Given:

• Population mean: $\mu = 30.00$
• Population standard deviation: $\sigma = 5$
• Confidence level: 90% (for part (a))

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Part (a): Ensuring the sample mean is between 25 and 35 with 90% confidence.

Step 1: Use the z-score corresponding to the 90% confidence level.

For a 90% confidence interval:

• 90% of the data lies between $-z$ and $z$.
• From standard normal tables or a calculator, $z = 1.645$.

Step 2: Determine the range for the sample mean.

The sample mean $\bar{x}$ follows a normal distribution with mean $\mu$ and standard error $\text{SE}$, given by: $\text{SE} = \frac{\sigma}{\sqrt{n}}$ The confidence interval for $\bar{x}$ is: $\mu - z \cdot \text{SE} \leq \bar{x} \leq \mu + z \cdot \text{SE}$ Substitute the values for the problem: $30 - 1.645 \cdot \text{SE} \leq \bar{x} \leq 30 + 1.645 \cdot \text{SE}$ For $\bar{x}$ to be between 25 and 35: $25 = 30 - 1.645 \cdot \text{SE}, \quad 35 = 30 + 1.645 \cdot \text{SE}$

Step 3: Solve for $\text{SE}$.

From $30 - 1.645 \cdot \text{SE} = 25$: $1.645 \cdot \text{SE} = 5 \quad \Rightarrow \quad \text{SE} = \frac{5}{1.645} \approx 3.04$

Step 4: Solve for $n$ using $\text{SE} = \frac{\sigma}{\sqrt{n}}$.

$\text{SE} = \frac{5}{\sqrt{n}} \quad \Rightarrow \quad \sqrt{n} = \frac{5}{\text{SE}} = \frac{5}{3.04} \approx 1.645$ $n = (1.645)^2 \approx 2.71$

Thus, a sample size of at least 3 is required.

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Part (b): Ensuring the standard error of the mean is 1.

The standard error is defined as: $\text{SE} = \frac{\sigma}{\sqrt{n}}$ Set $\text{SE} = 1$: $1 = \frac{5}{\sqrt{n}} \quad \Rightarrow \quad \sqrt{n} = 5$ $n = 5^2 = 25$

Thus, a sample size of 25 is required.

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Justification:

1. For part (a), the sample size ensures the sample mean falls within the desired confidence interval by controlling the standard error appropriately.
2. For part (b), the sample size directly determines the standard error, ensuring it achieves the required value.

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Summary:

• Part (a): $n \geq 3$
• Part (b): $n = 25$

Would you like further clarification or detailed explanations for specific steps? 😊

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Related Questions:

1. How do we interpret z-scores in confidence intervals?
2. Why does the standard error decrease with larger sample sizes?
3. What happens if the population standard deviation is unknown?
4. Can we apply this method if the population distribution is not normal?
5. How does increasing the confidence level affect the required sample size?

Tip:

When solving problems involving sample size, always confirm whether the population standard deviation ($\sigma$) is known or needs to be estimated using sample data.