Question

Schadek Silkscreen Printing Inc. purchases plastic cups on which to print logos for sporting events, proms, birthdays, and other special occasions. Zack Schadek, the owner, received a large shipment this morning. To ensure the quality of the shipment, he selected a random sample of 300 cups. He found 15 to be defective.
a. What is the estimated proportion defective in the population?
b. Develop a 95 percent confidenceinterval for the proportion defective.c. Zack has an agreement withhis supplier that he is to return lots that are 10 percent or moredefective. Should he return this lot? Explain your decision.

Answer 1

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

$CI = p \pm z\sqrt{\frac{p(1-p)}{n} }$

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

$CI = 0.05 \pm 1.96\sqrt{\frac{0.05(1-0.05)}{300} }$

$CI = 0.05 \pm 1.96(0.01258)$

$CI = 0.05 \pm 0.025$

$Upper = 0.05 + 0.025 = 0.075$

$Lower = 0.05 - 0.025 = 0.025$

$CI = (0.025, 0.075)$

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%