Hi Aubree,
The appropriate test for part (a) is somewhat of a judgment call. We do know standard deviation, and we know we can apply the Central Limit Theorem and use z for large n, but how large is large? My old textbook went with large n>=50, so we'll go with that for this question.
(a) We'll use the z-test for reasons described above. Formula to compute test statistic:
z=x-bar-mu-nought/(sigma/sqrt (n))
Let's break that down:
x-bar=sample mean
mu-nought=mean you are testing against
n=sample size
sigma=known population standard deviation
Now let's assign values to these variables:
x-bar=484
mu-nought=497
n=75
sigma=75
Therefore:
z=(484-497)/[(75/sqrt(75)]
z=(-13/8.660)
z= -1.50
Since the problem asked only for the test statistic, I will stop here without performing the actual test.
(b) This is more cut and dry regarding what test to use. We do not have known population standard deviation here, nor do we have particularly large sample size, so we have to use t.
Formula for t-test statistic:
t=(x-bar-mu-nought)/(s/sqrt(n))
Notice that this is the exact same formula we used for z with one key difference. We are now using the sample standard deviation s instead of the population standard deviation sigma. That's a requirement for t-tests since we don't have a known population standard deviation. Let's assign values again:
x-bar=484
mu-nought=497
s=77--remember we are using sample standard deviation
n=12
t=(484-497)/(77/sqrt(12))
t=-13/22.227
t= -0.58
Again, I'll stop here since we were asked only for the test statistic, not the actual results of the test. I hope this helps.
