Question

Assume that a sample is used to estimate a population mean. Find the margin of error that corresponds to a sample of size 9 with a mean of 79.5 and a standard deviation of 16.8 at a confidence level of 99.5%. Report margin of error to one decimal place.​

Answer 1

The margin of error is 18.8. (rounded to one decimal place).

To find the margin of error (M.E.) for a sample of size 9 with a mean of 79.5 and a standard deviation of 16.8 at a confidence level of 99.5%, we can use the formula:

`[M.E. = t_{\text{critical}} * \frac{\text{standard deviation}}{\sqrt{\text{sample size}}}]`

First, we need to find the critical t-value for a 99.5% confidence level with 8 degrees of freedom (sample size - 1). Using a t-distribution table or a calculator, the critical t-value is approximately 3.355.

Now, we can calculate the margin of error:

`[M.E. = 3.355 * (16.8)/(√(9))]`

`[M.E. = 3.355 * (16.8)/(3)]`

`[M.E. = 3.355 * 5.6]`

`[M.E. \approx 18.788]`

Rounding to one decimal place, the margin of error is approximately 18.8.