. Suppose that we want to estimate the true proportion of defectives in a very large shipment of adobe bricks, and that we want to be at least 95% confident that the error is at most 0.04. How large a sample will we need if:

A sample of 601 bricks will be needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

95% confidence level

So , z is the value of Z that has a pvalue of , so .

How large a sample will we need?

A sample of n bricks will be needed.

n is found when M = 0.04.

We have no estimate for the proportion of defective bricks, so we work with the worst case scenario, which is

Rounding up

A sample of 601 bricks will be needed.

b

step-by-step explanation:

the total is 23.97. 0.25% is off so you have to subtracting 25% from 100% and you will get 75% if you convert it to decimal you will get 0.75. you have to find the original because the formula is % times original = total.so you have the total and the % so you just have to plug it in.

168

step-by-step explanation:

i am pretty sure that the answer is 168