and imprints them with logos for sporting events, proms Schadek Silkscreen Print

Question

and imprints them with logos for sporting events, proms Schadek Silkscreen Printing Inc. purchases plastic cups birthdays, and other special occasions. Zack Schadek, the owner, recelved 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 a. What is the estimated proportion defective in the population? (Round your answer to 2 declmal places.) Develop a 95% confidence interval for the proportion defective. (Round your answers to 3 decimal places) c. Zack has an agreement with his supplier that if 10% or more of the cups are defective, he can return the order. Should he return this lot? No, the lot is not defective by 10% or more.

Explanation / Answer

a).

Consider the given problem here out of “300 cups” 15 are defective, => here the estimated proportional defective in the population is given by “15/300 = 0.05”, => p=0.05.

b).

Here we have to find out the “95%” confidence interval, => the level of significance is “a=5%”, => the critical value is given by, “ta/2=1.96” and “p=0.05”.

So, the “95%” confidence interval is given by, “p – [p(1-p)/n]^0.5*ta/2” and “p + [p(1-p)/n]^0.5*ta/2”.

=> p – [p(1-p)/n]^0.5*ta/2 = 0.05 – [0.05*(1-0.05)/300]^0.5*1.96 = 0.05 – 0.01258*1.96.

=> 0.05 – 0.02466 = 0.025.

Now, p + [p(1-p)/n]^0.5*ta/2 = 0.05 + [0.05*(1-0.05)/300]^0.5*1.96 = 0.05 + 0.02466 = 0.075.

So, here the “95%” confidence interval is given by, “0.025” to “0.075”.

c).

now, Zack has an agreement with his supplier that if “10%” or more of the cups are defective, then he can return the order, => here the “H0” is “P=10%=0.1” and the “H1” is “P > 0.1”.

=> under the “H0” the test statistic is given by, “t = n^0.5(p-0.1)/[0.1*(1-0.1)]^0.5”.

=> t = 300^0.5(0.05-0.1)/[0.1*0.9]^0.5 = 300^0.5(-0.05)/0.3 = 17.32*(-0.05)/0.3,

=> t = 17.32*(-0.05)/0.3 = (-2.8867), => t = (-2.8867).

Now, at “5%” level of significance the “H0” will be rejected is “t > ta = 1.645”. Here “t < ta”, => here “H0” is accepted and we can conclude that the lot is less than “10%” defective.

So, here the “1st“ be the correct option.

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