To determine which factors will make the confidence interval narrower, let's analyze the formula for a confidence interval for a population mean.

The two formulas are:

1. If the population standard deviation $\sigma$ is known: $\text{CI} = \bar{x} \pm Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$ Where:

• $\bar{x}$ is the sample mean,
• $Z_{\alpha/2}$ is the critical value from the standard normal distribution for the desired confidence level,
• $\sigma$ is the population standard deviation, and
• $n$ is the sample size.

2. If $\sigma$ is not known, and we use the sample standard deviation $s$, the formula becomes: $\text{CI} = \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ Where $t_{\alpha/2}$ is the critical value from the t-distribution.

Now, let’s evaluate the two options:

1. Decrease confidence level:

   • A lower confidence level (e.g., from 95% to 90%) reduces the critical value ($Z_{\alpha/2}$ or $t_{\alpha/2}$), which in turn decreases the margin of error, making the confidence interval narrower.

2. Decrease sample size:

   • Decreasing the sample size $n$ will increase the margin of error (since $\frac{1}{\sqrt{n}}$ becomes larger), making the confidence interval wider, not narrower.

Correct Answer:

• Decrease confidence level will make the confidence interval narrower.
• Decrease sample size will not make the interval narrower, it will make it wider.

Would you like further details or have any questions?

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Here are five related questions:

1. How does the critical value change with different confidence levels?
2. How does the width of a confidence interval depend on the sample standard deviation?
3. What is the impact of increasing the population size on the confidence interval?
4. How do you choose between the z-distribution and t-distribution for confidence intervals?
5. How does the shape of the t-distribution compare to the normal distribution?

Tip: Increasing the sample size is one of the most effective ways to narrow a confidence interval.