The z@score that corresponds to a 95% confidence level is 1.96. This implies that 95% of the area under the standard normal curve falls within 1.96 standard deviations of the mean, as shown in the figure below. 1You can approximate the distribution of the sample means with a normal curve by the Central Limit Theorem because n = 40 Ú 30.2 Using the values zc = 1.96, s = 2.4, and n = 40, E = zc s1 n = 1.96# 2.42 40 ≈ 0.7. Interpretation You are 95% confident that the margin of error for the population mean is about 0.7 hour. TRY IT YOURSELF 2 Use the data in Try It Yourself 1 and a 95% confidence level to find the margin of error for the mean number of hours spent per day on required athletic activities by all student-athletes in the conference. Assume the population standard deviation is 1.4 hours. Answer: Page A40 EXAMPLE 2 0.025 0.025 −zc = −1.96 zc = 1.96 z = 0 0.95 z Picturing the World A survey of a random sample of 1000 social media users found that the mean daily time spent on social media was 144 minutes. From previous studies, it is assumed that the population standard deviation is 21.4 minutes. (Adapted from TechJury) Frequency Minutes 200 172 144 116 88 x f 50 100 150 200 250 Daily Time Spent on Social Media For a 95% confidence interval, what would be the margin of error for the population mean daily time spent on social media?
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