The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 13-inch pizzas. She takes a random sample of 37 pizzas and records their mean and standard deviation as 13.50 inches and 1.90 inches, respectively. She subsequently computes the 95% confidence interval of the mean size of all pizzas as [12.89, 14.11]. However, she finds this interval to be too broad to implement quality control and decides to reestimate the mean based on a bigger sample. Using the standard deviation estimate of 1.90 from her earlier analysis, how large a sample must she take if she wants the margin of error to be under 0.5 inch

She must take a sample of 56.

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 Z-table as such z has a p-value of .

That is z with a pvalue of , so Z = 1.96.

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

Standard deviation estimate of 1.90

This means that

Wow large a sample must she take if she wants the margin of error to be under 0.5 inch?

This is n for which M = 0.5. So

Rounding up:

She must take a sample of 56.

#15. square

step-by-step explanation:

because quadrilatural (idk how to spell it) means somthing that is not 3d or like, equal sides (or somthing like that   you know what i meanz)

hope this halps!

bai!

b, i believe, but not sure.