How large a sample n would you need to estimate p with margin of error 0.04 with 95% confidence? Assume that you don’t know anything about the value of p . 1037 256 601 423

(b)  

Since we don't know about p we can assume . And replacing into equation (b) the values from part a we got:  

And rounded up we have that n=601

Step-by-step 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".  

estimation for the sample proportion

n sample size selected

Confidence =0.95 or 95%

The population proportion have the following distribution  

Solution to the problem

In order to find the critical values we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by and . And the critical values would be given by:  

The margin of error for the proportion interval is given by this formula:  

(a)  

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

(b)  

Since we don't know about p we can assume . And replacing into equation (b) the values from part a we got:  

And rounded up we have that n=601

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.