Question

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

nequals=154​,

x overbarxequals=30.5

​hg,

sequals=7.1

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

95​%

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

29.1

hgless than<muμless than<32.1

hg with only

16

sample​ values,

x overbarxequals=30.6

​hg, and

sequals=2.9

​hg?

Answer 1

Solution:

Given that,

n = 154

= 30.5

s = 7.1

Note that, Population standard deviation() is unknown. So we use t distribution.

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, d.f = n - 1 = 154 - 1 = 153

    =    =  _0.025,153 = 1.976

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n)

= 1.976 * (7.1 / 154)

= 1.1

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(30.5 - 1.1)   <   <  (30.5 + 1.1)

29.4 <   < 31.6

Now ,

compare this interval with given interval   29.1 <   < 32.1

Two intervals are different?

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