The J.R. Simplot Company produces frozen French fries that are then sold to customers such as McDonald's. The "prime" line of fries has an average length of 6.00 inches with a standard deviation of 0.50 inch. To make sure that Simplot continues to meet the quality standard for "prime" fries, they plan to select a random sample of n = 100 fries each day. The quality analysts will compute the mean length for the sample. They want to establish limits on either side of the 6.00 inch mean so that the chance of the sample mean falling within the limits is 0.99. What should these limits be?
A) Approximately ±0.13 inches
B) Within the approximate range of 5.87 inches to 6.13 inches
C) Within the range of about 4.71 inches to 7.29 inches
D) Approximately ±1.29 inches

Question

Quizplus · Accepted Answer

Given a 99% confidence level and a sample size of 100, we use the formula for the margin of error in a sample mean, which incorporates the standard deviation and the Z-score corresponding to the confidence level. For a 99% confidence level, the Z-score is approximately 2.58. The standard error (SE) is the standard deviation divided by the square root of the sample size (n), so SE = 0.50 / sqrt(100) = 0.05. The margin of error (ME) is Z * SE = 2.58 * 0.05 = 0.129, or approximately 0.13 inches. Therefore, the limits are 6 ± 0.13, which gives us the range of 5.87 inches to 6.13 inches.

The JR Simplot Company Produces Frozen French Fries That Are Then