Question

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 random 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?

Answer 1

Answer:

The margin of error at 90% confidence is 0.5133 gram.

Step-by-step explanation:

Margin of error (E) = (critical value × sample standard deviation) ÷ sqrt(n)

sample standard deviation = 1.5 grams

confidence level = 90% = 0.9

significance level = 1 - C = 1 - 0.9 = 0.1 = 10%

n (sample size) = 25

degree of freedom = n - 1 = 25 - 1 = 24

critical value (t) corresponding to 24 degrees of freedom and 10% significance level is 1.711

E = (1.711×1.5) ÷ sqrt(25) = 2.5665 ÷ 5 = 0.5133 gram