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 with mean weight of 15.26 grams and standard deviation of 2.56 grams. The margin of error (ME) to construct a 99% confidence interval for the true mean amount of cheese filling is

A) 2.064 B) 1.432 C) 3.745 D) 1.276 E) 1.797

b. 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 the 99% confidence interval of 14.16 to 15.84 grams. Which of the following is the correct interpretation?

I. We are 99% confident that the mean weight of cheese filling in all ravioli made by this process is between 14.16 and 15.84 grams.

II. 99% of all ravioli made by this process will have cheese filling weights between 14.16 and 15.84 grams.

III.We are 99% confident that the interval (14.16, 15.84) contains the mean weight of cheese filling in all ravioli made by this process.

IV. The weight of cheese filling in the ravioli is between 14.16 and 15.84 grams 99% of the time.

A) I only

B) III only

C)I and III.

D) II and III

E) I and IV

Answer 1

Answer: a. B) 1.432

              b. E) I and IV

Step-by-step explanation:

a. Margin of error to construct a 99% confidence interval is calculated as

$ME=t.\frac{s}{\sqrt{n}}$

where

t is the critical value expressed as t-score, which comes from t-distribution;

s is sample standard deviation

n is sample size

In this case, a t-score is used because the population standard deviation is unknown and the sample size is small, n < 30.

The way to find t-score is:

1) Define the degree of freedom, which is sample size minus 1: df = n - 1;

2) Compute alpha: $\alpha=1-\frac{confidence}{100}$;

3) Find the critical probability: $p=1-\frac{\alpha}{2}$

With the df and critical probability, consulte the t-distribution table and find the score.

For the cheese filling:

df = 24

p = 0.995

t-score = 2.797

Calculating margin of error:

$ME=2.797.\frac{2.56}{\sqrt{5}}$

ME = 1.432

The margin of error to construct a 99% confidence interval for the true mean amount of cheese filling is 1.432.

b. Confidence interval is an interval in which we have a percentage of certainty that the true population mean is. For example, with a 99% confidence interval of 14.16 to 15.84 grams, we are 99% confident that the true mean weight of cheese filling is between those numbers or that the weight of cheese filling is between 14.16 and 15.84 grams 99% of the time.