Question

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

nequals=187187​,

x overbarxequals=32.132.1

​hg,

sequals=7.37.3

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

9090​%

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

31.431.4

hgless than<muμless than<33.633.6

hg with only

2020

sample​ values,

x overbarxequals=32.532.5

​hg, and

sequals=2.82.8

​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 = = 32.1 hg

sample standard deviation = s = 7.3 hg

sample size = n = 187

Degrees of freedom = df = n - 1 = 187 - 1 = 186

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t0.05,186 = 1.653

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

= 1.653 * ( 7.3 / 187)

Margin of error = E = 0.9

The 90% confidence interval estimate of the population mean is,

- E < < + E

32.1 - 0.9 < < 32.1 + 0.9

31.2 hg < < 33.0 hg

Yes, because the confidence interval limits are not similar