Question

7 measured data points have a sample mean of 1403 and a standard deviation of 27. Determine the best estimate of the mean value at 95% probability level. Find a, where the best estimate of the mean value is expected to fall ±a about the sample mean.

Answer 1

Answer:

Step-by-step explanation:

Hello!

You have sample of n=7 with mean X[bar]= 1403 and standard deviation S=27 and are required to estimate the mean with a 95%CI.

Asuming this sample comes from a normal population I'll use a stuent t to estimate the interval (a sample of 7 units is too small for the standard normal to be accurate for the estimation):

[X[bar]±$t_{n-1;1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$]

$t_{n-1;1-\alpha /2} = t_{6;0.975}= 2.365$

[1403±2.365*$\frac{27}{\sqrt{7} }$]

[1378.87;1427.13]

The margin of error is the semiamplitude of the interval and you can calculate it as:

$d= \frac{Upbond-Lowbond}{2}= \frac{1427.13-1378.87}{2} = 24.13$

With a confidence level of 95% you'd expect that the real value of the mean is contained by the interval [1378.87;1427.13], the best estimate of the mean value is expected to be ± 24.13 of 1403.

I hope it helps!