Question

You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately σ = 79.5 σ=79.5 dollars. You would like to be 90% confident that your estimate is within 4 dollar(s) of average spending on the birthday parties. How many parents do you have to sample?

Answer 1

Answer:

$n=(\frac{1.64(79.5)}{4})^2 =1062.43 \approx 1063$

So the answer for this case would be n=1063 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=79.5$ represent the population standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$    (a)

And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.05;0;1)", and we got $z_{\alpha/2}=1.64$, replacing into formula (b) we got:

$n=(\frac{1.64(79.5)}{4})^2 =1062.43 \approx 1063$

So the answer for this case would be n=1063 rounded up to the nearest integer

Answer 2

Answer:

We need to sample at least 1069 parents.

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:

$\alpha = \frac{1-0.9}{2} = 0.05$

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 $1-0.05 = 0.95$, so $z = 1.645$

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 many parents do you have to sample?

We need to sample at least n parents.

n is found when $M = 4, \sigma = 79.5$. So

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

$4 = 1.645*\frac{79.5}{\sqrt{n}}$

$4\sqrt{n} = 1.645*79.5$

$\sqrt{n} = \frac{1.645*79.5}{4}$

$(\sqrt{n})^{2} = (\frac{1.645*79.5}{4})^{2}$

$n = 1068.92$

Rounding up

We need to sample at least 1069 parents.