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From the options given, the only option which would result on having a reduced error margin, is to increase the sample size.
<u>Recall</u><u> </u><u>:</u><u> </u>
<em>Margin of Error</em> = Zcritical × σ/√n
- Increasing the mean would have no impact on the margin of error as it is not a part of the factors which affects the margin of error value.
- Increasing the standard deviation, which is the Numerator will result in an increased margin of error value.
- By raising the confidence level, the critical value of Z will increase, hence widening the margin of error.
- The sample size, n being the denominator, would reduce the value of error margin.
Hence, only the sample size will cause a decrease in the margin of error of the interval.
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