Question

A manufacturer reported a sample mean = 22.0 g and a sample standard deviation = 2.5 g based on a sample of 20 of their products. They hope to design future quality control tests to satisfy several criteria: • The calculated error for a sample set (the error is the difference between the true mean and mean of the sample set) does not exceed 2.0 g. • To save money on those tests, they want to collect as few samples as possible. • They want more than 90% of the confidence intervals they calculate to include the true mean. What value of n would you recommend for future tests?

Answer 1

Answer:

Step-by-step explanation:

Given that a manufacturer reported a sample mean = 22.0 g and a sample standard deviation = 2.5 g based on a sample of 20 of their products.

The margin of error should not exceed 2.0 gm.

For 90% confidence interval to get margin of error 2 gm

we can say that margin of error = 1.645*std error

Std error = std deviation/square root of n

Hence if margin of error should not exceed 2 gm

we have

$1.645*\frac{s}{\sqrt{n} } \leq 2\\1.645*\frac{2.5}{\sqrt{n} } \leq 2\\\sqrt{n} \geq 2.05625\\n\geq 4.22$

n should be atleast 5.

But for using normal critical value n should be atleast 30.

So better to have sample size atleast 30.