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Chapter 7: Problem 11

Syracuse full-time students You'd like to estimate the proportion of the 14,201 (www.syr.edu/about/facts .html) undergraduate students at Syracuse University who are full-time students. You poll a random sample of 100 students, of whom 94 are full-time. Unknown to you, the proportion of all undergraduate students who are fulltime students is \(0.951 .\) Let \(X\) denote a random variable for which \(x=1\) denotes full-time student and for which \(x=0\) denotes part-time student. a. Describe the data distribution. Sketch a graph representing the data distribution. b. Describe the population distribution. Sketch a graph representing the population distribution. c. Find the mean and standard deviation of the sampling distribution of the sample proportion for a sample of size \(100 .\) Explain what this sampling distribution represents. Sketch a graph representing this sampling distribution.

Short Answer

Expert verified
a. Sample proportion is 0.94. b. Population proportion is 0.951. c. Mean = 0.951, SD = 0.0216.

Step by step solution

01

Understanding the Data Distribution

The data distribution is based on the sample of 100 students, where 94 are full-time, represented by 1, and 6 are part-time, represented by 0. Thus, the observed sample proportion (\(\hat{p}\)) of full-time students is \(\hat{p} = \frac{94}{100} = 0.94\). The distribution could be visualized as a bar chart with two bars: one at 1 with height 94% and another at 0 with height 6%.
02

Describing the Population Distribution

The population distribution involves all 14,201 undergraduate students. The true proportion of full-time students is known to be 0.951. Hence, if the data were collected for all students, 95.1% would be 1 (full-time) and 4.9% would be 0 (part-time). A bar chart for this distribution would have a bar at 1 with height 95.1% and another at 0 with height 4.9%.
03

Calculating Mean and Standard Deviation of Sampling Distribution

To find the mean of the sample proportion for a sample size of 100, we use the population proportion \(p = 0.951\). The mean of the sampling distribution (\(\mu_{\hat{p}}\)) is equal to \(p\), thus \(\mu_{\hat{p}} = 0.951\). The standard deviation is calculated as \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.951 \times 0.049}{100}} = 0.0216\). This sampling distribution shows how the sample proportion \(\hat{p}\) varies from sample to sample if we repeatedly take samples of size 100. A normal curve around \(0.951\) with standard deviation \(0.0216\) would represent this distribution visually.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportion Estimation
Understanding proportion estimation is key when you're trying to make educated guesses about larger groups, or populations, based on smaller groups, called samples. In the context of our exercise, we aim to estimate the proportion of full-time undergraduate students at Syracuse University. Using the sample of 100 students, where 94 are full-time, we compute the sample proportion, denoted as \(\hat{p}\). It's calculated as the number of full-time students divided by the total number of sampled students:
  • \(\hat{p} = \frac{94}{100} = 0.94\)
Here, \(\hat{p} = 0.94\) acts as our estimate of the population's proportion that is full-time.
Estimation becomes necessary when we cannot examine every individual in the population, due to constraints such as time or cost. This method allows us to make inferences about the population from which the sample is drawn, assuming that the sample is representative. Moreover, the sample proportion provides insight into what can be expected in the larger population, given that it closely reflects the true population proportion \(p = 0.951\) in this exercise.
Data Distribution
A data distribution describes how data is spread across different categories or values in a sample. In this case, we're concerned with a sample of 100 Syracuse University students.
The sample includes 94 students classified as full-time (denoted as 1) and 6 as part-time (denoted as 0). The data distribution can be shown as a simple bar chart:
  • One bar at '1' (full-time) with a height of 94%, representing the 94 full-time students.
  • Another bar at '0' (part-time) with a height of 6%, representing the 6 part-time students.
This bar chart vividly illustrates the concentration of full-time students in this sample, highlighting the emphasis on full-time status within the sampled population.
Such visualizations make it easy to analyze at a glance what proportion of your sample falls into each category. Data distributions give straightforward, summary information that helps in drawing initial conclusions about the sample.
Population Distribution
The population distribution, unlike the data distribution, considers every individual within the entire group of interest鈥攊n this case, all 14,201 undergraduate students. Knowing the true proportion, we see that 95.1% of all students are full-time.
When plotted, the population distribution similarly uses a bar chart:
  • A bar at '1' (full-time) with a height of 95.1%, representing the actual known proportion of full-time students.
  • A second bar at '0' (part-time) with a height of 4.9%, showing the rest.
Examining the population distribution helps validate our sample findings. With a known true proportion, comparing it against the sample proportion helps assess the sample's accuracy.
Population distributions offer complete data scenarios鈥攖hey lay the foundation for statistical inferences and aid in validating the assumptions made from sample data. They underscore the importance of sampling variability and serve as benchmarks when estimating and analyzing sample data.

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Most popular questions from this chapter

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Blood pressure Vincenzo Baranello was diagnosed with high blood pressure. He was able to keep his blood pressure in control for several months by taking blood pressure medicine (amlodipine besylate). Baranello's blood pressure is monitored by taking three readings a day, in early morning, at mid-day, and in the evening. a. During this period, the probability distribution of his systolic blood pressure reading had a mean of 130 and a standard deviation of \(6 .\) If the successive observations behave like a random sample from this distribution, find the mean and standard deviation of the sampling distribution of the sample mean for the three observations each day. b. Suppose that the probability distribution of his blood pressure reading is normal. What is the shape of the sampling distribution? Why? c. Refer to part b. Find the probability that the sample mean exceeds \(140,\) which is considered problematically high. (Hint: Use the sampling distribution, not the probability distribution for each observation.)

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