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1.) Using the 95 percent level of confidence, determine the confidence interval

ID: 3155869 • Letter: 1

Question

1.)

Using the 95 percent level of confidence, determine the confidence interval for . (Round your answers to 2 decimal places. Omit the "$" sign in your response.)

2.)

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.

What is the estimated proportion defective in the population? (Round your answer to 2 decimal places.)

What are the endpoints of a 95 percent confidence interval for the proportion defective. (Round your answers to 3 decimal places.)

Zack has an agreement with his supplier that he is to return lots that are 10 percent or more defective. Should he return this lot?

(Click to select)Yes, the lot is 10% defective.No, lot not 10% defective.Yes, 15 is outside the confidence interval.

3.)

We want to estimate the population mean within 5, with a 99 percent level of confidence. The population standard deviation is estimated to be 15. How large a sample is required? (Round up your answer to the next whole number.)

A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. They found the distribution of amounts spent per week followed the normal distribution with a population standard deviation of $5. A sample of 49 steady smokers revealed that = $20.

Explanation / Answer

1.

A)

Point estimate = sample mean = 20 [ANSWER]

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b)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    20          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    5          
n = sample size =    49          
              
Thus,              
Margin of Error E =    1.399974275          
Lower bound =    18.60002573          
Upper bound =    21.39997427          
              
Thus, the confidence interval is              
              
(   18.60002573   ,   21.39997427   ) [ANSWER]

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