Question

A random sample of 60 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact test, and on 19 of these helmets some damage was observed (a) Find a 95% two-sided confidence interval on the true proportion of helmets of this type that would show damage from this test. Round your answers to 3 decimal places. (b) Using the point estimate of p obtained from the preliminary sample of 60 helmets, how many helmets must be tested to be 95% confident that the error in estimating the true value of p is less than 0.02? (c) How large must the sample be if we wish to be at least 95% confident that the error in estimating p is less than 0.02, regardless of the true value of p?

Answer 1

Here

sample proportion of helmets where damage was observed $\hat{p} = \frac{19}{60} = 0.316667$

sample size n = 60

a 95% confidence interval on the true proportion of helmets where damage was observed

$=(\hat{p} - z_{0.025}*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}, \hat{p} + z_{0.025}*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}})$

$=(0.316667 - z_{0.025}*\sqrt{\frac{0.316667*(1-0.316667)}{60}}, 0.316667 + z_{0.025}*\sqrt{\frac{0.316667*(1-0.316667)}{60}})$$=(0.316667 - 1.95996*\sqrt{\frac{0.316667*(1-0.316667)}{60}}, 0.316667 + 1.95996*\sqrt{\frac{0.316667*(1-0.316667)}{60}})$

$=(0.198963,0.434370)$

b) To find n such that

Margin of error < 0.02

$\Rightarrow z_{0.025}*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}} < 0.02$

$\Rightarrow 1.95996*\sqrt{\frac{0.316667*(1-0.316667)}{n}} < 0.02$

$\Rightarrow \sqrt{n} > \frac{1.95996*\sqrt{0.316667*(1-0.316667)}}{0.02}$

$\Rightarrow \sqrt{n} > 45.5863$

$\Rightarrow n > 45.5863^2 = 2078.11$

$\Rightarrow n \approx 2079$