You're a quality control manager in a fruit juice company. You want to choose a sample size for a quality check on the cans used to package the juice. From
a batch of 30,000 cans, you want a sample size that will allow you to closely
approximate the population parameters for the entire batch. Which sample
size will most likely produce the least variability among different samples?

246.

Step-by-step explanation:

a) The 95% confidence level two-tail confidence interval for the mean value of the key index of this batch is between 98.22 and 99.78

b) The minimum sample size to achieve this is 246.

Step-by-step explanation:

We have that to find our  level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so  

Now, find the margin of error M as such

In which  is the standard deviation of the population(square root of the variance) and n is the size of the sample. So in this question,  

The lower end of the interval is the sample mean subtracted by M. So it is 99 - 0.78 = 98.22

The upper end of the interval is the sample mean added to M. So it is 99 + 0.78 = 99.78.

a) The 95% confidence level two-tail confidence interval for the mean value of the key index of this batch is between 98.22 and 99.78

(b) (5 points) If we want the sampling error to be no greater than 0.5, what is the minimum sample size to achieve this based on the same confidence level with part (a)

We need a sample size of n

n is found when  

Then

Rounding up

The minimum sample size to achieve this is 246.

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.

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