Here are summary statistics for randomly selected weights of newborn​ girls: nequals161​, x overbarequals32.8 ​hg, sequals7.2 hg. construct a confidence interval estimate of the mean. use a 90​% confidence level. are these results very different from the confidence interval 31.7 hgless thanmuless than34.5 hg with only 12 sample​ values, x overbarequals33.1 ​hg, and sequals2.7 ​hg? what is the confidence interval for the population mean mu​? 31.6 hgless thanmuless than 34.6 hg ​(round to one decimal place as​ needed.)

We have:
n = 161
xbar = 32.8hg
s = 7.2hg
cl = 90% = 0.90

Since we have only the sample standard deviation s, we need to use the t-distribution.
The degrees of freedom DF = 161 - 1 = 160
α = (1 - 0.90)/2 = 0.05

If you look at these values in a t-distribution table you find t = 1.65

Now we can build the confidence interval:
xbar +/- (t · s /√n) = 32.8 +/- (1.65 · 7.2 / √161)

Therefore:
31.86 < μ < 33.74

In order to understand if these values are different from the ones you are given, let's calculate the confidence interval for the latest:
n = 12
xbar = 33.1hg
s = 2.7hg
31.7 < μ < 34.5

Calculate the error: (34.5 - 33.1) = (33.1 - 31.7) = 1.4
We know t · s /√n = 1.4
and we can solve for t:
t = 1.4  · √n / s = 1.4 · √12 / 2.7 = 1.7962

Looking at a t-distribution table, we find α = 0.05
which brings to a confidence level of 90%, which is the same for the previous part.

Since the sample size is big enough, we can use the normal distribution and the z-score:
Looking at a normal distribution table, we find z-score = 1.645, which is very similar to the t-value found previously. We don't know the population standard deviation, but for such a big sample the sample standard deviation is a good estimate, therefore:
xbar +/- (z* · s /√n) = 32.8 +/- (1.645 · 7.2 / √161)
31.8 < μ < 33.7