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Assume that a sample is used to estimate a population mean μ μ. Find the margin of error M.E. that corresponds to a sample of size 19 with a mean of 30.8 and a standard deviation of 10.5 at a confidence level of 99.5%.

Respuesta :

Answer:

The margin of error is 30.22

Step-by-step explanation:

Sample size 19(lesser than 30), and we only have the sample standard deviation. So we use the t-distribution.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 19 - 1 = 18

Now, we have to find a value of T, which is found looking at the t table, with 18 degrees of freedom(y-axis) and a confidence level of 0.995([tex]t_{995}[/tex]). So we have T = 2.878

The margin of error is:

M = T*s = 10.5*2.878 = 30.22

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