Question 1175140
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Hi
1) Normal Distribution: the population standard deviation σ  = 12.1 ounces 
   Sample Size is 33: Sample mean = 44  
90% confidence Interval, find ME 
   ME = 1.645 (12.1)/√33 = 3.465
CI = 44 ± 3.465

2)  Normal Distribution:  Sample of 12    mean = 66.533   sd = 9.012  
80% CI  z = 1.362 df 11
ME ={{{ t*sigma/sqrt(n)}}}= {{{1.362(9.012)/sqrt(12)}}}= 3.543
80% C.I. = (62.99,70.08)

Wish You the Best in your Studies.
Confidence Level	80%	90%	95%	98%	99%
Two-sided test p-values	0.2	0.1	0.05	0.02	0.01
One-sided test p-values	0.1	0.05	0.025	0.01	0.005
Degrees of Freedom (df=n-1)					
               1	3.078	6.314	12.71	31.82	63.66
               2	1.886	2.92	4.303	6.965	9.925
               3	1.638	2.353	3.182	4.541	5.841
               4	1.533	2.132	2.776	3.747	4.604
               5	1.476	2.015	2.571	3.365	4.032
               6	1.44	1.943	2.447	3.143	3.707
               7	1.415	1.895	2.365	2.998	3.499
               8	1.397	1.86	2.306	2.896	3.355
               9	1.383	1.833	2.262	2.821	3.25
              10	1.372	1.812	2.228	2.764	3.169
              11	1.362	1.796	2.201	2.718	3.106

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