Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals192​, x overbarequals27.8 ​hg, sequals7.8 hg. Construct a confidence interval estimate of the mean. Use a 90​% confidence level. Are these results very different from the confidence interval 26.5 hgless thanmuless than29.9 hg with only 14 sample​ values, x overbarequals28.2 ​hg, and sequals3.5 ​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

a)

sample std dev ,    s =    7.8000
Sample Size ,   n =    192
Sample Mean,    x̅ =   27.8000

Level of Significance ,    α =    0.1          
degree of freedom=   DF=n-1=   191          
't value='   tα/2=   1.6529   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   7.800   / √   192   =   0.5629
margin of error , E=t*SE =   1.6529   *   0.563   =   0.930
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    27.80   -   0.930   =   26.8696
Interval Upper Limit = x̅ + E =    27.80   -   0.930   =   28.7304
90%   confidence interval is (   26.9 < µ <   28.7 )