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An article in Computers \& Electrical Engineering ['Parallel Simulation of Cellular Neural Networks" (1996, Vol. 22, pp. \(61-84\) ) considered the speed- up of cellular neural networks (CNN) for a parallel general-purpose computing architecture based on six transputers in different areas. The data follow: \(\begin{array}{llllll}3.775302 & 3.350679 & 4.217981 & 4.030324 & 4.639692 \\\ 4.139665 & 4.395575 & 4.824257 & 4.268119 & 4.584193 \\ 4.930027 & 4.315973 & 4.600101 & & \end{array}\) (a) Is there evidence to support the assumption that speed-up of CNN is normally distributed? Include a graphical display in your answer. (b) Construct a \(95 \%\) two-sided confidence interval on the mean speed-up. (c) Construct a \(95 \%\) lower confidence bound on the mean speed-up.

Short Answer

Expert verified
The CNN speed-up data is approximately normal, with a 95% CI of (3.9858, 4.5344) and a lower bound of 4.0289.

Step by step solution

01

Calculate the sample statistics

First, calculate the sample mean \( \bar{x} \) and sample standard deviation \( s \) for the provided data. There are 13 observations: \- Sample mean \( \bar{x} \) is given by: \[\bar{x} = \frac{1}{13} \sum_{i=1}^{13} x_i = 4.2601\]- Sample standard deviation \( s \) is given by: \[s = \sqrt{\frac{1}{12} \sum_{i=1}^{13} (x_i - \bar{x})^2} = 0.4542\]
02

Assess normality using graphical display

To assess if the data is normally distributed, create a Normal Probability Plot (Q-Q plot) of the data. This involves plotting the observed values against the expected quantiles from a normal distribution. If the points roughly form a straight line, then the data may be considered approximately normally distributed.
03

Construct 95% Confidence Interval

We construct the confidence interval using the sample statistics.The 95% confidence interval is given by:\[\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\]Where \( t_{\alpha/2} \) is the t-value for 12 degrees of freedom at the 95% confidence level. For \( \alpha = 0.05 \), \( t_{0.025} \approx 2.179 \). \[4.2601 \pm 2.179 \cdot \frac{0.4542}{\sqrt{13}}\]Calculating further:\[4.2601 \pm 0.2743 \Rightarrow (3.9858, 4.5344)\]
04

Construct 95% Lower Confidence Bound

The lower confidence bound is calculated by using only one side of the t-distribution:\[\bar{x} - t_{\alpha} \cdot \frac{s}{\sqrt{n}}\]Using \( t_{0.05} \approx 1.782 \):\[4.2601 - 1.782 \cdot \frac{0.4542}{\sqrt{13}} = 4.0289\]This gives a 95% lower confidence bound of 4.0289 for the mean speed-up.

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Key Concepts

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

Confidence Interval
In statistics, a confidence interval is a range of values that is likely to contain the true value of an unknown population parameter. It is expressed as a percentage, such as 95%, indicating the degree of certainty that the true parameter is within this range.
For example, in the context of the CNN speed-up, we found a 95% two-sided confidence interval to be approximately \(3.9858, 4.5344\). This means we are 95% confident that the true mean speed-up falls within this range.
  • To calculate this, use the sample mean , the sample standard deviation, and the t-value, which arises from the student t-distribution.
  • It's important to adjust the t-value according to the sample size; here, with 13 observations, we used the t-value for 12 degrees of freedom.
  • This adjustment accounts for variability and ensures our interval provides a good estimate.
Understanding confidence intervals helps in making conclusions beyond the sample data, allowing predictions and decisions with a stated level of certainty.
Normal Distribution
The normal distribution is a probability distribution that is symmetric about the mean, depicting that data near the mean are more frequent in occurrence than data far from the mean. In many real-world situations, the normal distribution can be used as an approximation for other types of distributions.
To determine if data, like the CNN speed-up values, follow a normal distribution, a graphical assessment such as a Q-Q plot can be helpful. The Q-Q plot compares the distribution of the sample data to a normal distribution:
  • If the sample data points lie approximately along a straight line, it suggests the data could be normally distributed.
  • Assessing normality is crucial for confidence interval calculations since many statistical tests assume normally distributed data.
  • If data do not align with normality, alternative methods or transformations may be required.
Through a Q-Q plot, we can visually affirm the adequacy of the normal distribution assumption, which aids in constructing reliable statistical inferences.
Sample Statistics
Sample statistics are numerical characteristics calculated from a sample, used to estimate population parameters. In our case, we are interested in the sample mean and sample standard deviation, which are key to further analyses such as confidence intervals.
The sample mean \( \bar{x} \) is simply the average of all observed values, serving as an estimate of the population mean. For our CNN speed-up data, this was calculated as 4.2601.
  • The sample mean is calculated by summing all the sample values and dividing by the number of observations.
  • The sample standard deviation \( s \) measures the dispersion or spread of the observations around the sample mean.
  • Calculated as \( s = \sqrt{\frac{1}{n-1} \sum (x_i - \bar{x})^2} \,\) the standard deviation helps understand variability in the dataset.
Every statistical analysis begins with understanding these basic measures of central tendency and variability, which form the foundation for more complex statistical modeling.

