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The standard deviation for a population is \(\sigma=6.30 .\) A random sample selected from this population gave a mean equal to \(81.90 .\) The population is known to be normally distributed. a. Make a \(99 \%\) confidence interval for \(\mu\) assuming \(n=16\). b. Construct a \(99 \%\) confidence interval for \(\mu\) assuming \(n=20\). c. Determine a \(99 \%\) confidence interval for \(\mu\) assuming \(n=25\). d. Does the width of the confidence intervals constructed in parts a through \(\mathrm{c}\) decrease as the sample size increases? Explain.

Short Answer

Expert verified
a. \(81.90 \pm 2.57 * (6.30 / \sqrt{16})\). b. \(81.90 \pm 2.57 * (6.30 / \sqrt{20})\). c. \(81.90 \pm 2.57 * (6.30 / \sqrt{25})\). d. Yes, the width of the confidence intervals decreases as the sample size increases. This is because increasing the sample size leads to a smaller standard error, which in turn leads to a smaller confidence interval.

Step by step solution

01

Calculate the Standard Error

First, calculate the standard error of the mean using the formula: \(\text{se} = \frac{\sigma}{\sqrt{n}}\), where \(\sigma = 6.30\) is the standard deviation and \(n\) is the sample size. This will vary for parts a, b, and c.
02

Determine the Z-Score

For a 99% confidence interval, the Z-score for a two-tailed test is approximately 2.57. This is found in standard z-tables under the column and row representing 0.995 (0.5 + 0.99/2).
03

Calculate the Confidence Interval

Now calculate the confidence interval using the formula: \(\text{CI} = \text{Mean} ± Z*\text{se}\), where Mean is the sample mean and Z is the z-score from the previous step.
04

Apply Steps 1-3 for Each Part

Now you apply the steps 1-3 to each part of the exercise. For example, in part a, \(n=16\) so \(\text{se} = 6.30 / \sqrt{16}\). Then calculate the confidence interval.
05

Compare the Confidence Intervals

Finally, to answer part d, compare the width (upper limit - lower limit) of the confidence intervals calculated in parts a through c. The smaller the width, the smaller the range of the interval.

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Key Concepts

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

Standard Deviation
The standard deviation is a measure that tells us how much individual data points in a set differ from the mean (average) of the set. In simpler terms, it shows the dispersion or spread of data values. If the standard deviation is large, it means that the data points are spread out over a wider range of values. Conversely, a smaller standard deviation indicates that the data points are closer to the mean.

For example, in our exercise, the population standard deviation is given as \( \sigma = 6.30 \). This means that, for the entire data set, the average distance between each data point and the mean is 6.30 units. The standard deviation helps provide context about the data's variability, which is especially useful when we create confidence intervals.

Confidence intervals rely heavily on understanding the spread represented by standard deviation, as it influences calculations of the standard error and, subsequently, the interval's width.
Sample Size
Sample size refers to the number of observations or data points we have in our sample. It's an essential component in statistics because it affects the reliability or precision of statistical measures. The more samples we have, the more accurately we can estimate the population parameters.

In the exercise, different sample sizes \( n = 16, 20, 25 \) are used to create confidence intervals. With a larger sample size, our estimation of the population mean typically becomes more precise.

The key reasons a larger sample size results in more reliable statistics are:
  • The sampling distribution of the sample mean becomes narrower, meaning less variability.
  • It decreases the standard error, which results in a narrower confidence interval.
Hence, as you increase your sample size, the width of the confidence interval decreases, indicating that you're more certain about the range in which the population mean lies.
Z-Score
In statistics, a Z-score indicates how many standard deviations an element is from the mean. This standardization helps compare data from different normal distributions.

In our context, a Z-score is particularly important when calculating confidence intervals for known population standard deviations. For instance, if we look at a 99% confidence interval, it closely associates with a Z-score of approximately 2.57 in a standard normal distribution. This value indicates that to cover 99% of the data using a two-tailed test, you need to go 2.57 standard deviations in either direction from the sample mean.

The Z-score helps determine the critical values for our confidence interval calculation, which provides insights into how extraordinary or typical a result is when considering the surrounding data and its deviation.
Standard Error
The standard error is a statistical term that measures the accuracy with which a sample distribution represents a population. It is essentially an estimate of the standard deviation of the sample mean, calculated as the population standard deviation \( \sigma \) divided by the square root of the sample size \( n \).

Mathematically, it's calculated using the formula:\[\text{SE} = \frac{\sigma}{\sqrt{n}}\]
For example, if \( \sigma = 6.30 \) and \( n = 16 \), then the standard error is \( \frac{6.30}{\sqrt{16}} = 1.575 \).

The standard error indicates how much variability one can expect in the sample mean from one sample to another. A smaller standard error means that the sample mean is closer to the population mean, leading to more precise confidence intervals. As sample size increases, the standard error tends to decrease, enhancing the precision of our confidence intervals.

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Most popular questions from this chapter

A researcher wants to determine a \(99 \%\) confidence interval for the mean number of hours that adults spend per week doing community service. How large a sample should the researcher select so that the estimate is within \(1.2\) hours of the population mean? Assume that the standard deviation for time spent per week doing community service by all adults is 3 hours.

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