Question

A sample of 56 randomly selected cellphones of a new model were checked for a particular defect. Of these, 38 of the cellphones were found to be defective. Find a 80% confidence interval for p. Use the Agresti-Coull method.

Answer 1

Answer:

Confidence Interval: (0.60, 0.76)

Step-by-step explanation:

The provided information is:

Sample Size (n) = 56

Consider X be the number of defective cellphones. Then X = 38

Proportion of defective cellphones is: $p=\dfrac{38}{56}=0.68$

At 80% confidence interval, value of z is 1.28

Thus, the 80% confidence interval for proportion is given as:

$p\pm z\sqrt{\frac{p(1-p)}{n}}}\\=0.68\pm1.28\sqrt{\frac{0.68\times0.32}{56}}\\=(0.60,0.76)$