Question

Assume that a sample is used to estimate a population proportion mu or mean. Find the margin of error M.E. that corresponds to a sample of size 18 with a mean of 20.1 and a standard deviation of 13.5 at a confidence level of 90%.

Answer 1

so i guess for this problem i have to use z(alpha/2) * (standard dev/sqrt(n)) right?

Answer 2

because i'm tryig to find my margin of error

Answer 3

A confidence interval for the mean (with a small sample) is calculated as:
\[\bar{x} \pm t * \frac{s}{\sqrt{n}}\] but can also be written as:
\[ \bar{x} \pm \text{margin of error}\]

so the margin of error is \[t* \frac{s}{\sqrt{n}} \] where \(t\) is the critical value for a 90% confidence interval from the t-distribution with degrees of freedom = \(n-1\), and \(s\) is the standard deviation and \(n\) is the sample size.

Answer 4

so, its 1.645 * (13.5/sqrt(18)) = 5.234

Answer 5

You can use a normal distribution, with a critical \(Z\) values instead of a critical \(t\) value if the standard deviation given is a known population standard deviation (rather than a sample standard deviation, but it's not "clear" from the question, and usually the pop. std. dev. is not given). Otherwise, the sample size is small, so usually small is considered n = 30 (but this might depend on your prof), in which case the t-distribution will be better

Answer 6

small is considered n <= 30*

Answer 7

oh, in my textbook it says N<30

Answer 8

n*

Answer 9

ok, then yes I suggest you use a t-table for the critical value (rather than using 1.645 which is form a normal distribution table)

Answer 10

so, 1.740

Answer 11

1.740 * (13.5/sqrt(18)) = 5.5

Answer 12

so anytime n is less than or equal to 30, i have to use the t table?

Answer 13

Yes, UNLESS the problem specifically tells you that they gave you a -population- standard deviation (rather than a -sample- standard deviation).

Answer 14

so you mean like the pop. is not normally distributed and n < or = to 30?

Answer 15

ty for the help.

Answer 16

Also another question, how would i know in the problem when to use the t distributions and the standard normal?

Answer 17

1) If the population standard deviation is given, and the sample size is large (n >= 30), use a normal distribution

2) If the population standard deviation is given, and the sample size is small (n < 30), use the normal distribution

3) If the population standard deviation is NOT given (the sample std. deviation is given instead), and the sample size is large (n >= 30), use the normal distribution (however, using the t distribution is actually more accurate to use), but the normal approximation is usually okay to use.

4) If the population standard deviation is NOT given (the sample std. deviation is given instead), and the sample size is small (n < 30), use the t distribution for degrees of freedom n-1

... note that cases 3 and 4 are usually what occur... it is very rare and unlikely that you will have population standard deviations at your disposal like in cases 1 and 2. But it's something to be aware of in the problem.

Answer 18

Awesome! thank you very much for the help and info. It helped a lot.

Answer 19

your welcome! :)

Answer 20

Maybe an easier summary for the conditions is: