/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 The alternating current (AC) bre... [FREE SOLUTION] | 91Ó°ÊÓ

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The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article "Testing Practices for the AC Breakdown Voltage Testing of Insulation Liquids" (IEEE Electrical Insulation Magazine, 1995: 21-26) gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions. \(\begin{array}{llllllllllllllll}62 & 50 & 53 & 57 & 41 & 53 & 55 & 61 & 59 & 64 & 50 & 53 & 64 & 62 & 50 & 68 \\ 54 & 55 & 57 & 50 & 55 & 50 & 56 & 55 & 46 & 55 & 53 & 54 & 52 & 47 & 47 & 55 \\ 57 & 48 & 63 & 57 & 57 & 55 & 53 & 59 & 53 & 52 & 50 & 55 & 60 & 50 & 56 & 58\end{array}\) a. Construct a boxplot of the data and comment on interesting features. b. Calculate and interpret a \(95 \%\) CI for true average breakdown voltage \(\mu\). Does it appear that \(\mu\) has been precisely estimated? Explain. c. Suppose the investigator believes that virtually all values of breakdown voltage are between 40 and 70 . What sample size would be appropriate for the \(95 \% \mathrm{Cl}\) to have a width of \(2 \mathrm{kV}\) (so that \(\mu\) is estimated to within \(1 \mathrm{kV}\) with \(95 \%\) confidence)?

Short Answer

Expert verified
a. Boxplot shows slight right skewness and possibly outliers. b. 95% CI for mean is [55.51, 58.79], fairly precise estimate. c. Sample size of 124 needed for 95% CI width of 2 kV.

Step by step solution

01

Organize the Data to Construct a Boxplot

First, organize the data. Identify the minimum, first quartile (Q1), median, third quartile (Q3), and maximum values of the dataset. Our dataset consists of the breakdown voltages given in the problem: \[62, 50, 53, 57, 41, 53, 55, 61, 59, 64, 50, 53, 64, 62, 50, 68, 54, 55, 57, 50, 55, 50, 56, 55, 46, 55, 53, 54, 52, 47, 47, 55, 57, 48, 63, 57, 57, 55, 53, 59, 53, 52, 50, 55, 60, 50, 56, 58\].Arrange these in ascending order, then calculate the quartiles.
02

Draw the Boxplot

Using the five-number summary (minimum, Q1, median, Q3, maximum), draw the boxplot. - Minimum: 41 - Q1: 51 - Median: 54 - Q3: 57 - Maximum: 68 The boxplot will have a box from Q1 to Q3, with a line at the median. Draw whiskers from the minimum to Q1 and from Q3 to the maximum. Note any outliers if applicable.
03

Comment on Boxplot Features

From the boxplot, observe that: - The distribution is slightly skewed to the right, as the median is closer to Q1 than Q3. - There are potential outliers, particularly at the higher end of the data, indicated by values beyond the upper whisker.
04

Calculate the Sample Mean and Standard Deviation

The sample mean \( \bar{x} \) is calculated by summing all observations and dividing by the number of data points.\[ \bar{x} = \frac{\text{sum of all values}}{n} = \frac{2743}{48} \approx 57.15 \]The sample standard deviation \( s \) is calculated using the formula:\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \]Substitute the values to get \( s \approx 5.67 \).
05

Compute the 95% Confidence Interval for \( \mu \)

The 95% CI for the mean is given by:\[ \bar{x} \pm Z_{0.025} \left( \frac{s}{\sqrt{n}} \right) \]Where \( Z_{0.025} \approx 1.96 \) is the critical value from the standard normal distribution for a 95% CI. Using the calculated mean and standard deviation:\[ 57.15 \pm 1.96 \left( \frac{5.67}{\sqrt{48}} \right) \]Calculating this gives an interval of approximately \([55.51, 58.79]\). This indicates the range within which the true mean \( \mu \) is expected to lie with 95% confidence.
06

Evaluate Precision of the Estimate

The 95% CI of \([55.51, 58.79]\) shows a relatively narrow range, indicating a reasonably precise estimate of \( \mu \). The calculation does not suggest significant variability, given the reasonable sample size.
07

Determine the Required Sample Size for CI Width of 2 kV

To achieve a desired confidence interval width \( w \) of 2 kV:\[ n = \left( \frac{Z_{0.025} \times s}{w/2} \right)^2 \]Substitute \( Z_{0.025} = 1.96 \), \( s = 5.67 \), and \( w = 2 \):\[ n = \left( \frac{1.96 \times 5.67}{1} \right)^2 \approx 123.29 \]Rounding up, a sample size of 124 observations is required to achieve the desired interval width.

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Key Concepts

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

Boxplot
A boxplot is a graphical representation that shows key statistical measures of a dataset, such as the median, quartiles, and potential outliers. It's a useful tool to visually assess the distribution of data.

To create a boxplot, start by organizing the data and calculating important statistics:
  • Minimum: The smallest data point.
  • First Quartile (Q1): The midpoint of the lower half of the data.
  • Median: The middle value when the data is sorted.
  • Third Quartile (Q3): The midpoint of the upper half of the data.
  • Maximum: The largest data point.
With these calculations, a box is drawn from Q1 to Q3 with a line through the median. This highlights the spread and the central tendency of the data.

Whiskers extend from the edges of the box to the minimum and maximum values, excluding outliers, which are marked separately. Using this dataset:
  • Minimum: 41
  • Q1: 51
  • Median: 54
  • Q3: 57
  • Maximum: 68
The boxplot shows a right skew, as the median (54) is closer to Q1 than Q3. This means there are more data values on the higher end of the range.

Outliers, such as values beyond the whiskers, suggest variations that might not reflect the central data trends. Identifying these can be crucial in understanding data distribution and the potential influence of extreme values.
Confidence Interval
A Confidence Interval (CI) provides a range of values that likely contain a population parameter, such as the mean, with a specified level of confidence.

The 95% CI means that we can be 95% confident that the true population mean falls within this range. It's calculated using the sample mean, standard deviation, and the size of the sample.

The formula for a 95% CI is:\[\bar{x} \pm Z_{0.025} \left( \frac{s}{\sqrt{n}} \right)\]where:
  • \( \bar{x} \) is the sample mean.
  • \( Z_{0.025} \approx 1.96 \) for a 95% confidence level.
  • \( s \) is the standard deviation.
  • \( n \) is the sample size.
The sample mean here is approximately 57.15, with a standard deviation of 5.67 over 48 samples. Calculating the CI gives us a range of approximately \[ [55.51, 58.79] \].

Since the interval is relatively narrow, it suggests a precise estimation of the population mean. In practical terms, it's a focused estimate indicating where the real average breakdown voltage likely lies. This level of precision implies confidence in predicting behaviors based on the sample data.
Sample Size Determination
Sample size determination is crucial when planning experiments or studies. It ensures that the estimates of parameters, like means or proportions, are precise and reliable.

Given a specific confidence interval width, we calculate the needed sample size using: \[n = \left( \frac{Z_{0.025} \times s}{w/2} \right)^2\]where:
  • \( Z_{0.025} \) is approximately 1.96 for a 95% confidence level.
  • \( s \) is the estimated standard deviation.
  • \( w \) is the desired total width of the CI.
In this context, the goal was a CI width of 2 kV, meaning \( w/2 = 1 \) kV. An estimated \( s \) of 5.67 translates into needing approximately 124 observations.

This process ensures the CI is narrow enough to provide useful insights. A larger sample size typically results in a more precise CI, reducing uncertainty and providing stronger evidence for decision-making. It's always better to balance resource availability and desired precision when determining sample size.

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Most popular questions from this chapter

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