Question

Here are summary statistics for randomly selected weights of newborn​ girls:

nequals=231231​,

x overbarxequals=30.130.1

​hg,

sequals=7.97.9

hg. Construct a confidence interval estimate of the mean. Use a

9999​%

confidence level. Are these results very different from the confidence interval

27.427.4

hgless than<muμless than<33.033.0

hg with only

1414

sample​ values,

x overbarxequals=30.230.2

​hg, and

sequals=3.53.5

​hg?

What is the confidence interval for the population mean

muμ​?

nothing

hgless than<muμless than<nothing

hg ​(Round to one decimal place as​ needed.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 30.1 hg

sample standard deviation = s = 7.9 hg

sample size = n = 231

Degrees of freedom = df = n - 1 = 231 - 1 = 230

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t0.005,230 = 2.597

Margin of error = E = t/2,df * (s /n)

= 2.597 * ( 7.9 / 231)

Margin of error = E = 1.3

The 99% confidence interval estimate of the population mean is,

- E < < + E

30.1 - 1.3 < < 30.1 + 1.3

( 28.8 hg < < 31.4 hg )

Yes, because the confidence interval limits are not similar