Mathematics High School

suppose net gain, in dollars, of the departments for an industry per day are normally distributed and have a known population standard deviation of 325 dollars and an unknown population mean. a random sample of 20 departments is taken and gives a sample mean of 1640 dollars. find the confidence interval for the population mean with a 98% confidence level. round your answer

Answers

Answer 1

The 98% confidence interval for the population mean net gain of the departments is 1640 ± 2.33 * 72.672 = (1470.67 dollars , 1809.33 dollars).

To calculate the confidence interval, we'll use the formula:

Confidence Interval = Sample Mean ± (Critical Value) * (Standard Deviation / √Sample Size)

The critical value for a 98% confidence level can be obtained from the standard normal distribution table, and in this case, it is 2.33 (approximately).

Plugging in the values, we have:

Confidence Interval = 1640 ± 2.33 * (325 / √20)

Calculating the standard error (√Sample Size) first, we get √20 ≈ 4.472.

we can calculate the confidence interval:

Confidence Interval = 1640 ± 2.33 * (325 / 4.472)

Confidence Interval = 1640 ± 2.33 * 72.672

Confidence Interval ≈ (1470.67 dollars , 1809.33 dollars)

Therefore, with a 98% confidence level, we can estimate that the population mean net gain of the departments falls within the range of 1470.67 to 1809.33.

To know more about confidence interval, refer here:

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