Were the values used correct when calculating the interval?
Error: The wrong percentage was substituted in the formula. Correction:
Practice makes perfect
In a random sample of 2500 consumers, 61 % prefer Game A over Game B. We want to describe and correct the error in giving an interval that is likely to contain the exact percent of all consumers who prefer Game A over Game B.
Margin of error = ± 1/sqrt(n)
Now, this means that if the percent of the sample responding a certain way is p, in this case 61 %, then the percent of the population who would respond the same way is likely to be less than the margin of error away from p. Therefore, it is likely to be between the two following values.
p- 1/sqrt(n) and p+ 1/sqrt(n)
If we now take a look at the solution, we see that an incorrect value must have been used for the margin of error. Let's substitute the correct value, which is 0.02= 2 %, and see what we get.
