Question

a student believes that a certain number cube is unfair and is more likely to land with a six facing up. the student rolls the number cube 45 times and the cube lands with a six facing up 12 times. assuming the conditions for inference have been met, what is the 99% confidence interval for the true proportion of times the number cube would land with a six facing up?
0.27 ± 2.58√(0.27(1 - 0.27)/45)
0.73 ± 2.33√(0.73(1 - 0.73)/45)
0.27 ± 2.33√(0.27(1 - 0.27)/45)
0.73 ± 2.58√(0.73(1 - 0.73)/45)

Explanation:

Step1: Calculate the sample proportion

The sample proportion \(\hat{p}=\frac{x}{n}\), where \(x = 12\) (number of successes, i.e., number of times 6 is rolled) and \(n=45\) (number of trials). So \(\hat{p}=\frac{12}{45}\approx0.27\)

Step2: Determine the z - score for 99% confidence interval

For a 99% confidence interval, the z - score \(z\) is \(z = 2.58\) (from the standard normal distribution table)

Step3: Recall the formula for the confidence interval of a proportion

The formula for the confidence interval of a proportion is \(\hat{p}\pm z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)

Substitute \(\hat{p}=0.27\), \(z = 2.58\) and \(n = 45\) into the formula

Answer:

\(0.27\pm2.58\sqrt{\frac{0.27(1 - 0.27)}{45}}\)