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Assume that a sample is used to estimate a population mean . Find the margin of error M.E. that corresponds to a sample of size 5 with a mean of 75.2 and a standard deviation of 21.2 at a confidence level of 98% Report ME accurate to one decimal place because the sample statistics are presented with this accuracy M.E. Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Respuesta :

We have the following:

[tex]\begin{gathered} df=n-1 \\ =5-1 \\ df=4 \end{gathered}[/tex]

therefore:

[tex]\begin{gathered} ME=t_{critical}\cdot\frac{s}{\sqrt{n}} \\ ME=3\text{.}747\cdot\frac{21.2}{\sqrt{5}} \\ ME=35.52 \end{gathered}[/tex]

The margin the error that corresponds to a sample of size of 5 with mean 75.2 and a standard deviation of 21.2 at a confidence level of 98% is 35.52

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