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.
[tex]p=X/n=97/1750=0.055[/tex]
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.055*0.945}{1750}}\\\\\\ \sigma_p=\sqrt{0.00003}=0.005[/tex]
The critical z-value for a 99% confidence interval is z=2.576.
The margin of error (MOE) can be calculated as:
[tex]MOE=z\cdot \sigma_p=2.576 \cdot 0.005=0.014[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=p-z \cdot \sigma_p = 0.055-0.014=0.041\\\\UL=p+z \cdot \sigma_p = 0.055+0.014=0.069[/tex]
The 99% confidence interval for the population proportion is (0.041, 0.069).
