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 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?

A. 0.3 grams

B. 0.5133 grams

C. 0.8391 grams

D. 1.5 grams

E. 0.06 grams

Answer 1

A.0.39 grams

margin of error at 90% confidence intervals is

M.E = 0.39 grams

Explanation:

Explanation:-

Given a sample of size n = 25

mean of the sample x⁻ = 15 grams

Standard deviation of the sample 'S' = 1.5 grams

margin of error at 90% confidence intervals is determined by

`Margin error = (t_(0.90) S.D)/(√(n) )`

The degrees of freedom ν = n-1 = 25-1 =24

The tabulated value t₀.₉₀ = 1.318

`Margin error = (1.318X 1.5)/(√(25) )`

Margin of error = 0.3954

Conclusion:-

Margin of error at 90% confidence intervals = 0.3954