Answer:
The 95% confidence interval for the mean of mid-upper arm circumference based on your sample is between 139.29 mm and 160.71 mm.
Step-by-step explanation:
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,which is the sample size subtracted by 1. So
df = 4 - 1 = 3
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 3 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 3.1824
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 3.1824\frac{6.73}{\sqrt{4}} = 10.71[/tex]
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 150 - 10.71 = 139.29 mm
The upper end of the interval is the sample mean added to M. So it is 150 + 10.71 = 160.71 mm
The 95% confidence interval for the mean of mid-upper arm circumference based on your sample is between 139.29 mm and 160.71 mm.
