Question

Schadek Silkscreen Printing Inc. purchases plastic cups and imprints them with 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 and inspected them for defects. He found 15 to be defective.

a. What is the estimatedproportion 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.

Answer 1

Answer:

(a) The estimated proportion of defective in the population is 0.05.

(b) The 95% confidence interval for the proportion defective cups is (2.5%, 7.5%).

(c) Zack does not needs to return the lots.

Step-by-step explanation:

Let X = number of defective cups.

The random sample of cups selected is of size, n = 300.

The number of defective cps in the sample is, X = 15.

(a)

The proportion of the defective cups in the population can be estimated by the sample proportion because the sample selected is quite large.

The sample proportion of defective cups is:

$\hat p=\frac{X}{n}=\frac{15}{300}=0.05$

Thus, the estimated proportion of defective in the population is 0.05.

(b)

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p \pm z_{\alpha/2}\times\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Compute the critical value of z for 95% confidence level as follows:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval for p as follows:

$CI=\hat p \pm z_{\alpha/2}\times\sqrt{\frac{\hat p(1-\hat p)}{n}}$

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

     $=0.05\pm 0.025\\=(0.025, 0.075)$

Thus, the 95% confidence interval for the proportion defective cups is (2.5%, 7.5%).

(c)

It is provided that Zack has an agreement with his supplier that he is to return lots that are 10% or more defective.

The 95% confidence interval for the proportion defective is (2.5%, 7.5%). This implies that 95% of the lots have 2.5% to 7.5% defective items.

Thus, Zack does not needs to return the lots.

Answer 2

The estimated proportion of defective in the population is 0.05 and the 95 percent confidence interval for the proportion defective is (0.025,0.075).

Given :

• 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 and inspected them for defects.
• He found 15 to be defective.

a) The formula given below is used in order to determine the estimated proportion of defective in the population.

$\hat{p} = \dfrac{X}{n}$

$\hat{p} = \dfrac{15}{300}$

$\hat{p} = 0.05$

So, the estimated proportion of defective in the population is 0.05.

b) The below formula is used in order to determine the 95 percent confidence interval for the proportion defective.

$CI =\hat{p}\pm z_{\alpha /2}\times \sqrt{ \dfrac{\hat{p}(1-\hat{p})}{n}}$

Now, substitute the known terms in the above expression.

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

$CI = 0.05\pm 0.025$

So, the 95 percent confidence interval for the proportion defective is (0.025,0.075).

c) According to the given data, Zack has an agreement with his supplier that he is to return lots that are 10 percent or more defective.

So, from the above calculation, it can be concluded that he did not have to return the lots.

For more information, refer to the link given below:

https://brainly.com/question/10951564