A toy company would like to estimate the proportions of toys produced by their younger workers. In a random sample of 300 toys, they found that 75 were defective. Construct a 98% confidence interval for the population proportion of the toys that are defective.
Can someone explain how I would go about this?
I recall that the formula for developing a confidence interval is (point estimate)$\pm$(critical value)(standard error).
Since we are working with a 98% confidence interval, I'm pretty sure that the critical value is 2.326.
Am I on the right track? How do I proceed?
$\endgroup$ 11 Answer
$\begingroup$Yes, you are on the right track. The approximated confidence interval is
$$\large{\left[\hat p-z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}, \ \hat p+z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}\right]}$$
With $\alpha=1-0.98=0.02 \Rightarrow 1-\frac{\alpha}{2}=0.99$. And $z_{0.99}=2.326$. The other values are $\hat p=\frac{75}{300}=0.25$ and $n=300.$
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