44. A sample of 100 employees of an entity had a mean weight of 170 pound with a standard deviation of 5 pounds. Which of the following sample size will result in a confidence interval of the mean $$ \pm 1 $$ pound for this data at the $$ 90 \% $$ confidence interval? A) 3 B) 280 C) 20 D) 68

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To determine the required sample size $$ n $$ for a confidence interval of the mean $$ \pm 1 $$ pound at a $$ 90\% $$ confidence level, we can use the formula for the margin of error (ME) in a confidence interval:

$$
\text{ME} = z^* \left( \frac{s}{\sqrt{n}} \right)
$$

Where:
- $$ z^* $$ is the z-score corresponding to the desired confidence level.
- $$ s $$ is the sample standard deviation.
- $$ n $$ is the sample size.

**Given:**
- $$ s = 5 $$ pounds
- Desired $$ \text{ME} = 1 $$ pound
- Confidence level = $$ 90\% $$, so $$ z^* \approx 1.645 $$

**Setting up the equation:**

$$
1.645 \left( \frac{5}{\sqrt{n}} \right) = 1
$$

**Solving for $$ n $$:**

1. Divide both sides by $$ 1.645 $$:

$$
\frac{5}{\sqrt{n}} = \frac{1}{1.645} \approx 0.607
$$

2. Multiply both sides by $$ \sqrt{n} $$ and then divide by 0.607:

$$
\sqrt{n} = \frac{5}{0.607} \approx 8.23
$$

3. Square both sides to solve for $$ n $$:

$$
n \approx (8.23)^2 \approx 67.7
$$

Since the sample size must be a whole number, we round up to the nearest whole number, resulting in $$ n \approx 68 $$.

**Answer:**  
68
A sample size of 68 is needed to achieve a $$ \pm 1 $$ pound confidence interval at a $$ 90\% $$ confidence level.

44. A sample of 100 employees of an entity had a mean weight of 170 pound with a standard deviation of 5 pounds. Which of the following sample size will result in a confidence interval of the mean \( \pm 1 \) pound for this data at the \( 90 \% \) confidence interval? A) 3 B) 280 C) 20 D) 68

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