Question

In: Statistics and Probability

Exercise 8-24 Schadek Silkscreen Printing, Inc., purchases plastic cups on which to print logos for sporting...

Exercise 8-24

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 540 cups. He found 19 to be defective.

a. What is the estimated proportion defective in the population? (Round the final answer to 3 decimal places.)

Estimated proportion defective ___.

b. What are the endpoints of a 90% confidence interval for the proportion defective. (Round the final answers to 3 decimal places.)

Endpoints____ , ___

Exercise 8-16

The Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 28 people reveals the mean yearly consumption to be 28 kilograms (kg) with a standard deviation of 11 kg. Assume a normal population.

a-2. What is the best estimate of this value?

Estimate value           

c. For a 98% confidence interval, what is the value of t? (Round the final answer to 3 decimal places.)

Value of t           

d. Develop the 98% confidence interval for the population mean. (Round the final answers to 3 decimal places.)

Confidence interval for the population mean is  and  .

Solutions

Expert Solution

1)

a)

sample proportion, = 0.035

b)


sample size, n = 540
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.035 * (1 - 0.035)/540) = 0.0079

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

Margin of Error, ME = zc * SE
ME = 1.64 * 0.0079
ME = 0.013

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.035 - 1.64 * 0.0079 , 0.035 + 1.64 * 0.0079)
CI = (0.022 , 0.048)


2)

a)

Estimate value = 28

b)

sample mean, xbar = 28
sample standard deviation, s = 11
sample size, n = 28
degrees of freedom, df = n - 1 = 27

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.473


ME = tc * s/sqrt(n)
ME = 2.473 * 11/sqrt(28)
ME = 5

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (28 - 2.473 * 11/sqrt(28) , 28 + 2.473 * 11/sqrt(28))
CI = (22.859 , 33.141)


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