Question

Use the confidence level and sample data to find a confidence interval for estimating the population mu. round your answer to the same number of decimal places as the sample mean.
test scores: n = 92, mean = 90.6, sigma = 8.9; 99% confidence

Answer 1

The confidence level and sample data to find a confidence interval for estimating the population mu. round your answer to the same number of decimal places as the sample mean 88.2 < μ < 93.0.

What are the confidence level?

The confidence stage refers to the proportion of probability, or certainty, that the self-belief c language could include the real populace parameter whilst you draw a random pattern many times.

1. We have this given info:
2. n = 92 = pattern length xbar = 90.6 = pattern imply
3. sigma =8.! = populace popular deviation
4. C=99% self-assurance level
5. Because n > 30 and due to the fact we recognize sigma, this permits us to apply the Z distribution (aka popular ordinary distribution).
6. At 99% self-assurance, the z essential price is more or less z = 2.576 use a reference sheet, table, or calculator to decide this.
7. The decrease sure of the self assurance c programming language (L) is more or less L = xbar - z*sigma/sqrt(n) L = 90.6-2.576*8.9/sqrt(92) perp= 88.209757568781 perp= 88.2
8. The top sure (U) of this self assurance c programming language is more or less U = xbar + z*sigma/sqrt(n) L = 88.209757568781 L = 88.2.
9. The top sure (U) of this self assurance c programming language is more or less U=xbar+z^ * sigma/sqrt(n) U = 90.6 + 2.576*8.9/sqrt(92) U = 92.990242431219
10. U = 93.0.
11. Therefore, the self assurance c programming language withinside the format (L, U) is about (88.2, 93.zero).

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