Question

In: Math

Systematic random samples are often used to choose a sample of apartments in a large building...

Systematic random samples are often used to choose a sample of apartments in a large building or dwelling units in a block at the last stage of a multistage sample. An example will illustrate the idea of a systematic sample. Suppose that we must choose 5 addresses out of 125. Because 125/5=25, we can think of the list as five lists of 25 addresses. Choose 1 of the first 25 at random, using software or Table B. The sample contains this address and the addresses 25, 50, 75, and 100 places down the list from it. If 13 is chosen, for example, then the systematic random sample consists of the addresses numbered 13, 38, 63, 88, and 113.

(a) A study of dating among college students wanted a sample of 200 of the 8000 single male students on campus. The sample consisted of every 40th name from a list of the 8000 students. Explain why the survey chooses every 40th name.

(b) Use software or Table B at line 112 to choose the starting point for this systematic sample.

Solutions

Expert Solution

(a)

200 students needed to be selected out of 8000. So, 8000/200 =40. Thus, the survey chooses every 40th name.

(b)

The starting point for this survey is: 04, i.e., 4th name is selected first and then every 40th name is to be selected. So, 4, 44, 84, 124,........

Last name (200th student: n =200) to be selected following this sequence which is in Arithmetic Progression is: tn =a+(n-1)d =4+(200 - 1)*40 =4+199(40) =4+7960 =7964th name.

So, the sequence is: 4, 44, 84, 124,...........,7964.

(since in Table B below, along the line 112, we have first two digits of 59 in first column which is greater than 40, we have 88 in the second column >40 and then in the third column, we have 04 <40. So, we selected 4th name as a starting point).

(If we have selected 59th name first. Then, our last term (n =200) of the sequence will be: a+(n-1)d =59+119(40) =59+7960 =8019th name but we have only 8000 students. So, we need to select 40 or less as the starting point because if select 40, last term will be: 40+199(40) =8000 and less than 40 as a starting point leads to last term of less than 8000).

(So, we can also use random number generator between 1 and 40(inclusive) to select the starting point).


Related Solutions

Monthly rental rate data were collected from a large random sample of apartments in Manhattan. The...
Monthly rental rate data were collected from a large random sample of apartments in Manhattan. The mean rate was $3156.50 per month. The distribution of sample means was normal with a standard error of $92.801. Construct a 95% confidence interval to estimate the mean rate in the population of all Manhattan apartments. [2974.610, 3338.390] [3003.855, 3309.145] [3138.311, 3174.689] [3151.681, 3158.319]
How to draw a simple random sample of size 9 and a systematic sample of size...
How to draw a simple random sample of size 9 and a systematic sample of size 9 from 45 samples size. can you explain the steps for each selected sample for me . thanks
A systematic random sample was taken from the set of all Presidents of the United States....
A systematic random sample was taken from the set of all Presidents of the United States. The data file potus heights.csv random sample includes the height (in inches) of each sampled President. (a) From this data, estimate the average height of United States Presidents. Calculate two error bounds for your estimate, one using the usual SRS formula, and one using the successive difference variance estimator. (b) Which variance estimator is more appropriate for these data? Briefly explain president hgt Van...
You randomly choose some unfurnished one-bedroom apartments from a large number of advertisements in your local...
You randomly choose some unfurnished one-bedroom apartments from a large number of advertisements in your local newspaper. You calculated their mean monthly rent is $570 and their standard deviation is $150. Construct the following confidence intervals for the mean monthly rent of all advertised one-bedroom apartments. Confidence interval with C=90%, n=10:
Your local newspaper contains a large number of advertisements for unfurnished one-bedroom apartments. You choose 16...
Your local newspaper contains a large number of advertisements for unfurnished one-bedroom apartments. You choose 16 at random and calculate that their mean monthly rent is $508 and that the standard deviation of their rents is $78. Please show all work and write legibly Construct a 90% confidence interval for the mean monthly rent of all advertised one-bedroom apartments. (Fill in the blanks below and give your answers to 2 decimal places.) The 90% 90%  confidence interval is ($__________ , $...
What are the commonly used probability sampling techniques? Briefly explain random sampling and systematic random sampling...
What are the commonly used probability sampling techniques? Briefly explain random sampling and systematic random sampling and compare their differences.
Two random samples are selected from two independent populations. A summary of the samples sizes, sample...
Two random samples are selected from two independent populations. A summary of the samples sizes, sample means, and sample standard deviations is given below: n1=37,n2=44,x¯1=58.9,x¯2=74.7,s1=5.5s2=10.1 n 1 =37, x ¯ 1 =58.9, s 1 =5.5 n 2 =44, x ¯ 2 =74.7, s 2 =10.1 Find a 95.5% confidence interval for the difference μ1−μ2 μ 1 − μ 2 of the means, assuming equal population variances. Confidence Interval
Two random samples are selected from two independent populations. A summary of the samples sizes, sample...
Two random samples are selected from two independent populations. A summary of the samples sizes, sample means, and sample standard deviations is given below: n1=39,n2=48,x¯1=52.5,x¯2=77.5,s1=5s2=11 Find a 97.5% confidence interval for the difference μ1−μ2 of the means, assuming equal population variances. Confidence Interval =
Explain in detailS the systematic random sampling. According to the following information, define the sample elements:...
Explain in detailS the systematic random sampling. According to the following information, define the sample elements: No. of elements in population: 17024 No. of desired sample: 19 The first random number of samples: 112
Q2) Explain in detail the systematic random sampling. According to the following information, define the sample...
Q2) Explain in detail the systematic random sampling. According to the following information, define the sample elements (20 points): No. of elements in population: 17024 No. of desired sample: 19 The first random number of samples: 112
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT