Question

Suppose a computer engineer is interested in determining the average weight of a motherboard manufactured by a certain company. A summary of a large sample provided to the engineer suggest a mean weight of 11.8 ounces and an estimated standard deviation, sigma = 0.75. How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4? How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?

Answer 1

Answer:

We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.

We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,

Step-by-step explanation:

How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4?

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.005 = 0.995$, so $z = 2.575$

Now, find the margin of error(width) as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

For this item, we have:

$M = 0.4, \sigma = 0.75$. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.4 = 2.575*\frac{0.75}{\sqrt{n}}$

$0.4\sqrt{n} = 1.93125$

$\sqrt{n} = \frac{1.93125}{0.4}$

$\sqrt{n} = 4.828125$

$\sqrt{n}^{2} = (4.828125)^{2}$

$n = 23$

We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.

How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find the margin of error(width) as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

For this item, we have:

$M = 0.5, \sigma = 0.75$. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.5 = 1.96*\frac{0.75}{\sqrt{n}}$

$0.5\sqrt{n} = 1.47$

$\sqrt{n} = \frac{1.47}{0.5}$

$\sqrt{n} = 2.94$

$\sqrt{n}^{2} = (2.94)^{2}$

$n \cong 9$

We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,