Mathematics, 26.07.2019 15:00 omar9844

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.)

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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 from: jialala

Refer the below solution for better understanding.

Step-by-step explanation:

Given :

$\rm \bar{x} = 32.8 \; hg$

n = 161

s = 7.2 hg

Confidence Level = 90% = 0.90

Solution :

Degree of freedom = 161 - 1 = 160

$\alpha = \dfrac{1 - 0.90}{2} =0.05$

In a t-distribution table you find t = 1.65

Now the confidence interval is,

xbar +/- (t · s /√n) = 32.8 +/- (1.65 · 7.2 / √161)

$\bar{x} \pm (t.\dfrac{s}{\sqrt{n} })= 32.8\pm (1.65\times\dfrac{7.2}{\sqrt{161} })$

Therefore,

31.86

To find that these values are different from given values, calculate the confidence interval for the latest,

n = 12

$\rm \bar{x} = 33.1\; hg$

s = 2.7 hg

31.7

Now calculate the error,

(34.5 - 33.1) = (33.1 - 31.7) = 1.4

We know that

$\rm t\times\dfrac{s}{\sqrt{n} } = 1.4$

$\rm t = 1.4 \times \dfrac{\sqrt{n} }{s}$

$\rm t = 1.4\times \dfrac{\sqrt{12} }{2.7} = 1.7962$

Now looking at t-distribution table, we find $\alpha = 0.05$ , which brings to a confidence level of 90%.

Now, looking at normal distribution table, we find z-score = 1.645, which is very similar to the t-value found previously. Therefore,

$\bar{x}\pm (z\times\dfrac{s}{\sqrt{n} }) = 32.8\pm(1.645\times\dfrac{7.2}{\sqrt{161} })$

31.8

For more information, refer the link given below

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Answer from: Quest

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Answer from: Quest

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Here are summary statistics for randomly selected weights of newborn girls: nequals161, x overbareq...

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