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The breaking strengths of cables produced by a certain manufacturer have a standard deviation of 80 pounds. A random sample of 50 newly manufactured cables has a mean breaking strength of 1700 pounds. Based on this sample, find a 90% confidence interval for the true mean breaking strength of all cables produced by this manufacturer.

Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.

User Putnam
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Final answer:

Using the provided standard deviation and sample mean, along with the z-score for a 90% confidence interval, the true mean breaking strength of the cables is estimated to be between 1681.3 pounds and 1718.7 pounds.

Step-by-step explanation:

To find a 90% confidence interval for the true mean breaking strength of all cables produced by a certain manufacturer, where the standard deviation is known to be 80 pounds, and based on a random sample of 50 cables with a mean breaking strength of 1700 pounds, we use the formula for a confidence interval for the mean:

Confidence Interval = μ ± (z* × (σ/√n))

Here, μ is the sample mean, z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size.

First, we find the z-score for a 90% confidence level, which is approximately 1.645. Then, we calculate the margin of error:

Margin of Error = 1.645 × (80/√50) = 1.645 × (80/7.071) = 18.749 pounds (rounded to three decimal places)

The 90% confidence interval is thus:

1700 ± 18.749 = (1681.3, 1718.7) pounds

Therefore, we are 90% confident that the true mean breaking strength of the cables lies between 1681.3 pounds and 1718.7 pounds, when rounded to one decimal place.

User Christophe P
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