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Sagot :
Answer:
95%: (3278.354 ; 3270.083)
99% : (3221.646 ; 3278.354)
Step-by-step explanation:
Given :
Sample size, n = 12
Mean, xbar = 3250
Sample standard deviation = √1000
The 95% confidence interval :
Mean ± Margin of error
Margin of Error = Tcritical * s/√n
Tcritical at 0.05, df=12-1 = 11 ;
Tcritical at 95% = 2.20
Hence,
Margin of Error = (2.20 * √1000/√12) = 20.083
Confidence interval : 3250 ± 20.083
Lower boundary = 3250 - 20.083 = 3229.917
Upper boundary = 3250 + 20.083 = 3270.083
2.)
The 99% confidence interval :
Mean ± Margin of error
Margin of Error = Tcritical * s/√n
Tcritical at 0.01, df=12-1 = 11 ;
Tcritical at 99% = 3.106
Hence,
Margin of Error = (3.106 * √1000/√12) = 28.354
Confidence interval : 3250 ± 28.354
Lower boundary = 3250 - 28.354 = 3221.646
Upper boundary = 3250 + 28.354 = 3278.354
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