Question

Assume that a sample is used to estimate a population mean μ . Find the 99.9% confidence interval for a sample of size 811 with a mean of 56.8 and a standard deviation of 15.3. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

Answer 1

ANSWER:

Given that,

Point estimate = sample mean = = 56.8

sample standard deviation = s = 15.3

sample size = n = 811

Degrees of freedom = df = n - 1 = 811-1= 810

At 99.9% confidence level the t is ,

= 1 - 999% = 1 - 0.999 = 0.001

/ 2 = 0.001 / 2 = 0.0005

t /2,df = t0.0005,24 = 3.492

Margin of error = E = t/2,df * (s /n)

= 3.492* (15.3 / 811)

= 1.9

The 99.9% confidence interval estimate of the population mean is

56..8 - 1.9 < < 56.8 + 1.9

n mean is,

- E < < + E

54.9< < 58.7

99.9% C.I. = (54.9 , 58.7)

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