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A sample of 230 observations is selected from a normal population with a population standard deviation of 26. The sample mean is 18. (Use  t Distribution Table & z Distribution Table.)

Question

A sample of 230 observations is selected from a normal population with a population standard deviation of 26. The sample mean is 18. (Use  t Distribution Table & z Distribution Table.)

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Solution

You didn't specify a question, but I'm assuming you want to calculate the confidence interval for the population mean. Here's how you can do it:

  1. Identify a sample size or number of observations: n = 230
  2. Identify a standard deviation or standard error: σ = 26
  3. Identify a sample mean: x̄ = 18

Since the sample size is large (n > 30), we can use the z-distribution to calculate the confidence interval.

  1. Choose your desired confidence level. For example, if you want a 95% confidence level, the z-value from the z-distribution table is 1.96.

  2. Calculate the standard error of the mean (SE). SE = σ/√n = 26/√230 ≈ 1.71

  3. Calculate the margin of error (E). E = z * SE = 1.96 * 1.71 ≈ 3.35

  4. Calculate the confidence interval. The lower limit is x̄ - E = 18 - 3.35 = 14.65 and the upper limit is x̄ + E = 18 + 3.35 = 21.35.

So, the 95% confidence interval for the population mean is (14.65, 21.35).

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