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Most popular questions from this chapter

A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is normally distributed with standard deviation \(\sigma=20\). (a) Find a \(95 \% \mathrm{CI}\) for \(\mu\) when \(n=10\) and \(\bar{x}=1000\). (b) Find a \(95 \%\) CI for \(\mu\) when \(n=25\) and \(\bar{x}=1000\). (c) Find a \(99 \%\) CI for \(\mu\) when \(n=10\) and \(\bar{x}=1000\). (d) Find a \(99 \%\) CI for \(\mu\) when \(n=25\) and \(\bar{x}=1000\). (e) How does the length of the CIs computed above change with the changes in sample size and confidence level?

Suppose that \(n=100\) random samples of water from a freshwater lake were taken and the calcium concentration (milligrams per liter) measured. A \(95 \%\) CI on the mean calcium concentration is \(0.49 \leq \mu \leq 0.82\) (a) Would a \(99 \%\) CI calculated from the same sample data be longer or shorter? (b) Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between 0.49 and \(0.82 .\) Is this statement correct? Explain your answer. (c) Consider the following statement: If \(n=100\) random samples of water from the lake were taken and the \(95 \%\) CI on \(\mu\) computed, and this process were repeated 1000 times, 950 of the CIs would contain the true value of \(\mu\). Is this statement correct? Explain your answer.

A rivet is to be inserted into a hole. A random sample of \(n=15\) parts is selected, and the hole diameter is measured. The sample standard deviation of the hole diameter measurements is \(s=0.008\) millimeters. Construct a \(99 \%\) lower confidence bound for \(\sigma^{2}\).

An article in Knee Surgery, Sports Traumatology, Arthroscopy ["Arthroscopic Meniscal Repair with an Absorbable Screw: Results and Surgical Technique" ( \(2005,\) Vol. \(13,\) pp. \(273-279\) ) \(]\) showed that only 25 out of 37 tears \((67.6 \%)\) located between 3 and \(6 \mathrm{~mm}\) from the meniscus rim were healed. (a) Calculate a two-sided \(95 \%\) confidence interval on the proportion of such tears that will heal. (b) Calculate a \(95 \%\) lower confidence bound on the proportion of such tears that will heal.

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a continuous probability distribution with median \(\tilde{\mu}\). (a) Show that $$ \begin{aligned} P\left\\{\min \left(X_{i}\right)<\tilde{\mu}<\max \right.&\left.\left(X_{i}\right)\right\\} \\ &=1-\left(\frac{1}{2}\right)^{n-1} \end{aligned} $$ Hint: The complement of the event \(\left[\min \left(X_{i}\right)<\tilde{\mu}\right.\) \(\left.<\max \left(X_{i}\right)\right]\) is \(\left[\max \left(X_{i}\right) \leq \tilde{\mu}\right] \cup\left[\min \left(X_{i}\right) \leq \tilde{\mu}\right],\) but \(\max \left(X_{i}\right) \leq \tilde{\mu}\) if and only if \(X_{i} \leq \tilde{\mu}\) for all \(\left.i\right]\) (b) Write down a \(100(1-\alpha) \%\) confidence interval for the median \(\tilde{\mu}\), where $$ \alpha=\left(\frac{1}{2}\right)^{n-1} $$

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