Question

Chocolate bars produced by a certain machine are labeled 8.0 ounces. The distribution of the actual weights of these chocolate bars is claimed to be Normal with a mean of 8.1 ounces and a standard deviation of 0.1 ounces. A quality control manager initially plans to take a simple random sample of size n from the production line. If he were to double his sample size (to 2n), by what factor would the standard deviation of the sampling distribution of change?

Answer 1

Answer:

The standard deviation of the sampling distribution would change by a factor of $\frac{1}{\sqrt{2}}$

Step-by-step explanation:

We use the central limit theorem to solve this question.

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

The standard deviation of the population is $\sigma = 0.1$

Sample size n

$s_{n} = \frac{0.1}{\sqrt{n}}$

Sample size 2n

$s_{2n} = \frac{0.1}{\sqrt{2n}}$

What factor would the standard deviation of the sampling distribution of change?

$F = \frac{s_{2n}}{s_{n}} = \frac{\frac{0.1}{\sqrt{2n}}}{\frac{0.1}{\sqrt{n}}} = \frac{\sqrt{n}}{\sqrt{2n}} = \frac{1}{\sqrt{2}}$

So the standard deviation of the sampling distribution would change by a factor of $\frac{1}{\sqrt{2}}$