Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals176​, x overbarequals32.7 ​hg, sequals6.8 hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 31.3 hgless thanmuless than34.5 hg with only 18 sample​ values, x overbarequals32.9 ​hg, and sequals2.4 ​hg?

Answer 1

Level of Significance ,    α =    0.01
sample std dev ,    s =    6.8000
Sample Size ,   n =    176
Sample Mean,    x̅ =   32.700

degree of freedom=   DF=n-1=   175  
't value='   tα/2=   2.6042   [Excel formula =t.inv(α/2,df) ]
          
Standard Error , SE =   s/√n =   0.5126  
margin of error ,   E=t*SE =   1.335  
99% confidence interval is           
Interval Lower Limit=   x̅ - E =    31.4   
Interval Upper Limit=   x̅ + E =    34.0

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with 18 samples and  xbar =32.9 ​hg, and s =2.4 ​hg
confidence interval is           
Interval Lower Limit=   x̅ - E =    31.3
Interval Upper Limit=   x̅ + E =    34.5

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these results are not very different from the confidence interval (31.3,34.5).

NO, because confidence interval limits are similar.