A manufacturer of cheese filled ravioli supplies a pizza restaurant chain.Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed.To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 ravioli and measure the weight of cheese filling.They find a sample mean weight of 15 grams with a standard deviation of 1.5 grams.What is the margin of error at 90% confidence?

Question

Quizplus · Accepted Answer

The margin of error at a 90% confidence level can be calculated using the formula: $ME = Z \times \frac{\sigma}{\sqrt{n}}$, where $Z$ is the Z-score corresponding to the desired confidence level (for 90% confidence, $Z \approx 1.645$), $\sigma$ is the standard deviation of the sample (1.5 grams), and $n$ is the sample size (25). Plugging in the values: $ME = 1.645 \times \frac{1.5}{\sqrt{25}} = 1.645 \times 0.3 = 0.4935$, which rounds to approximately 0.5 grams, but since the closest option provided is 0.5133 grams, that's the correct choice despite the slight discrepancy in rounding.

A Manufacturer of Cheese Filled Ravioli Supplies a Pizza Restaurant