1 2 3 4 5 6 7 8 48:50 All else being equal, if you cut the sample size in half, how does this affect the margin of error when using the sample to make a statistical inference about the mean of the normally distributed population from which it was drawn? \[ M E=\frac{z \bullet s}{\sqrt{n}} . \] The margin of error is multiplied by \( \sqrt{0.5} \). The margin of error is multiplied by \( \sqrt{2} \). The margin of error is multiplied by 0.5 . The margin of error is multiplied by 2 .

When you cut the sample size in half, the new sample size \( n' \) becomes \( n/2 \). Plugging this into the margin of error formula \( ME = \frac{z \cdot s}{\sqrt{n'}} \) gives us \( ME = \frac{z \cdot s}{\sqrt{n/2}} = \frac{z \cdot s \cdot \sqrt{2}}{\sqrt{n}} \). That means the margin of error is multiplied by \( \sqrt{2} \). So your inference gets sloppier—like trying to hit a target with half the darts! In real-world terms, this is crucial for researchers and businesses. If you’re conducting a survey and decide to halve your sample size to save resources, your margin of error will increase. This means any conclusions drawn could be less reliable, leading to poor decision-making. So, always remember: quality in sample size counts!