College officials want to estimate the percentage of students who carry a gun, knife, or other such weapon. How many randomly selected students must be surveyed in order to be 96% confident that the sample percentage has a margin of error of 1.5 percentage points?
(a) Assume that there is no available information that could be used as an estimate of p̂.
(b) Assume that another study indicated that 4% of college students carry weapons.
I have done the following
a) alpha = 1 - 0.96
Margin of error = 0.015
p hat = 0.04
q hat = 0.04
z asterisk = z alpha/2 = 0.02 = 0.5080
I used the formula
n = p hat * q hat * (z asterisk/margin of error)^2
= 0.04 * 0.04 * (0.5080 / 0.015)^2
= 1.6129
b) I did the same formulas and answered this question assuming that the 4% = 0.04 of college students are p hat, which is the same as the previous question
This seems really wrong, what are the correct answers and solutions?
Suppose you sample $n$ students at random and find that $X$ carry weapons. It seems that the style of 96% CI you are using is as follows.
$$\hat p \pm 2.05\sqrt{\frac{\hat p(1-\hat p)}{n}},$$
where $\hat p = X/n$ and 2.05 cuts 2% of the area from the upper tail of the standard normal distribution.
Let's do part (b) first. You want the margin of error to be 1.5% = 0.015. From the previous survey we have the estimate $p \approx 0.04.$ So you set $$2.05\sqrt{\frac{\hat p(1-\hat p)}{n}} \approx 2.05\sqrt{\frac{0.04(0.96)}{n}} = 0.015$$ and solve the equality for $n.$
Now for part (a). Because you know nothing about the value of $p$ you want to protect against the 'worst case scenario', which is when the margin of error is the largest. That is when $p = 1/2.$ So in this case, you set $$2.05\sqrt{\frac{\hat p(1-\hat p)}{n}} \le 2.05\sqrt{\frac{0.5(1-0.5)}{n}} = 0.015$$ and solve the equality for $n,$ which will be considerably larger than in part (b). [Several thousand.]
In case you are wondering why I chose $p = 1/2$ as a way to maximize the margin of error, the crucial part is $p(1=p)$ and I have graphed it below. You can see that the maximum is at $p = 1/2.$