Question

a sample of size n=77 is drawn from a population whose standard deviation is o=24. find the margin of error for 95% confidence interval for u. round the answer to at least three decimal places

Answer 1

The margin of error for a 95% confidence interval for the population mean is approximately 5.35 (rounded to three decimal places).

The margin of error for a 95% confidence interval for the population mean
`(\(\mu\))` can be calculated using the formula:

`\[ \text{Margin of Error} = Z * (\sigma)/(√(n)) \]`

Where:

`\( Z \)` is the Z-score corresponding to the desired confidence level (for a 95% confidence interval,
`\( Z \approx 1.96 \))`,

σ is the population standard deviation,

`\( n \)` is the sample size.

Let's plug in the values:

`\[ \text{Margin of Error} = 1.96 * (24)/(√(77)) \]`

Now, let's calculate it:

`\[ \text{Margin of Error} \approx 1.96 * (24)/(√(77)) \]`

`\[ \text{Margin of Error} \approx 1.96 * (24)/(8.775) \]`

`\[ \text{Margin of Error} \approx 1.96 * 2.735 \]`

`\[ \text{Margin of Error} \approx 5.35 \]`

Therefore, the margin of error for a 95% confidence interval for the population mean is approximately 5.35 (rounded to three decimal places).