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

`n=((1.640(2.5))/(2))^2 =4.2 \approx 5`

So the answer for this case would be n=5 rounded up to the nearest integer

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=22` represent the sample mean

`\mu` population mean (variable of interest)

s=2.5 represent the sample standard deviation

n represent the sample size

ME = 2 represent the margin of error accepted

Solution to the problem

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (2)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

`n=((z_(\alpha/2) s)/(ME))^2` (3)

We can use as estimator for the population deviation the sample deviation
`\hat \sigma = s`

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got
`z_(\alpha/2)=1.640`, replacing into formula (3) we got:

`n=((1.640(2.5))/(2))^2 =4.2 \approx 5`

So the answer for this case would be n=5 rounded up to the nearest integer