Answer:
The 95% confidence interval for µ based on these data is (408,692).
Explanation:
Sample mean and standard deviation:
The first step to solve this question is finding the sample mean and standard deviaton.
![\overline{x} = (550+620+480)/(3) = 550](https://img.qammunity.org/2022/formulas/mathematics/high-school/fua9x69vob46joffnphprld99rkup5qpo3.png)
![s = \sqrt{((550-550)^2+(620-550)^2+(480-550)^2)/(3)} = 57.1548](https://img.qammunity.org/2022/formulas/mathematics/high-school/7h50m1a8kb6250x8x72lyg9xzuigqupl7j.png)
Confidence interval:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
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 = 3 - 1 = 2
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 2 degrees of freedom(y-axis) and a confidence level of
. So we have T = 4.3027
The margin of error is:
![M = T(s)/(√(n)) = 4.3027(57.1548)/(√(3)) = 142](https://img.qammunity.org/2022/formulas/mathematics/high-school/g6693nu75njetr30f1dvyvbot0l5pjw3bt.png)
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 550 - 142 = 408
The upper end of the interval is the sample mean added to M. So it is 550 + 142 = 692
The 95% confidence interval for µ based on these data is (408,692).