A sample of $40$ cows is drawn to estimate the mean weight of a large herd of cattle. If the standard deviation of the sample is $96$ kg, what is the maximum error in a $90\text{%}$ confidence interval estimate?
Asked
Active
Viewed 5,754 times
1
-
Welcome to math stack exchange! The standard devitation for the weight of the $40$ cows is $\sqrt{40}\cdot 96kg$. If we assume a normal distribution, we can take $1.645\cdot\sqrt{40}\cdot 96kg$ as the maximal allowed deviation. – Peter Jul 24 '16 at 20:44
1 Answers
2
You use t-distribution since you have sample standard deviation $s = 96$, and with $n = 40,\alpha = 0.1,df = 39, t_{\frac{\alpha}{2}}$ = ? ( can you look it up the table A-3 in Triola book ? ) . Thus $ E = t_{\frac{\alpha}{2}}\cdot \dfrac{s}{\sqrt{n}} $ is your formula.
DeepSea
- 77,651