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Chapter 6: Q3E (page 338)

Question: A random sample of n measurements was selected from a population with unknown mean渭and known standard deviation蟽2. Calculate a 95% confidence interval for伪for each of the following situations:

a. n = 75, X = 28,蟽2= 12

b. n = 200, X= 102, 蟽2= 22

c. n = 100, X= 15,蟽2=.3

d. n = 100, X= 4.05, 蟽2= .83

e. Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a鈥揹? Explain.

Short Answer

Expert verified

Answer

a.(28.4卤27.6)b.(102.33卤101.67)c.(100.0588卤99.9412)d.(100.16268卤99.83732)e.No

Step by step solution

01

Step 1: Computing μ  when n is 75, x is 28 and σ2 is 12

a.

As the confidence interval is 95%, the significance level will be 5% which means 0.05.

Therefore,伪2=0.052

=0.025

. Now the value of z伪2can be found from the z table and so the values is 1.96.

The margin of error is calculated below:

The calculation of the confidence intervals of the boundaries are calculated below:

02

Step 1: Computing μ  when n is 200, x is 102 and σ2 is 21

b.

As the confidence interval is 95%, the significance level will be 5% which means 0.05.

Therefore,伪2=0.052

=0.025

. Now the value of can be found from the z table and so the values is 1.96.

The margin of error is calculated below:

The calculation of the confidence intervals of the boundaries are calculated below:

03

Step 1: Computing μ  when n is 100, x is 15 and σ2 is 0.3

c

As the confidence interval is 95%, the significance level will be 5% which means 0.05.

Therefore,伪2=0.052

=0.025

. Now the value of can be found from the z table and so the values is 1.96.

The margin of error is calculated below:

The calculation of the confidence intervals of the boundaries are calculated below:

04

Step 1: Computing μ  when n is 100, x is 4.05 and σ2 is 0.83

d.

As the confidence interval is 95%, the significance level will be 5% which means 0.05.

Therefore,伪2=0.052

=0.025

.

Now the value ofZ伪2can be found from the z table and so the values is 1.96.

The margin of error is calculated below:

The calculation of the confidence intervals of the boundaries are calculated below:

05

Determination of the validity of the assumption 

e.

The answer is no in this case. According to the theory of central limit theorem, the sample mean remains exactly normal. This applies specifically for the large samples where the size remains above 30.

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Most popular questions from this chapter

Water pollution testing. The EPA wants to test a randomlyselected sample of n water specimens and estimate themean daily rate of pollution produced by a mining operation.If the EPA wants a 95% confidence interval estimatewith a sampling error of 1 milligram per liter (mg/L),how many water specimens are required in the sample?Assume prior knowledge indicates that pollution readingsin water samples taken during a day are approximately

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The vessel arrived at a Massachusetts port with 11,000 bags of scallops, from which the harbormaster randomly selected 18 bags for weighing. From each such bag, his agents took a large scoopful of scallops; then, to estimate the bag鈥檚 average meat per scallop, they divided the total weight of meat in the scoopful by the number of scallops it contained. Based on the 18 [numbers] thus generated, the harbormaster estimated that each of the ship鈥檚 scallops possessed an average of 139 of a pound of meat (that is, they were about seven percent lighter than the minimum requirement). Viewing this outcome as conclusive evidence that the weight standard had been violated, federal authorities at once confiscated 95 percent of the catch (which they then sold at auction). The fishing voyage was thus transformed into a financial catastrophe for its participants. The actual scallop weight measurements for each of the 18 sampled bags are listed in the table below. For ease of exposition, Barnett expressed each number as a multiple of of a pound, the minimum permissible average weight per scallop. Consequently, numbers below 1 indicate individual bags that do not meet the standard. The ship鈥檚 owner filed a lawsuit against the federal government, declaring that his vessel had fully complied with the weight standard. A Boston law firm was hired to represent the owner in legal proceedings, and Barnett was retained by the firm to provide statistical litigation support and, if necessary, expert witness testimony.

0.93

0.88

0.85

0.91

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  1. Recall that the harbormaster sampled only 18 of the ship鈥檚 11,000 bags of scallops. One of the questions the lawyers asked Barnett was, 鈥淐an a reliable estimate of the mean weight of all the scallops be obtained from a sample of size 18?鈥� Give your opinion on this issue.
  2. As stated in the article, the government鈥檚 decision rule is to confiscate a catch if the sample mean weight of the scallops is less than 136 of a pound. Do you see any flaws in this rule?
  3. Develop your own procedure for determining whether a ship is in violation of the minimum-weight restriction. Apply your rule to the data. Draw a conclusion about the ship in question.

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c. Give a practical interpretation of the interval, part c.

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A random sample of 70 observations from a normally distributed population possesses a sample mean equal to 26.2 and a sample standard deviation equal to 4.1.

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e. Would your confidence intervals of parts a and c be valid if the distribution of the original population was not normal? Explain
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