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.)
Accepted Solution
A:
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