Hi Vi,
This is somewhat of a tough call in terms of which confidence interval to use, z or t, so I'll post both solutions with working. Typically, for known sigma (5.6 here), we can use z, but sample size (10) is a little small.
- z first. For a z-confidence interval, the formula is: CI= x-bar +/- z*SE
Breaking this down:
x-bar=sample mean
z*=z-critical value
SE=standard error
First, we must compute standard error as well, which is:
SE=sigma/sqrt(n) where:
sigma=standard deviation
n=sample size
Now, let's compute that for this problem:
sigma=5.6
n=10
SE=5.6/sqrt(10)
SE=1.77
Now, back to the original formula:
CI=x-bar +/- z*SE
x-bar=61
z*=1.96, available at z-table but I recommend memorizing this critical value for 95% CIs
SE=1.77
CI=61 +/- (1.96*1.77)
CI=61 +/- 3.47
Margin of Error is back end of confidence interval:
MoE= (z*SE)
MoE=1.96*1.77
MoEz=3.47
That's the margin of error for a z-confidence interval. Now, for a t-confidence interval, everything remains the same in our calculation except for the critical value. Formula is:
CI=x-bar +/- t*SE
x-bar=sample mean
t*=t-critical value, explained below
SE=standard error
For this problem,
x-bar=61
SE=1.77
To get t*, you need the degrees of freedom. Formula is df=n-1, so:
df=10-1=9
Now, we check the t-table for 9 degrees of freedom and 95% confidence:
t*=2.262
CI=61 +/- (1.77*2.262)
CI=61 +/- 4.00
MoE=SE*t*
MoE= (1.77*2.262)
MoEt=4.00
2.Everything remains the same except our confidence level and, therefore, our critical values, so:
CI=x-bar +/- z* SE
x-bar=61
z*=2.575, available at z-table
SE=1.77
CI=61 +/- (2.575*1.77)
MoEz=z*SE
MoEz=4.56
For t, again, everything is the same except critical value:
x-bar=61
SE=1.77
df=9
Go to t-table for 9 df and 99% confidence level:
t*=3.250
CI=61 +/- (3.250*1.77)
MoEt=(t*SE)
MoEt=(3.250*1.77)
MoEt=5.75
I hope this helps.
