STAT 6304 HOMEWORK ONE

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1.1 a. The population of interest is the weight of the shrimp maintained on the specific diet for a period of 6 months. b. The sample is the 100 shrimp selected from the pond and maintained on the specific diet for a period of 6 months. c. The weight gain of the shrimp over 6 months. d. Since the sample is only a small proportion of the whole population, it is necessary to evaluate what the mean weight may be for any other randomly selected 100 shrimps.

1.5 a. The population of interest is the population of those who would vote in the 2004 senatorial campaign. b. The population from which the sample was selected is registered voters in this state. c. The sample will adequately represent the population, unless there is a difference between registered voters in the state and those who would vote in the 2004 senatorial campaign. d. The results from a second random sample of 5,000 registered voters will not be exactly the same as the results from the initial sample. Results vary from sample to sample. With either sample we hope that the results will be close to that of the views of the population of interest.

2.7 The list of registered voters could be used as the sampling frame for selecting the persons to be included in the sample. 2.11 a. Simple random sampling. b. Stratified sampling. c. Cluster sampling.

STAT 6304 HOMEWORK ONE

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STAT 6304 HOMEWORK ONE

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STAT 6304 HOMEWORK ONE

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3.7 a. The separate relative frequency histograms for the two treatments appear below.

b. The histogram for the New Therapy begins and ends with bins that are slightly higher than the bins in the histogram for the Standard Therapy. This would indicate that the New Therapy generates a few more large values than the Standard Therapy. However, there is not convincing evidence that the New Therapy generates a longer survival time.

STAT 6304 HOMEWORK ONE

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3.13 All three homeownership distributions are unimodal and skewed to the left (since they are skewed, they are not symmetric). The homeownership distributions each have outliers (37% in 1985, 40% in 1996, and 44% in 2002).

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Exercises from R Code Ch3 :

Use the data of fourth grade presented in Table 3.1, O/L (page 60) to do the following:

(a) Make a histogram of N (school size) by R. Explain your choice of intervals. Include your R code.

Answer:

The range is 171-36=135 and there are 22 observations in the fourth grade. Since square root of

22 is about 5, I choose 5 intervals. The width of interval should be at least 135/5 =27, rounding

to 30. Therefore, the selected intervals are [30, 60), [60, 90), [90, 120), [120, 150), [150, 180).

Here is the R code and the histogram:

> fourth.n<-tb3dot1$N[23:44]

> n<-sort(fourth.n)

> brk<-seq(30,180,30)

> hist(n,breaks=brk)

(b) Find the sample mean and median of N. On the horizontal axis (bottom scale) of your histogram,

mark the position of the sample mean and median. At which point, the histogram seems to

more "balance"?

Answer:

The sample mean is 82.64 and sample median is 78. The histogram is skewed to the right and

82.64 looks like a bit farther from the center/balance point. Therefore, median is a better choice

of center in this case.

Histogram of n

n

Fre

quen

cy

50 100 150

02

46

8

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(c) Find the interval that extends two sample standard deviations on either side of the sample mean.

Sketch this interval on the bottom scale of your histogram. What proportion of all observations

actually falls in this interval? (You may find the sample mean and standard deviation either with

software or with a calculator.)

Answer:

The standard deviation is 31.85. The mean +/- 2 standard deviation is therefore (18.94, 146.34). The

proportion is 21/22=95%, which matches with the Empirical rule but the histogram shows non-

normal.