b) If we change \epsilon =0.1 to \epsilon =0.1/k, for k\geq 1, the value of the upper bound does not change from the value calculated in part (a) if we increase n by a factor of k^2.
This is based on the argument in the last video, which states that the probability of the sample mean deviating from the true mean can be controlled by the variance and sample size. Since \epsilon = 0.1/k is k times smaller than \epsilon = 0.1, we need to decrease the probability bound by a factor of k in order to maintain the same level of accuracy. Since the probability bound is inversely proportional to the sample size n, we need to increase n by a factor of k^2 to maintain the same value of the upper bound.
By the argument in the last video, if the X_ i are i.i.d. with mean \mu and variance \sigma ^2, and if M_ n=(X_1+\cdots +X_ n)/n, then we have an inequality of the form
\mathbf{P}\big ( |M_ n-\mu | \geq \epsilon \big ) \leq \frac{a\sigma ^2}{n},
for a suitable value of a.
a) If \epsilon =0.1, then the value of a is:
unanswered
b) If we change \epsilon =0.1 to \epsilon =0.1/k, for k\geq 1 (i.e., if we are interested in k times higher accuracy), how should we change n so that the value of the upper bound does not change from the value calculated in part (a)?
n should
stay the same
increase by a factor of k
increase by a factor of k^2
decrease by a factor of k
none of the above
1 answer
1 answer