Hi Kaitlyn,
Once again, we have different definitions by different statisticians for "large sample size" when deciding whether to use z or t. I looked into this a bit for you: z requires known population standard deviation and "Large n," but some call large n 30 and others say 50 or 75, etc. Since this is not clear, I will compute both a z and a t confidence interval.
For z:
CI= x-bar +/- z*SE
x-bar=sample mean
z*=z-critical value
SE=standard error, computed below
We have a sample, so we need standard error, computed as:
SE=sigma/sqrt(n) where:
n=sample size
sigma=standard deviation
Now, let's get our standard error for this problem. Note that standard error is the same formula for both z and t, so we won't need to compute again. Substituting:
sigma=8
n=41
SE=8/sqrt(41)
SE=1.25
Now, back to our original confidence interval formula for z:
CI=x-bar +/- z*SE
x-bar=70
z*=1.96, always the z-critical value for 95% 2-sided confidence intervals
SE=1.25
CI=70 +/- (1.96*1.25)
CI= 70+/- 2.45
CI=(67.55, 72.45)
Now, let's do t just in case our researcher thinks their "n" is too low for z. Formula is:
CI=x-bar +/- t*SE
x-bar=sample mean
t*=t-critical value, which we calculate below
SE=standard error as computed above
Now, let's get t*. First, we need degrees of freedom--df=n-1.
df=41-1
df=40
Now, we go to the t-table and check the row for 40 degrees of freedom and the column for 95% confidence:
t*=2.021
Now, we have all we need for the t-confidence interval
CI=x-bar +/- t*SE
x-bar=70
t*=2.021
SE=1.25
CI=70 +/- (2.021*1.25)
CI=70 +/- 2.53
CI=(67.47, 72.53)
In this case, not much difference between z and t confidence intervals. I hope this helps.
