Sample Mean, x̅ = ΣX/n = 30.5833
sample std dev , s = √(Σ(X- x̅ )²/(n-1) )
= 9.3200
Sample Size , n = 24
Level of Significance , α =
0.05
degree of freedom= DF=n-1= 23
't value=' tα/2= 2.07 [Excel
formula =t.inv(α/2,df) ]
Standard Error , SE = s/√n = 9.32/√24=
1.9024
margin of error , E=t*SE = 2.0687
* 1.9024 = 3.94
confidence interval is
Interval Lower Limit = x̅ - E = 30.58
- 3.9355 = 26.65
Interval Upper Limit = x̅ + E = 30.58
- 3.9355 = 34.52
95% confidence interval is ( 26.65 < µ
< 34.52 )
There is 95% confidence that
true mean lies within confidence interval
.................
Level of Significance, α =
0.05
Number of Items of Interest, x =
12
Sample Size, n = 24
Sample Proportion , p̂ = x/n =
0.5000
z -value = Zα/2 = 1.96 [excel formula
=NORMSINV(α/2)]
Standard Error , SE = √[p̂(1-p̂)/n] =
0.102062
margin of error , E = Z*SE = 1.960
* 0.10206 = 0.2000
95% Confidence Interval is
Interval Lower Limit = p̂ - E = 0.50000
- 0.20004 = 0.3000
Interval Upper Limit = p̂ + E = 0.50000
+ 0.20004 = 0.7000
..............
Please let me know in case of any doubt.
Thanks in advance!
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