Question

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

nequals=221

x overbar x=28.8

​hg,

s=7.9

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

99%

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

25.0

hg less than<muμless than<31.4

hg with only

12

sample​ values,

x overbarxequals=28.2

​hg, and

sequals=3.6

​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.)

Are the results between the two confidence intervals very​ different?

A.

​No, because the confidence interval limits are similar.

B.

​Yes, because the confidence interval limits are not similar.

C.

​No, because each confidence interval contains the mean of the other confidence interval.

D.

​Yes, because one confidence interval does not contain the mean of the other confidence interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 28.8

sample standard deviation = s = 7.9

sample size = n = 221

Degrees of freedom = df = n - 1 = 221 - 1 = 220

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t_0.005,220= 2.598

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

= 2.598* ( 7.9 / 221)

Margin of error = E = 1.4

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

- E < < + E

28.8 - 1.4 < < 28.8 + 1.4

(27.4 < < 30.2)

correct option is = C.

​No, because each confidence interval contains the mean of the other confidence interval.