**Reducing the margin of error while keeping the sample size constant will decrease the confidence.****The margin of error will be smaller for a bigger sample size if the confidence level is constant.****The confidence will increase for a larger sample size if the margin of error is fixed.****If the sample size is doubled while the confidence level is kept the same, the margin of error will be halved.**

This question aims to find the confidence interval for different scenarios in the statistical data.

The concepts required for this question are confidence-interval value, the margin of error, sample mean, and confidence level. The confidence interval is the certainty value of statistical data while the confidence level is the percentage value of how confident you are about a survey’s outcome. The margin of error tells us how much error can occur in the confidence interval value.

The confidence interval is given as:

\[ CI = \overline{x} \pm z \frac{\sigma}{\sqrt{n}} \]

**Expert Answer:**

**1)** If we reduce the margin of error for a given sample size, it should increase the confidence. As the margin of error increases, the uncertainty increases with it. Mathematically, we can also prove that by reducing the margin of error, our confidence interval will be more precise. Hence, the given statement is $false$.

**2)** $z$ is the confidence value while $n$ is the sample size with $\sigma$ as the standard deviation. If we increase the sample size, it will reduce the margin of error as the sample size is in inverse relation. Hence, the statement is $true$.

**3)** Fixing the margin of error while increasing the sample is an ambiguous statement because the margin of error is dependent on the sample size and its standard deviation. We can fix the confidence value and standard deviation while we increase the sample size. This will increase the certainty of the confidence interval. Hence, the statement is $true$.

**4)** This statement is $false$, as we can see in the formula of the confidence interval that the sample size is under the square root. To halve the margin of error, we would require a sample size that is $4$ times larger.

**Numerical Results:**

If we change the sample size to $n=4n$, the margin of error becomes half.

\[ CI = \overline{x} \pm z \frac{\sigma}{\sqrt{4n}} \]

\[ CI = \overline{x} \pm \dfrac{1}{2} (z \frac{\sigma}{\sqrt{n}}) \]

**Example:**

A survey of $400$ people found the mean weight to be $67 kg$ with a standard deviation of $8.6$ at a $95\%$ confidence level. Find the confidence interval.

\[ n = 400, \sigma = 8.6, \overline{x} = 67 \]

The $z$ value of $95\%$ confidence level is $1.96$ from the $z-table$.

\[ CI = 67 \pm 1.96 \frac{8.6}{\sqrt{400}} \]

\[ CI = 67 \pm 0.843 \]

The confidence interval for this survey lies in $66.16 kg$ to $67.84 kg$ with a confidence level of $95\%$.