Question

A smart phone manufacturer is interested in constructing a 99% confidence interval for the proportion of smart phones that break before the warranty expires. 97 of the 1750 randomly selected smart phones broke before the warranty expired. Round your answers to three decimal places. A. With 99% confidence the proportion of all smart phones that break before the warranty expires is between and .

Answer 1

Answer:

With 99% confidence the proportion of all smart phones that break before the warranty expires is between 0.041 and 0.069.

Step-by-step explanation:

We have to calculate a 99% confidence interval for the proportion.

The sample proportion is p=0.055.

$p=X/n=97/1750=0.055$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.055*0.945}{1750}}\\\\\\ \sigma_p=\sqrt{0.00003}=0.005$

The critical z-value for a 99% confidence interval is z=2.576.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=2.576 \cdot 0.005=0.014$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.055-0.014=0.041\\\\UL=p+z \cdot \sigma_p = 0.055+0.014=0.069$

The 99% confidence interval for the population proportion is (0.041, 0.069).