The chamber of commerce in a beach resort town wants to estimate the proportion of visitors who are repeat visitors. From previous experience they believe the portion is in the vicinity of 0.5 and they want to estimate the proportion to within ± 0.03 percentage points with 95 percent confidence. The sample size they should use is: Question 30 options: n = 95 n = 545 n = 1068 n = 33

And rounded up we have that n=1068

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".  

Solution to the problem

In order to find the critical value 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 value 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)  

And replacing into equation (b) the values from part a we got:

And rounded up we have that n=1068

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23