Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals220​, x overbarequals30.3 ​hg, sequals6.3 hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 29.0 hgless thanmuless than31.8 hg with only 18 sample​ values, x overbarequals30.4 ​hg, and sequals2.1 ​hg? What is the confidence interval for the population mean mu​? nothing hgless thanmuless than nothing hg ​(Round to one decimal place as​ needed.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 30.3

sample standard deviation = s = 6.3

sample size = n = 220

Degrees of freedom = df = n - 1 = 220 - 1 = 219

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t_0.005,219 = 2.598

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

= 2.598 * ( 6.3/ 220)

Margin of error = E = 1.1

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

- E < < + E

30.3 - 1.1 < < 30.3 + 1.1

(29.2 < < 31.4)