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

`n=((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

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)(s)/(√(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=((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=((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

We need to sample at least 1069 parents.

Explanation:

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

`\alpha = (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*(\sigma)/(√(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*(\sigma)/(√(n))`

`4 = 1.645*(79.5)/(√(n))`

`4√(n) = 1.645*79.5`

`√(n) = (1.645*79.5)/(4)`

`(√(n))^(2) = ((1.645*79.5)/(4))^(2)`

`n = 1068.92`

Rounding up

We need to sample at least 1069 parents.