Question

Find a \(99 \%\) lower confidence bound for the binomial proportion \(p\) when a random sample of \(n=400\) trials produced \(x=196\) successes.

Short answer

Answer: The 99% lower confidence bound for the binomial proportion is approximately 0.4415.

Step-by-step solution

Calculate the Sample Proportion

Firstly, calculate the sample proportion \(\hat{p}\) by dividing the number of successes \(x\) by the number of trials \(n\): \(\hat{p} = \frac{x}{n} = \frac{196}{400} = 0.49\)

Compute the Standard Error

Next, compute the standard error \(SE\) of the sample proportion using the following formula: \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.49 \times 0.51}{400}} \approx 0.0249\)

Find the Critical Value for a 99% Confidence Level

Since we want a 99% lower confidence bound, we need a critical value of \(z\) for a standard normal distribution. By checking a standard normal table or using statistical software that finds percentiles for the standard normal distribution, we can find the critical value \(z\) for a one-tailed test with a 99% confidence level. This critical value is approximately \(z = -2.33\) (note the negative sign as it is a lower bound).

Calculate the 99% Lower Confidence Bound

Lastly, we will now calculate the 99% lower confidence bound using the sample proportion, critical value, and standard error: \(LCB = \hat{p} + z \times SE = 0.49 + (-2.33) \times 0.0249 \approx 0.4415\) Thus, a 99% lower confidence bound for the binomial proportion \(p\) is approximately \(0.4415\).