Question

A sample of 4 children was drawn from a population of rural Indian children aged 12 to 60 months. The sample mean of mid-upper arm circumference was 150 mm with a standard deviation of 6.73. What is a 95% confidence interval for the mean of mid-upper arm circumference based on your sample

Answer 1

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.

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
`1 - (1 - 0.95)/(2) = 0.975`. So we have T = 3.1824

The margin of error is:

`M = T(s)/(√(n)) = 3.1824(6.73)/(√(4)) = 10.71`

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.