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For all job positions in a company, assume that, a few years ago, the average time to fill a job position was 37 days with a standard deviation of 12 days. For the purpose of comparison, the manager of the hiring department selected a random sample of 100 of today's job positions. He observed a sample mean of 39 days and a standard deviation of 13 days. a. Describe the center and variability of the population distribution. What shape does it probably have? Explain. b. Describe the center and variability of the data distribution. What shape does it probably have? Explain. c. Describe the center and variability of the sampling distribution of the sample mean for \(n=100 .\) What shape does it have? Explain. d. Explain why it would not be unusual to observe a job position that would take more than 55 days to fill, but it would be highly unusual to observe a sample mean of more than 50 days for a random sample size of 100 job positions.

Short Answer

Expert verified
a) Population mean is 37 days; normal shape. b) Sample mean of 39 days; assumed normal for large n. c) Sampling mean is 39 days; normal shape with standard error 1.3. d) Individual >55 days is possible; sample mean >50 days is unlikely.

Step by step solution

01

Understanding the Population Distribution

The average time to fill a job position a few years ago is a population mean of 37 days (). The population standard deviation was 12 days (4). Assuming that the data is normally distributed based on past observations, the shape of the population distribution is approximately normal. Normal distributions have symmetrical shapes around the mean.
02

Analyzing the Current Data Distribution

Today's sample of 100 positions has a sample mean of 39 days and a sample standard deviation of 13 days. The shape of this current sample distribution is unknown without more advanced statistical tests (e.g., normality tests) but for a large enough sample size (such as 100), the Central Limit Theorem allows assumptions of normality for the sample mean.
03

Evaluating the Sampling Distribution

The sampling distribution of the sample mean with a sample size of 100 () would have a mean of 39 days. The standard deviation of the sampling distribution, known as the standard error, can be calculated as 8/6 = 859 where 8 is the sample standard deviation, and 6 is the sample size. Hence, the standard error is 79 4 = 1.3 days. By the Central Limit Theorem, the sampling distribution will be approximately normal.
04

Comparing Probability of Observations

For individual job positions, observing a job fill time greater than 55 days is feasible given the population standard deviation is 12 and could include long tails in a population distribution. For the sample mean, the standard error is small (1.3 days), which makes extreme sample means (e.g., more than 50 days) improbable to occur, based on standard normal z-scores and probabilities.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics that helps us understand the distribution of sample means. Even if the population distribution is not perfectly normal, the theorem tells us that the distribution of the sample mean will approach a normal distribution as the sample size increases.

This is crucial when we deal with large samples, as it allows us to make inferences about the population. In practical terms, this means:
  • When we take a large enough sample (usually 30 or more is considered large), the mean of the sample will be approximately normally distributed around the population mean.
  • This approximation to normality helps in calculating probabilities and making predictions about the population based on the sample.
In the context of the original exercise, the sample size of 100 is sufficiently large to apply the CLT. This ensures that the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution.
Population Distribution
Population distribution refers to the way in which a set of data is spread out across different values. In the given exercise, the population distribution is characterized by a mean of 37 days, with a standard deviation of 12 days for time to fill a job position.

Assuming past records suggest normality, the distribution is likely bell-shaped and symmetrical around the mean.

Key aspects to know about population distribution include:
  • The mean provides a central value around which the data is centred.
  • The standard deviation communicates how spread out the data is around the mean. A higher standard deviation indicates a wider spread.
Understanding the population distribution is essential when making predictions or assumptions about individual data points or for comparing to sample data.
Sample Mean
The sample mean is a calculation of the average value in a sample set of data. It's a critical number for statisticians as it serves as an estimate of the population mean.

To compute it, you add up all the sample observations and divide by the number of observations in the sample.
  • The formula is: \( \overline{x} = \frac{\sum x_i}{n} \), where \( \overline{x} \) is the sample mean, \( x_i \) is each individual observation, and \( n \) is the sample size.
  • In the exercise, the sample mean is 39 days for the 100 job positions.
This provides a snapshot of current conditions, allowing comparisons with past data. However, remember that while "it is a reliable estimator of the population mean, it might vary with different samples."
Standard Error
The standard error measures the variability or precision of the sample mean. It tells us how much the sample mean would vary if we took many samples from the population.

