4. A company produces precision 1 meter (1000 mm) rulers. The actual distribution of lengths of the rulers produced by this company is Normal, with mean $\mu$ and standard deviation $\sigma = 0.02$ mm. Suppose I select a simple random sample of four of the rulers produced by the company and I measure their lengths in mm. The sample yields $\bar{x} = 1000$. The margin of error for a 90% confidence interval for $\mu$ is a. 0.0082. b. 0.0115. c. 0.0165. d. 0.0196.
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02$. We want to find the margin of error for a 90% confidence interval for the population mean $\mu$. Show more…
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A company produces precision 1-meter (1000 mm) rulers. The actual distribution of lengths of the rulers produced by this company is normal with mean μ and standard deviation 0.02 mm. Suppose we select a simple random sample of four of the rulers produced by the company and measure their lengths in millimeters. The results of these four measurements are as follows: 1000.01 999.98 1000.00 1000.01 We will use a 90% confidence interval to estimate the mean length of 1m rulers. We will use a Z-procedure to calculate the margin of error and the confidence interval to five decimal places. Table C shows that we are 90% confident that the mean length of 1m rulers is between mm and mm.
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A company produces precision 1000-millimeter rulers. The actual distribution of the lengths of the rulers produced by this company is normal, with mean μ and standard deviation σ = 0.02 millimeters. Suppose we select a simple random sample of four of the rulers produced by the company and measure their lengths in millimeters. The sample yields a z-value of 1.96. A 90% confidence interval for μ is: 1000 ± 0.0082 1000 ± 0.0115 1000 ± 0.0165 1000 ± 0.0196
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