Let's determine the sample size used in a study that resulted in a 95% confidence interval for the population mean being 160 to 170, given the population standard deviation is 25.

Understanding Confidence Intervals

A confidence interval provides a range of values within which the true population parameter (in this case, the population mean $\mu$) is estimated to lie with a certain level of confidence. The general formula for a confidence interval for the population mean when the population standard deviation $\sigma$ is known is:

$$\text{CI} = \bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$

Where:

• $\bar{x}$ is the sample mean.
• $z_{\alpha/2}$ is the z-score corresponding to the desired confidence level ($\alpha$ is the significance level, $1 - \alpha$ is the confidence level).
• $\sigma$ is the population standard deviation.
• $n$ is the sample size.
• The term $z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ is called the Margin of Error (ME).

Calculating the Margin of Error

The given confidence interval is [160, 170]. The width of the confidence interval is the difference between the upper and lower bounds:

$$\text{Width} = \text{Upper Bound} - \text{Lower Bound} = 170 - 160 = 10$$

The margin of error is half of the width:

$$\text{ME} = \frac{\text{Width}}{2} = \frac{10}{2} = 5$$

Finding the Z-score for 95% Confidence

For a 95% confidence level, the significance level $\alpha = 1 - 0.95 = 0.05$. We need the z-score $z_{\alpha/2}$, which is $z_{0.05/2} = z_{0.025}$. The z-score that leaves 0.025 probability in the upper tail (and 0.025 in the lower tail) is 1.96.

So, $z_{0.025} = 1.96$.

Solving for Sample Size (n)

We know the margin of error (ME), the z-score ($z_{\alpha/2}$), and the population standard deviation ($\sigma$). We can use the margin of error formula to solve for the sample size $n$:

$$\text{ME} = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$

Substitute the known values:

$$5 = 1.96 \times \frac{25}{\sqrt{n}}$$

Now, we rearrange the equation to solve for $\sqrt{n}$:

$$\sqrt{n} = \frac{1.96 \times 25}{5}$$

$$\sqrt{n} = \frac{49}{5}$$

$$\sqrt{n} = 9.8$$

To find $n$, we square both sides of the equation:

$$n = (9.8)^2$$

$$n = 96.04$$

The calculated sample size is 96.04. Since the sample size must be a whole number, and the option 96 is provided, this is the value that matches the calculation closely, considering potential rounding or the options available.

Therefore, the size of the sample used in this study is 96.

Revision Table: Key Concepts

Additional Information on Sample Size Calculation

When determining the minimum sample size required to achieve a specific margin of error for a confidence interval, the formula is derived from the ME formula:

$$n = \left(\frac{z_{\alpha/2} \sigma}{\text{ME}}\right)^2$$

In our case, substituting the values:

$$n = \left(\frac{1.96 \times 25}{5}\right)^2$$

$$n = \left(\frac{49}{5}\right)^2$$

$$n = (9.8)^2$$

$$n = 96.04$$

Conventionally, when calculating the *minimum* required sample size, you should always round the result *up* to the next whole number to ensure the desired margin of error is not exceeded. In this specific calculation, 96.04 rounded up would be 97. However, 97 is not among the provided options. The option 96 is the closest integer to the calculated value of 96.04, and matching one of the given options is necessary.