Question

Polokwane Butchery supplies Vienna sausages in the entire KZN province. An inspector wants to investigate the marked mass shown on Vienna sausages. A pilot study showed a mean of 11.8kg per pack and a variance of 0.49kg. How many packs should the inspector sample to be 99% confident that the sample mean will differ by at most 0.2kg?

a. 53
b. 73
c. 87
d. 105

Answer 1

To calculate the sample size required to be 99% confident that the sample mean will differ by at most 0.2kg, use the formula n = (Z * σ / E) ^ 2, where n is the sample size, Z is the Z-score corresponding to the confidence level, σ is the standard deviation, and E is the maximum allowable error. Plugging in the given values, the sample size is 73.

Step-by-step explanation:

To calculate the sample size required to be 99% confident that the sample mean will differ by at most 0.2kg, we can use the formula:

n = (Z * σ / E) ^ 2

Where:

• n is the sample size
• Z is the Z-score corresponding to the confidence level
• σ is the standard deviation
• E is the maximum allowable error (0.2kg in this case)

Given that the mean is 11.8kg, the standard deviation is 0.49kg, and the desired confidence level is 99%, we can calculate the Z-score using a Z-table or calculator. The Z-score for a 99% confidence level is approximately 2.576. Plugging the values into the formula:

n = (2.576 * 0.49 / 0.2) ^ 2 = 73 (rounded)

Therefore, the inspector should sample 73 packs of Vienna sausages to be 99% confident that the sample mean will differ by at most 0.2kg.