Question

We want to estimate the population mean within 8, with a 80% level of confidence. the population standard deviation is estimated to be 21. how large a sample is required? (round the intermediate calculation to 2 decimal places. round the final answer to the nearest whole number.)

sample size:

Answer 1

To estimate the population mean within 8, with an 80% level of confidence and a population standard deviation of 21, a sample size of approximately 42 is required.

Step-by-step explanation:

In order to determine the required sample size, we use the formula for the margin of error in estimating the population mean:

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

Here,
`\( Z \)` is the Z-score corresponding to the desired confidence level. For an 80% confidence level,
`\( Z \)` is approximately 1.28. The population standard deviation
`\( \sigma \)` is given as 21, and
`\( n \)` is the required sample size.

The formula can be rearranged to solve for the sample size
`(\( n \))`:

`\[ n = \left(\frac{Z * \sigma}{\text{Margin of Error}}\right)^2 \]`

Substituting the provided values:

`\[ n = \left((1.28 * 21)/(8)\right)^2 \]`

Calculating this yields a sample size of approximately 42. Thus, a sample size of 42 is needed to estimate the population mean within 8, with an 80% level of confidence and a population standard deviation of 21.

In summary, a larger sample size provides a more precise estimate of the population mean. The Z-score accounts for the desired confidence level, and the margin of error ensures the accuracy of the estimation within the specified range. The calculations, rounded to two decimal places in intermediate steps, lead to a final answer rounded to the nearest whole number for practical application.