Given Information:
Number of defective cups = 15
Total number of cups = 300
Required Information:
a) defective proportion = p = ?
b) 95% confidence interval of defective proportion = ?
Answer:
a) defective proportion = p = 0.05 = 5%
b) 95% confidence interval of defective proportion = (2.5%, 7.5%)
Step-by-step explanation:
The estimated defective proportion is given by
p = Number of defective cups/Total number of cups
p = 15/300
p = 0.05
p = 5%
The confidence interval of defective proportion is given by
![CI = p \pm z\sqrt{\frac{p(1-p)}{n} }](https://tex.z-dn.net/?f=CI%20%3D%20p%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D)
Where p is the defective proportion, z is the z-score corresponding to 95% confidence level and n is the total number of cups.
The z-score corresponding to 95% confidence level is 1.96
![CI = 0.05 \pm 1.96\sqrt{\frac{0.05(1-0.05)}{300} }](https://tex.z-dn.net/?f=CI%20%3D%200.05%20%5Cpm%201.96%5Csqrt%7B%5Cfrac%7B0.05%281-0.05%29%7D%7B300%7D%20%7D)
![CI = 0.05 \pm 1.96(0.01258)](https://tex.z-dn.net/?f=CI%20%3D%200.05%20%5Cpm%201.96%280.01258%29)
![CI = 0.05 \pm 0.025](https://tex.z-dn.net/?f=CI%20%3D%200.05%20%5Cpm%200.025)
![Upper = 0.05 + 0.025 = 0.075](https://tex.z-dn.net/?f=Upper%20%3D%200.05%20%2B%200.025%20%3D%200.075)
![Lower = 0.05 - 0.025 = 0.025](https://tex.z-dn.net/?f=Lower%20%3D%200.05%20-%200.025%20%3D%200.025)
![CI = (0.025, 0.075)](https://tex.z-dn.net/?f=CI%20%3D%20%280.025%2C%200.075%29)
CI = (2.5%, 7.5%)
So that means we are 95% confident that the defective proportion is between 2.5% to 7.5%.
Zack should not return this lot since the defective percentage is below 10%