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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?

Short Answer

Expert verified
Population mean is $900; standard deviation is $300. Sample mean is $980; standard deviation is $276. Sampling mean $ 50at $9, stand . deviation $113.38.

Step by step solution

01

Identify Population Distribution Parameters

The mean of the population distribution is given as \( \mu = \\(900 \). The standard deviation of the population distribution is \( \sigma = \\)300 \). These values represent the average and the variability of daily sales in the entire population.
02

Identify Sample Data Parameters

The mean of the sample data (the seven days of daily sales) is given as \( \\(980 \), and the standard deviation of the sample data is \( \\)276 \). The sample mean tells us the average sales for these specific seven days, while the standard deviation indicates how much the sales varied around the mean in these seven days.
03

Find Sampling Distribution Parameters

For a sample size \( n = 7 \), the mean of the sampling distribution of the sample mean is equal to the population mean, \( \mu_{\bar{x}} = \mu = \\(900 \). The standard deviation of the sampling distribution of the sample mean is given by the formula \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \). Therefore, \( \sigma_{\bar{x}} = \frac{300}{\sqrt{7}} \approx \\)113.38 \). This standard deviation, known as the standard error, describes how much the sample means vary from the population mean if we were to take multiple samples of size 7.

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Key Concepts

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

Population Distribution
In statistics, the term **population distribution** refers to the spread of values in an entire population from which samples may be drawn. Imagine if we were to map out the daily sales for Aunt Erma's Restaurant across a long period of time. This long-term data would create a distribution with a certain shape. In our example, the population distribution is described by its mean \(\mu = 900\) and its standard deviation \(\sigma = 300\).

These values give us an idea of:
  • The average daily sales, which is represented by the mean \(\mu\).
  • The variability of daily sales, which is expressed by the standard deviation \(\sigma\).


By knowing the mean and standard deviation of the population distribution, we gain insights into the central tendency and spread of the entire data set representing daily sales.
Sample Mean
The **sample mean** refers to the average of a subset or a sample taken from a larger population. In our specific case, we're referring to the seven days of sales at Aunt Erma's Restaurant, considered as a random sample.

The sample mean was calculated as \(980\), which indicates:
  • For those seven days, the average daily sales were higher than the population mean.
This shows how sample statistics can sometimes differ from population parameters, depending on which days are sampled.
Standard Deviation
The **standard deviation** is a fundamental concept in statistics that provides insight into the spread of data points in a dataset. For Aunt Erma's sample of seven days, a standard deviation of \(276\) helps us understand the variability or dispersion in sales for that sampling period.

This value tells us how much individual sales figures tended to fluctuate around the mean of \(980\). A smaller standard deviation would imply sales were generally close to the average, while a larger standard deviation suggests a wider spread of sales amounts.
Standard Error
The **standard error** quantifies the amount of variability or dispersion in sample means, rather than in individual data points. It's essentially the standard deviation of the sampling distribution and is calculated using the formula:

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]

Applying this to Aunt Erma's restaurant, where \(\sigma = 300\) and the sample size \(n = 7\), the standard error is approximately \(113.38\).

This tells us:
  • If we took many samples of seven days each, we'd expect the average sales to vary around the true population mean of \(900\) by roughly \(113.38\).
Thus, standard error provides a measure of precision for our sample mean, showing potential variation from the population mean.

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Most popular questions from this chapter

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.

In a class of 150 students, the professor has each person toss a fair coin 50 times and calculate the proportion of times the tosses come up heads. Roughly \(95 \%\) of students should have proportions between which two numbers? a. 0.49 and 0.51 b. 0.05 and 0.95 c. 0.42 and 0.58 d. 0.36 and 0.64 e. 0.25 and 0.75 Explain your answer.

The sampling distribution of a sample mean for a random sample size of 100 describes a. How sample means tend to vary from random sample to random sample of size 100 . b. How observations tend to vary from person to person in a random sample of size 100 . c. How the data distribution looks like the population distribution when the sample size is larger than 30 . d. How the standard deviation varies among samples of size 100 .

The scores on the Psychomotor Development Index (PDI), a scale of infant development, have a normal population distribution with mean 100 and standard deviation 15. An infant is selected at random. a. Find the \(z\) -score for a PDI value of 90 . b. A study uses a random sample of 225 infants. Using the sampling distribution of the sample mean PDI, find the \(z\) -score corresponding to a sample mean of 90 . c. Explain why a PDI value of 90 is not surprising, but a sample mean PDI score of 90 for 225 infants would be surprising.

According to the U.S. Census Bureau, Current Population Survey, Annual Social and Economic Supplement, the average income for females was \(\$ 28,466\) and the standard deviation was \(\$ 36,961\) in \(2015 .\) A sample of 1,000 females was randomly chosen from the entire United States population to verify if this sample would have a similar mean income as the entire population. a. Find the probability that the mean income of the females sampled is within two thousand of the mean income for all females. (Hint: Find the sampling distribution of the sample mean income and use the central limit theorem). b. Would the probability be larger or smaller if the standard deviation of all females' incomes was \(\$ 25,000 ?\) Why?

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