The 2000 census "long form" asked the total 1999 income of the householder, the person in whose name the dwelling unit was owned or rented. This census form was sent to a random sample of 17% of the nation's households. Suppose that the households that returned the long form are an SRS of the population of all households in each district. In Middletown, a city of 40,000 persons, 2621 householders reported their income. The mean of the responses was x = $33,453, and the standard deviation was s = $8721. The sample standard deviation for so large a sample will be ve

Question

anonymous · Answer

very close to the population standard deviation s. Use these facts to give an approximate 99% confidence interval for the 1999 mean income of Middletown householders who reported income.

aihberkhan · Answer

Okay. I am actually working on these problems right now haha. Well I can help! :)

So what this problem is working with, is confidence interval and is also involving the z - distribution. 

@hannah_hayward

aihberkhan · Answer

So the formula for the income is $X$~$ N(\mu, \sigma^2) $

aihberkhan · Answer

Now we have an unknown population variance here. However, $n = 40000$ is very large.

anonymous · Answer

okay...

anonymous · Answer

so we would need to take a sample?

aihberkhan · Answer

The population variance is estimated using the sample variance, which is $s^2$

The sample mean, xbar $= 33,453$ and $s = 8721$

We have to use the z-distribution. 

So we know that $99$% Cl is given due to:

xbar $\pm ~z(a/2) ~\sigma/\sqrt{n}$ 

OR

xbar $\pm ~ z(0.005)(8721/\sqrt40000)$

Now we will just substitute everything into that equation:

$= 33,453 \pm 2.576 (8721/\sqrt40000)$

$= 33,453 \pm 112$

That is it! :)

aihberkhan · Answer

Hope this helped! Have a great day! :)

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anonymous · Answer

It did, thank you so much!