Question

Suppose that the standard deviation of an observation in a given population is known to be 15. How many observations in a sample are needed to estimate the mean of the sample within ±5 of the population mean with 95% confidence?

Answer 1

To estimate the population mean within ±5 with 95% confidence and a known standard deviation of 15, a total of 35 observations are needed.

Step-by-step explanation:

To estimate the population mean within ±5 with 95% confidence when the standard deviation (σ) is known to be 15, we use the formula for the sample size (n) in relation to the margin of error (E), standard deviation, and the z-score corresponding to the desired confidence level:

n = (z * σ / E)^2

For a 95% confidence level, the z-score is 1.96. Plugging the numbers into the formula:

n = (1.96 * 15 / 5)^2

n = (1.96 * 3)^2

n = (5.88)^2

n = 34.5744

Since we can't have a fraction of an observation, we round up to the nearest whole number. Therefore:

n = 35 observations are needed to be 95% confident that the sample mean is within ±5 of the population mean.

Answer 2

To estimate the mean of the sample within ±5 of the population mean with 95% confidence, approximately 34 observations in the sample are needed.

Step-by-step explanation:

The formula for the margin of error
`(\(E\))` in a confidence interval is given by:

`\[ E = Z * \left((\sigma)/(√(n))\right) \]`

where:

-
`\(Z\)` is the z-score corresponding to the desired level of confidence,

-
`\(\sigma\)` is the population standard deviation,

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

For a 95% confidence interval,
`\(Z\)` is approximately 1.96 (assuming a normal distribution). Given that
`\(\sigma = 15\)` and
`\(E = 5\)`, we can rearrange the formula to solve for
`\(n\)`:

`\[ n = \left((Z * \sigma)/(E)\right)^2 \]`

Substituting the values, we get:

`\[ n = \left((1.96 * 15)/(5)\right)^2 \approx 34 \]`

Therefore, approximately 34 observations in the sample are needed to estimate the mean of the sample within ±5 of the population mean with 95% confidence.