Question

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

nequals=229​,

x overbarxequals=28.8

​hg,

sequals=7.7

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

95​%

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

26.8

hgless than<muμless than<30.2

hg with only

18

sample​ values,

x overbarxequals=28.5

​hg, and

sequals=3.5

​hg?

What is the confidence interval for the population mean

muμ​?

nothing

hgless than<muμless than<nothing

hg ​(Round to one decimal place as​ need

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 28.8 hg

sample standard deviation = s = 7.7 hg

sample size = n = 229

Degrees of freedom = df = n - 1 = 229 - 1 = 228

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,228 = 1.970

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

= 1.970 * ( 7.7 / 229)

Margin of error = E = 1.0

The 95% confidence interval estimate of the population mean is,

- E < < + E

28.8 - 1.0 < < 28.8 + 1.0

( 27.8 hg < < 29.8 hg )

Yes, because the confidence interval limits are not similar