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]±[tex]t_{n-1;1-\alpha /2}[/tex] * [tex]\frac{S}{\sqrt{n} }[/tex]]
[tex]t_{n-1;1-\alpha /2} = t_{6;0.975}= 2.365[/tex]
[1403±2.365*[tex]\frac{27}{\sqrt{7} }[/tex]]
[1378.87;1427.13]
The margin of error is the semiamplitude of the interval and you can calculate it as:
[tex]d= \frac{Upbond-Lowbond}{2}= \frac{1427.13-1378.87}{2} = 24.13[/tex]
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!
