Question

A sample of the daily iron intake of 22 women had mean 14.6 mg and standard deviation 4.2 mg. Assuming daily iron intake is normally distributed, a 90% confidence interval for the mean daily iron intake of all women is about (a) [13.4, 15.8] (b) [12.1, 17.1] (c) [14.5, 14.7] (d) [13.1, 16.1] (e) [12.8, 16.4]

Answer 1

By using t-distribution ,

df=degrees of freedom=n-1=22-1=21

Now , the critical value is , $t_{df,\alpha/2}=t_{21,0.10/2}=1.721$ ; From t-table

Therefore , the 90% confidence interval for the mean daily iron intake of all women is,

$\bar{X}\pm t_{df,\alpha/2}\frac{s}{\sqrt{n}}$

$14.6\pm 1.721*\frac{4.2}{\sqrt{22}}$

$14.6\pm 1.5411$