  • The standard error is calculated using the formula: \( SE = \frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation and \( n \) is the sample size.
In our exercise, with a sample standard deviation of 13 days and sample size of 100, the standard error is calculated to be \( \frac{13}{\sqrt{100}} = 1.3 \) days.

This value helps determine how accurately the sample mean represents the population mean:
  • A smaller standard error, like 1.3 days, implies that the sample mean is likely close to the population mean.
  • A larger standard error suggests more variability and less certainty about the sample mean being a good estimate of the population mean.
This small standard error highlights why it would be unusual to observe a sample mean that is significantly different from the population mean, such as more than 50 days in the exam question example.

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Most popular questions from this chapter

Access the Sampling Distribution of the Sample Mean (discrete variable) web app on the book's website. Enter the probabilities \(\mathrm{P}(X=x)\) of 0.1667 for the numbers 1 through 6 to specify the probability distribution of a fair die. (This is a discrete version of the uniform distribution shown in the first column of Figure 7.11.) The resulting population distribution has \(\mu=3.5\) and \(\sigma=1.71\) a. In the box for the sample size \(n,\) enter 2 to simulate rolling two dice. Then, press the Draw Sample(s) button several times and observe how the histogram for the simulated sampling distribution for the mean number shown on two rolls is building up. Finally, simulate rolling two dice and finding their average 10,000 times by selecting the corresponding option. Describe (shape, center, spread) the resulting simulated sampling distribution of the sample mean, using the histogram of the 10,000 generated sample means. (Note: Statistics for the simulated sampling distribution are reported in the tile of its plot.) b. Are the reported mean and standard deviation of the simulated sampling distribution close to the theoretical mean and standard deviation for this sampling distribution? Compute the theoretical values and compare. c. Repeat part a, but now simulate rolling \(n=30\) dice an finding their average face value. What are the major changes you observe in the simulated sampling distribution?

In Aunt Erma's Restaurant, the daily sales follow a probability distribution that has a mean of \(\mu=\$ 900\) and a standard deviation of \(\sigma=\$ 300\). This past week the daily sales for the seven days had a mean of \(\$ 980\) and a standard deviation of \(\$ 276\). Consider these seven days as a random sample from all days. a. Identify the mean and standard deviation of the population distribution. b. Identify the mean and standard deviation of the data distribution. What does the standard deviation describe? c. Identify the mean and the standard deviation of the sampling distribution of the sample mean for samples of seven daily sales. What does this standard deviation describe?

In college basketball, a shot made from beyond a designated arc radiating about 20 feet from the basket is worth three points instead of the usual two points given for shots made inside that arc. Over his career, University of Florida basketball player Lee Humphrey made \(45 \%\) of his three-point attempts. In one game in his final season, he made only 3 of 12 three-point shots, leading a TV basketball analyst to announce that Humphrey was in a shooting slump. a. Assuming Humphrey has a \(45 \%\) chance of making any particular three-point shot, find the mean and standard deviation of the sampling distribution of the proportion of three-point shots he will make out of 12 shots. b. How many standard deviations from the mean is this game's result of making 3 of 12 three-point shots? c. If Humphrey was actually not in a slump but still had a \(45 \%\) chance of making any particular three-point shot, explain why it would not be especially surprising for him to make only 3 of 12 shots. Thus, this is not really evidence of a shooting slump.

In one lottery option in Canada (Source: Lottery Canada), you bet on a six- digit number between 000000 and \(999999 .\) For a \(\$ 1\) bet, you win \(\$ 100,000\) if you are correct. The mean and standard deviation of the probability distribution for the lottery winnings are \(\mu=0.10\) (that is, 10 cents) and \(\sigma=100.00\). Joe figures that if he plays enough times every day, eventually he will strike it rich, by the law of large numbers. Over the course of several years, he plays 1 million times. Let \(\bar{x}\) denote his average winnings. a. Find the mean and standard deviation of the sampling distribution of \(\bar{x}\). b. About how likely is it that Joe's average winnings exceed \(\$ 1,\) the amount he paid to play each time? Use the central limit theorem to find an approximate answer.

The central limit theorem implies a. All variables have approximately bell-shaped data distributions if a random sample contains at least about 30 observations. b. Population distributions are normal whenever the population size is large. c. For sufficiently large random samples, the sampling distribution of \(\bar{x}\) is approximately normal, regardless of the shape of the population distribution. d. The sampling distribution of the sample mean looks more like the population distribution as the sample size increases.

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