Confidence Intervals for Population Means with Known Std Dev

5.1 Confidence Interval for Population Mean (Population Standard Deviation σ is Known) Suppose we want to know the mean (average) for a large population but have difficulty obtaining all the population data. We may conduct a survey to obtain sample data to estimate the population mean. Confidence level For the standard normal distribution, 95% confidence level corresponds the middle 95% area as follows: The total area is 100% or 1. Then the sum of the remaining area (two tails) is: α = 1 – 0.95 = 0.05. Hence Each tail area is α/2 = 0.025. In general, the middle area can be denoted as CL (confidence level). The two tail areas are denoted α/2:

Critical Values For CL = 95%, we can use the z-table to find the corresponding z-scores (or critical values) -1.96 and 1.96 as shown in the following: z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 : -1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 The often-used CL are 90%, 95% and 99%. Their critical values are as follows: z 0.05 (or 0.04) z 0.06 z 0.08 (or 0.07) -1.6 0.05 -1.9 0.025 -2.5 0.005 In general, we use -z α/2 and z α/2 to denote the critical values:

Confidence Interval Suppose a sample of size n is randomly selected from a normally distributed population, and the sample mean is x . If the population standard deviation σ is known, then the confidence interval for the population mean µ is between: x − z α / 2 σ √ n and x + z α / 2 σ √ n We can also combine the above formulas as x± z α / 2 σ √ n Example A polling company in a city wants to know the average exercise time for adults in the city. It conducts a survey of 1000 adults. The sample mean x = 26.9 minutes. Suppose the population standard deviation σ = 11.4 minutes. (a) What is the population? What is the sample? (b) Construct a 90% confidence interval for the population mean exercise time. Solution (a) The population is all the adults in the city. The sample is the 1000 adults. (b) CL = 90% α = 1 – 0.9 = 0.1 α/2 = 0.1/2 = 0.05 Apply the formula x± z α / 2 σ √ n with x = 26.9, σ = 11.4, and z α/2 = 1.65: x − z α / 2 σ √ n = 26.9 − 1.65 × 11.4 √ 1000 = 26.9 − 0.59 = 26.31 x + z α / 2 σ √ n = 26.9 + 1.65 × 11.4 √ 1000 = 26.9 + 0.59 = 27.49 We may use one of following to explain the result (required):  The 90% confidence interval is: _________(___________|_____________)_________ 26.31 26.9 27.49 Or

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 The population mean is between 26.31 minutes and 27.49 minutes. Example A survey wants to estimate the average daily social media time for people ages 16-25. A sample of 1500 people ages 16-25 are surveyed. The sample mean is 6.6 hours. Suppose the population standard deviation σ = 2.1 hours. (a) What is the population? What is the sample? (b) Construct a 95% confidence interval for the population mean. Solution (a) The population is all the people ages 16-25. The sample is the 1500 people surveyed. (b) CL = 95% = 0.95 α = 1 – 0.95 = 0.05 α/2 = 0.05/2 = 0.025 x − z α / 2 σ √ n = 6.6 − 1.96 × 2.1 √ 1500 = 6.6 − 0.11 = 6.49 x + z α / 2 σ √ n = 6.6 + 1.96 × 2.1 √ 1500 = 6.6 + 0.11 = 6.71  The 95% confidence interval is: _________(___________|_____________)_________ 6.49 6.6 6.71 Or  The population mean is between 6.49 hours and 6.71 hours. 0.95 0.025 0.025 − 1.96 1.96

5.1 Confidence Interval for Population Mean (Population standard deviation Known)

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