Question

How large a sample should be selected to provide a 95% confidence interval with a margin of error of 10? Assume that the population standard deviation is 40. (Round your answer up to the nearest whole number.)

Answer 1

Answer:

62

Step-by-step explanation:

Given that in a hypothesis 95% confidence interval had a margin of error of 10.

Also given that the population standard deviation is known and equal to 40

Since population standard deviation is known, for calculating confidence interval we can use Z critical value.

For 95% confidence interval z critical value used is 1.96

Margin of error = 1.96* std error = 10

Hence standard error = $\frac{10}{1.96} \\=5.102$

Standard error is also equal to

$\frac{\sigma}{\sqrt{n} }\\ =\frac{40}{\sqrt{n} } =5.102\\\sqrt{n} =7.84\\n = 61.46=62$

Sample size should be 62

Answer 2

Answer:

We need a sample of at least 62.

Step-by-step explanation:

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 M 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.

How large a sample should be selected to provide a 95% confidence interval with a margin of error of 10?

We need a sample of at least n.

n is found when $M = 10, \sigma = 40$

So

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

$10 = 1.96*\frac{40}{\sqrt{n}}$

$10\sqrt{n} = 40*1.96$

Simplifying by 10

$\sqrt{n} = 4*1.96$

$(\sqrt{n})^{2} = (4*1.96)^{2}$

$n = 61.47$

Rounding up

We need a sample of at least 62.