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The following data are from a simple random sample. \\[ \begin{array}{llllll} 5 & 8 & 10 & 7 & 10 & 14 \end{array} \\] a. What is the point estimate of the population mean? b. What is the point estimate of the population standard deviation?

Short Answer

Expert verified
The point estimate of the population mean is 9, and the point estimate of the population standard deviation is approximately 3.098.

Step by step solution

01

List the Sample Data

First, consider the given data as a sample: \(5, 8, 10, 7, 10, 14\).
02

Calculate the Sample Mean

The point estimate for the population mean is the sample mean. To find this, sum up all the sample values and then divide by the number of values.\[ \text{Sample Mean} = \frac{5 + 8 + 10 + 7 + 10 + 14}{6} \]Calculate the sum:\[ 5 + 8 + 10 + 7 + 10 + 14 = 54 \]Now, divide the sum by the number of values:\[ \frac{54}{6} = 9 \]So, the sample mean is 9.
03

Calculate Each Deviation from the Mean

Subtract the sample mean from each data value to find the deviation for each point.- Deviation for 5: \(5 - 9 = -4\)- Deviation for 8: \(8 - 9 = -1\)- Deviation for 10: \(10 - 9 = 1\)- Deviation for 7: \(7 - 9 = -2\)- Deviation for 10: \(10 - 9 = 1\)- Deviation for 14: \(14 - 9 = 5\)
04

Square Each Deviation

Square each of the deviations from the mean calculated in the previous step to get rid of negative signs and emphasize larger errors.- \((-4)^2 = 16\)- \((-1)^2 = 1\)- \(1^2 = 1\)- \((-2)^2 = 4\)- \(1^2 = 1\)- \(5^2 = 25\)
05

Sum the Squared Deviations

Add all the squared deviations:\[ 16 + 1 + 1 + 4 + 1 + 25 = 48 \]
06

Calculate the Sample Variance

Compute the sample variance by dividing the sum of squared deviations by the number of data points minus one (the sample size minus one):\[ \text{Sample Variance} = \frac{48}{6 - 1} = \frac{48}{5} = 9.6 \]
07

Calculate the Sample Standard Deviation

To find the standard deviation, take the square root of the sample variance:\[ \text{Sample Standard Deviation} = \sqrt{9.6} \approx 3.098 \]
08

Conclusion

We now have both the point estimates calculated: - The point estimate of the population mean is 9. - The point estimate of the population standard deviation is approximately 3.098.

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Key Concepts

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

Sample Mean
The sample mean is an essential concept in statistics, often serving as a point estimate for the population mean. To calculate the sample mean, you sum up all the sample values and divide this sum by the number of data points. For the sample
  • 5
  • 8
  • 10
  • 7
  • 10
  • 14
we first calculate the sum: \[5 + 8 + 10 + 7 + 10 + 14 = 54\]
Then, divide this sum by the number of data points, which is 6:\[\text{Sample Mean} = \frac{54}{6} = 9\]

Why Use the Sample Mean?

The sample mean is a straightforward measure of the central tendency of the data. It is particularly useful because it serves as a good unbiased estimate of the population mean, especially when dealing with simple random samples.

Importance and Limitations

Although the sample mean is a helpful estimate, it is affected by outliers and may not fully represent a population with skewed distributions. Nonetheless, it remains a foundational statistic when conducting point estimation.
Population Standard Deviation
The population standard deviation provides a measure of the spread or variability in a dataset. When dealing with a sample, we use the sample standard deviation as a point estimate for the population standard deviation. This requires a few steps starting from calculating deviations to squaring those deviations.

Calculating Deviations

Subtract the sample mean from each data point to find individual deviations:
  • 5 - 9 = -4
  • 8 - 9 = -1
  • 10 - 9 = 1
  • 7 - 9 = -2
  • 10 - 9 = 1
  • 14 - 9 = 5

Squaring and Summing Deviations

Square each deviation:
  • (-4)^2 = 16
  • (-1)^2 = 1
  • 1^2 = 1
  • (-2)^2 = 4
  • 1^2 = 1
  • 5^2 = 25
Sum these squared deviations:\[16 + 1 + 1 + 4 + 1 + 25 = 48\]

Finding the Sample Standard Deviation

The sample standard deviation is derived by taking the square root of the sample variance:\[\text{Sample Standard Deviation} = \sqrt{\text{Sample Variance}} = \sqrt{9.6} \approx 3.098\] This value serves as the point estimate for the population standard deviation.
Sample Variance
Sample variance measures the average of the squared deviations from the sample mean, providing insight into the data's spread. After calculating the deviations and squaring them, the next step is to compute the variance.

Calculating the Sample Variance

To find sample variance, divide the sum of squared deviations by the sample size minus one. This adjustment (n-1 instead of n) corrects the bias in estimating a population parameter from a sample:\[\text{Sample Variance} = \frac{48}{6-1} = \frac{48}{5} = 9.6\]
This concept is crucial in statistics, as it helps estimate population variability when only sample data is available.

Significance of Sample Variance

Sample variance helps quantify the biological or physical variability inherent in data. Understanding this can be particularly vital in fields such as quality control, where knowing how much data fluctuates can inform process improvements.By understanding sample variance, we can better grasp how much our sample reflects potential deviations in the broader population.

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Most popular questions from this chapter

Assume that the population proportion is \(.55 .\) Compute the standard error of the proportion, \(\sigma_{\bar{p}},\) for sample sizes of \(100,200,500,\) and \(1000 .\) What can you say about the size of the standard error of the proportion as the sample size is increased?

Roper ASW conducted a survey to learn about American adults' attitudes toward money and happiness (Money, October 2003 ). Fifty-six percent of the respondents said they balance their checkbook at least once a month. a. Suppose a sample of 400 American adults were taken. Show the sampling distribution of the proportion of adults who balance their checkbook at least once a month. b. What is the probability that the sample proportion will be within ±.02 of the population proportion? c. What is the probability that the sample proportion will be within ±.04 of the population proportion?

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