lower limit $ upper limit $ moe $ Solution (a) xbar=58940, s=18980, n=25 =0.01,
ID: 2954579 • Letter: L
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lower limit $ upper limit $ moe $Explanation / Answer
(a) xbar=58940, s=18980, n=25 =0.01, Z(0.005)=2.58 (check normal table) The 99% CI is xbar ± Z*s/n --> 58940 ± 2.58*18980/sqrt(25) --> (49146.32, 68733.68) The margin of error is s/n = 18980/sqrt(25) = 3796 -------------------------------------------------------------------------------------------------------------------- (b) xbar=58940, s=18980, n=65 =0.01, Z(0.005)=2.58 (check normal table) The 99% CI is xbar ± Z*s/n --> 58940 ± 2.58*18980/sqrt(65) --> (52866.22, 65013.78) The margin of error is s/n = 18980/sqrt(65) = 2354.179 ---------------------------------------------------------------------------------------------------------------------- (c) xbar=58940, s=18980, n=118 =0.01, Z(0.005)=2.58 (check normal table) The 99% CI is xbar ± Z*s/n --> 58940 ± 2.58*18980/sqrt(118) --> (54432.09, 63447.91) The margin of error is s/n = 18980/sqrt(118) = 1747.251 --------------------------------------------------------------------------------------------------------------------------- (d) As the sample size increases, the margin of errordecreases. --------------------------------------------------------------------------------------------------------------------------- (e) As the sample sizeincreases, the confidence interval decreases in length.
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