Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals165​, x overbarequals29.1 ​hg, sequals7.2 hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 26.0 hgless thanmuless than31.2 hg with only 12 sample​ values, x overbarequals28.6 ​hg, and sequals2.9 ​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

Answer:

Given,

xbar = 28.6

standard deviation s = 2.9

sample = 12

degree of freedom = n - 1

= 12 - 1

= 11

t(alpha/2 , df) = t(0.005,11) = 3.11

99% CI = xbar +/- t*s/sqrt(n)

substitute values

= 28.6 +/- 3.11*2.9/sqrt(12)

= 28.6 +/- 2.60

= (26.0 , 31.2)

26.0 < u < 31.2

One can be 99% confident that the true population mean of new born girls is lies between 26.0 and 31.2

Now,

n = 165 , xbar = 29.1 , s = 7.2

degree of freedom = n - 1 = 165-1 = 164

t(alpha/2,df) = t(0.005,164) = 2.61

99% CI = xbar +/- t*s/sqrt(n)

substitute values

= 29.1 +/- 2.61*7.2/sqrt(165)

= 29.1 +/- 1.46

= (27.64 , 30.56)

The 95% confident that the true population mean is 27.64 < u < 30.56. The confidence intervals are different.