Question

After checking the assumptions, someone has computed the following 93% confidence interval for a set of sample data: 38< μ<57.

What is the margin of error for this interval?
What is the sample mean?

Answer 1

im not sure how to go about this?

Answer 2

The equation to use is as follows:
\[\bar{x}-1.81\frac{\sigma}{\sqrt{n}}<\mu < \bar{x}+1.81\frac{\sigma}{\sqrt{n}}\]
You are given the values of the expressions on either side of < mu < so you can set up a pair of equations to solve for the value of
\[\frac{\sigma}{\sqrt{n}}\]
The margin of error is found from calculating to value of
\[1.81\times \frac{\sigma}{\sqrt{n}}\]

Answer 3

all the examples in my book give me the n and standard deviation how am i to find those values?

Answer 4

based on the limited information, I believe the mean of the sample is the middle of 38 and 57: 47.5
And, the margin of error is the difference between the sample mean (47.5) and the top of the interval (57): 9.5

Answer 5

\[\bar{x}+1.81\frac{\sigma}{\sqrt{n}}=57\ ..........(1)\]
\[\bar{x}-1.81\frac{\sigma}{\sqrt{n}}=38\ ...........(2)\]
Subtracting equation (2) from equation (1) gives
\[3.62\frac{\sigma}{\sqrt{n}}=19\]
Therefore
\[\frac{\sigma}{\sqrt{n}}=\frac{19}{3.62}=5.2486\]
and the margin of error is 5.2486 * 1.81 = 9.5
The sample mean is found by adding equations (1) and (2) giving
\[2\times \bar{x}=95\]
\[\bar{x}=\frac{95}{2}=47.5\]