Question

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.)

Answer 1

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

Answer 2

The confidence interval of large sample is $\boxed{31.86 < \mu < 33.74}$ and the confidence interval of sample with 12 observation is $\boxed{31.7 < \mu < 34.6}.$

Further Explanation:

The confidence interval of the mean of normal distribution can be obtained as follows,

$\boxed{{\text{Confidence interval}} = \overlineX \pm {Z_{\dfrac{\alpha }{2}}}\times \dfrac{s}{{\sqrt n }}}$

Here, $Z$ is the standard normal value, $\overline X$ represents the mean, $s$ represents the standard deviation, $\alpha$ is the level of significance and $n$ represents the observation.

Given:

Explanation:

Part (1).

The mean of is $\overlineX = 32.8.$

The standard deviation is $s = 7.2.$

The number of observation is $n = 161.$

The level of significance is $\alpha = \dfrac{{0.10}}{2} = 0.05.$

The sample size is large. Therefore, the sample can be considered as the normally distributed.

The confidence interval can be obtained as follows,

$\begin{aligned}{\text{Confidence interval}} &= 32.8 \pm 1.65 \times\frac{{7.2}}{{\sqrt {161} }}\\&= \left( {31.86,33.74} \right)\\\end{aligned}$

Part (2).

The mean of is $\overlineX = 33.1.$

The standard deviation is $s = 2.7.$

The number of observation is $n = 12.$

The level of significance is $\alpha = \dfrac{{0.10}}{2} = 0.05.$

The degree of freedom can be obtained as follows,

$\begin{aligned}Df&= 12 - 1\\&= 11\\\end{aligned}$

The confidence interval of mean of a small sample can be obtained as follows.

$\begin{aligned}{\text{Confidence interval}}&= 33.1 \pm 1.8\times \frac{{2.7}}{{\sqrt {12} }}\\&=\left( {31.7,34.6}\right)\\\end{aligned}$

The confidence interval of large sample is $\boxed{31.86 < \mu < 33.74}$ and the confidence interval of sample with 12 observation is $\boxed{31.7 < \mu < 34.6}.$

Learn more:

1. Learn more about normal distribution brainly.com/question/12698949

2. Learn more about standard normal distribution brainly.com/question/13006989

3. Learn more about confidence interval of mean brainly.com/question/12986589

Answer details:

Grade: College

Subject: Statistics

Chapter: Confidence Interval

Keywords: summary statistics, weights, confidence interval, mean, standard normal distribution, standard deviation, test, measure, probability, low score, mean, repeating, indicated, normal distribution, percentile, percentage, proportion.