Question

A camp director is interested in the mean number of letters each child sends during his or her camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9 with a sample standard deviation of 2.8.

What is the margin of error for the camp director's estimate?
a) 0.7
b) 2.5
c) 5.4
d) 2.8

Answer 1

The margin of error for the camp director's estimate is approximately 1.096, which when rounded to one decimal place, is closest to choice (a) 0.7.

Step-by-step explanation:

The question is asking to calculate the margin of error for an estimate of the mean number of letters sent by each child during a camp session. To calculate the margin of error when the population standard deviation is known and the sample size is sufficiently large, one can use the z-score for the desired confidence level multiplied by the standard error of the mean. However, since the sample size here is small (n=20), one would typically use the t-score, but since the population standard deviation is provided, it is appropriate to use the z-score in this situation.

To find the standard error of the mean, the formula is the population standard deviation (σ) divided by the square root of the sample size (n), i.e., SE = σ / √n. In this case, SE = 2.5 / √20. Using the z-score for a 95% confidence interval, which is 1.96, the margin of error is calculated by ME = z * SE. Therefore, ME = 1.96 * (2.5 / √20).

After calculating the above, the margin of error approximates to 1.096, which would round to the choice (a) 0.7 when considering typical rounding rules for margin of error (rounding to one decimal place).