Solution:
Given data is ,
random sample n = 11
sample mean (x) = 920
standard deviation (s) = 25
a) 95% Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 11- 1 ) = 2.228
920 ± t(0.05/2, 11 -1) * 25/√(11)
Lower Limit = 920 - t(0.05/2, 11 -1) 25/√(11)
Lower Limit = 903.205
Upper Limit = 920 + t(0.05/2, 11 -1) 25/√(11)
Upper Limit = 936.795
95% Confidence interval is ( 903.205 , 936.795
)
b) 95% Confidence Interval is
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 11- 1 ) = 2.228
920 ± t(0.05/2, 11 -1) * 50/√(11)
Lower Limit = 920 - t(0.05/2, 11 -1) 50/√(11)
Lower Limit = 886.410
Upper Limit = 920 + t(0.05/2, 11 -1) 50/√(11)
Upper Limit = 953.590
95% Confidence interval is ( 886.410 , 953.590
)
c) 95% Confidence Interval is
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 11- 1 ) = 2.228
920 ± t(0.05/2, 11 -1) * 100/√(11)
Lower Limit = 920 - t(0.05/2, 11 -1) 100/√(11)
Lower Limit = 852.819
Upper Limit = 920 + t(0.05/2, 11 -1) 100/√(11)
Upper Limit = 987.181
95% Confidence interval is ( 852.819 , 987.181
)
a random sample of 11 items is drawn from a population whose standard deviation is unknown....
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