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A random sample of electronic components had the following operational times before failure, in hours,

322 350 346 347 335 323 341 355 329

Assume the population standard deviation is 36 and that the population is approximately normal. Construct a 90% confidence interval for the population mean operational time before failure.
O 18.9, 358.4)
O (279.4, 397.9)
O (332.1, 3453)
O 313.346.1)

User Wrozwad
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1 Answer

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Final answer:

To construct the 90% confidence interval for the population mean operational time before failure, the sample mean is calculated to be approximately 349.33 hours. With a Z-value of 1.645 for a 90% confidence level and a known population standard deviation, the margin of error is found and the interval is approximately (329.59, 369.07) hours.

Step-by-step explanation:

To construct a 90% confidence interval for the population mean operational time before failure using the provided sample data, we use the formula for the confidence interval for a population mean when the population standard deviation is known:

CI = μ ± (Z * (σ / √n))

μ is the sample mean.

Z is the Z-value from the standard normal distribution for the given confidence level.

σ is the population standard deviation.

n is the sample size.

First, we calculate the sample mean (μ). Add up all the sample values and divide by the number of samples (n).

Sample mean: μ = (322 + 350 + 346 + 347 + 335 + 323 + 341 + 355 + 329) / 9 = 3144 / 9 ≈ 349.33

Next, we look up the Z-value for a 90% confidence level, which is 1.645 for one-tailed (or 90/2 = 45% for two-tailed).

Now, we can calculate the margin of error:

Margin of error: E = Z * (σ / √n) = 1.645 * (36 / √9) ≈ 1.645 * 12 ≈ 19.74

Finally, we construct the confidence interval:

CI = 349.33 ± 19.74 ≈ (329.59, 369.07)

Thus, the 90% confidence interval for the population mean operational time before failure is approximately (329.59, 369.07) hours.

User Krokoziabla
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