90% Confidence Interval, Find Margin of Error

[cadag]

Problem is:

study of 100 people reveals mean amount spend on gifts is for a given person is $38. The population standard deviation is not known but is estimated to be $12.
A) for 90% Confidence Interval, find the Margin of Error in estimating the mean

B) Using results from a, what is 90% CI for the mean?

C) How many individuals need to be sampled to have a Margin of error of $1.2 at 90% CI?

My teacher has really confused me here, she told us that for this problem we are to use the T table, since it is NOT distributed normally. I have the T table with the degrees of freedom etc, but I the formula I see to use for that, requires the "S" value, and I am not sure what value this question is giving. Is the estimated value the of the population SD, the #12, the S value? I have assumed so, and done it using the formula.

for A)
I did:

Sample Mean +- T(a/2) S / SQRT(n)

which ends up being
38 +- t.05 12/ SQRT(100)

which is 38 +- 1.992

so then B) the CI is 36.008 ---> 39.992


now for C) our teacher told us to use this forumala which I cannot find in the book and do not know if it is right:

N = (t, n-1)[sup:wxytfl2t]2[/sup:wxytfl2t] s[sup:wxytfl2t]2[/sup:wxytfl2t] / E[sup:wxytfl2t]2[/sup:wxytfl2t]

I did this and ended up with 199.2

I'd like to know if this seems correct, I am confused as how to use the T value formula correctly, if I even have it correct.

 

[galactus]

I don't know why your teacher said to use the t-table. Yes, s is the standard deviation given here as 12.

minimum sample size is \(\displaystyle n=\left(\frac{z\cdot \sigma}{E}\right)^{2}\)

\(\displaystyle n=\left(\frac{1.645(12)}{1.2}\right)^{2}=271\)

Confidence interval is \(\displaystyle E=z\cdot\frac{\sigma}{\sqrt{n}}=1.974\)

add it to and subtract from the mean for the C.I.

\(\displaystyle 36.026\leq {\mu}\leq 39.974\)

In this case, n>30. If we look in the t-table, the degrees of freedom in most of them only go up to 29.

Below 29 is infinity, which has a value of 1.645, Same as the z table. That's because of the large value of 100.

As the degrees of freedom increase, the t-distribution approaches the normal distribution. After 30 d.f. the t distribution is very close the normal

distribution. That is why, with a large value like 100, I wonder why the teacher told you to use the t table.

We mostly use the t table when n< 30.

Using \(\displaystyle E=t_{c}\cdot\frac{s}{\sqrt{n}}\) we get the same value as before.