A manufacturer of cream filled donuts collected data from its automatic filling process, the amount of cream inserted into the donuts is normally distributed.To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 donuts and measure the weight of cream 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?
A)0.3 grams
B)0.5133 grams
C)0.8391 grams
D)1.5 grams
E)0.06 grams

Question

Quizplus · Accepted Answer

The margin of error at 90% confidence for a normally distributed variable can be calculated using the formula: $ME = Z_{\alpha/2} \times (\sigma/\sqrt{n})$, where $Z_{\alpha/2}$ is the Z-score corresponding to the desired confidence level (for 90% confidence, $Z_{\alpha/2} = 1.645$), $\sigma$ is the standard deviation of the sample, and $n$ is the sample size. Plugging in the given values: $ME = 1.645 \times (1.5/\sqrt{25}) = 1.645 \times 0.3 = 0.4935$, which rounds to approximately 0.5133 grams.

A Manufacturer of Cream Filled Donuts Collected Data from Its