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Consider a normal population distribution with the value of \(\sigma\) known. a. What is the confidence level for the interval \(\bar{x} \pm\) \(2.81 \sigma / \sqrt{n}\) ? b. What is the confidence level for the interval \(\bar{x} \pm\) \(1.44 \sigma / \sqrt{n} ?\) c. What value of \(z_{\alpha / 2}\) in the CI formula (7.5) results in a confidence level of \(99.7 \%\) ? d. Answer the question posed in part (c) for a confidence level of \(75 \%\).

Short Answer

Expert verified
a. 99.50%; b. 95.02%; c. 2.96; d. 1.15.

Step by step solution

01

Understanding the Confidence Interval

We begin by recognizing that the confidence interval formula for a normal distribution with known standard deviation (\(\sigma\)) is generally given by \(\bar{x} \pm z_{\alpha/2} (\sigma/\sqrt{n})\). Here, \(z_{\alpha/2}\) corresponds to the critical value for the desired confidence level.
02

Relating Critical Value to Confidence Level for (a)

We are given an interval \(\bar{x} \pm 2.81 (\sigma/\sqrt{n})\). This implies that the critical value \(z_{\alpha/2} = 2.81\). We need to find the confidence level corresponding to this \(z\)-value using the standard normal distribution table, which shows the area under the curve.
03

Finding Confidence Level for \(z=2.81\) (a)

Using the standard normal distribution table, the cumulative probability up to \(z=2.81\) is approximately 0.9975. This includes both tails of the distribution, so we calculate the confidence level by finding the complement of the extreme tails. Hence the confidence level is \(99.50\%\).
04

Relating Critical Value to Confidence Level for (b)

For part (b), the interval is \(\bar{x} \pm 1.44 (\sigma/\sqrt{n})\). Thus, \(z_{\alpha/2} = 1.44\). We will similarly use the standard normal distribution table to find the related confidence level.
05

Finding Confidence Level for \(z=1.44\) (b)

The cumulative probability for \(z=1.44\) is approximately 0.9251. To obtain the confidence level, we double this probability to account for both tails: \(0.9502\), indicating a confidence level of \(95.02\%\).
06

Solving Part (c) for \(99.7\%\) Confidence Level

The confidence level of \(99.7\%\) corresponds to covering \(0.003 / 2 = 0.0015\) in each tail. The cumulative probability needed is \(0.9985\) on each side, which corresponds to a \(z\)-value of approximately \(2.96\).
07

Solving Part (d) for \(75\%\) Confidence Level

For a \(75\%\) confidence level, the tails would cover \(0.25 / 2 = 0.125\). The cumulative probability required is \(0.875\), corresponding to a \(z\)-value of approximately \(1.15\).

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Key Concepts

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

Normal Distribution
The normal distribution is one of the most commonly discussed probability distributions, especially in statistics. It's shaped like a symmetrical bell curve, with data points concentrated around the mean. - The mean, median, and mode in a normal distribution are equal.- The curve is symmetric around the mean.- It’s defined by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)).In practical terms, many natural phenomena and test scores can be modeled using a normal distribution. This makes it incredibly useful for statistical analysis and inference. For example, when we talk about "bell curve" grading in exams, we are referring to distribution following this pattern. When working with normal distributions, it's crucial to understand how spread out the values are. The standard deviation determines the width of the curve: a small standard deviation indicates a steep curve, while a larger one results in a wider distribution.
Critical Value
The critical value is an essential concept when creating a confidence interval. It represents the point(s) on the standard normal distribution curve that marks the boundary for the top \( \alpha/2 \)n percentage of the data. This helps determine how "extreme" a value should be to be considered statistically significant. - A higher critical value generally results in a broader confidence interval.- It hinges on the desired confidence level, which represents the likelihood that the sample mean falls within this interval.For many standard statistical processes, tables or statistical software are used to find these values. For example:- If you're aiming for a 95% confidence interval, the critical value is often around 1.96.- The critical value for a 99% confidence level sits typically around 2.576.Knowing these values allows statisticians and researchers to set robust boundaries for their inferences or predictions.
Confidence Level
The confidence level is a vital statistic that tells us how certain we are that the true population parameter is within the confidence interval. A higher confidence level means more certainty that you are capturing the actual mean, but also results in a wider interval. - Common confidence levels include 90%, 95%, and 99%. - A 95% confidence level asserts that if you were to take 100 different samples, about 95 of the sample means would lie within the calculated interval. However, increasing the confidence level without increasing sample size can lead to less precision, as the interval might become excessively wide. Choosing the appropriate confidence level depends on the field of study and desired precision. Researchers need to balance the size of the interval against how sure they wish to be of their estimates, considering both statistical significance and practical applications.
Standard Normal Distribution
A special type of normal distribution, the standard normal distribution, emphasizes simplicity by standardizing data. It converts your normal distribution into a form that has a mean of 0 and a standard deviation of 1. This simplification is achieved through standardization:- Each data point is transformed using: \[ Z = \frac{X - \mu}{\sigma} \]Here,- \( X \) is the original data value,- \( \mu \) is the mean of the distribution,- \( \sigma \) is the standard deviation.nThe resulting \( Z \)-scores allow for easier comparison across different data sets and assessments of where a particular value stands in relation to the distribution.Standard normal distributions are key when using tables for critical values, because they simplify what would otherwise be complex calculations in unique data sets. Most importantly, they make it possible to apply statistical procedures universally across different types of data, by focusing on how far individual observations deviate from the mean.

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Most popular questions from this chapter

The alternating current \((\mathrm{AC})\) 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 \((\mathrm{kV})\) 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 \(95 \%\) CI to have a width of \(2 \mathrm{kV}\) (so that \(\mu\) is estimated to within \(1 \mathrm{kV}\) with \(95 \%\) confidence)?

The article "Distributions of Compressive Strength Obtained from Various Diameter Cores" (ACI Materials \(J ., 2012: 597-606\) ) described a study in which compressive strengths were determined for concrete specimens of various types, core diameters, and lengthto-diameter ratios. For one particular type, diameter, and \(/ / d\) ratio, the 18 tested specimens resulted in a sample mean compressive strength of 64.41 MPa and a sample standard deviation of \(10.32 \mathrm{MPa}\). Normality of the compressive strength distribution was judged to be quite plausible. a. Calculate a confidence interval with confidence level \(98 \%\) for the true average compressive strength under these circumstances. b. Calculate a \(98 \%\) lower prediction bound for the compressive strength of a single future specimen tested under the given circumstances. [Hint: \(t_{.01217}=\) 2.224.]

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streaming is gaining in popularity. The Harris poll reported on November 13,2012 , that \(53 \%\) of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device. a. Calculate and interpret a confidence interval at the \(99 \%\) confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time. b. What sample size would be required for the width of a \(99 \%\) CI to be at most. 05 irrespective of the value of \(\hat{p}\) ?

Ultra high performance concrete (UHPC) is a relatively new construction material that is characterized by strong adhesive properties with other materials. The article "Adhesive Power of Ultra High Performance Concrete from a Thermodynamic Point of View" \((J\). of Materials in Civil Engr., 2012: 1050-1058) described an investigation of the intermolecular forces for UHPC connected to various substrates. The following work of adhesion measurements (in \(\mathrm{mJ} / \mathrm{m}^{2}\) ) for UHPC specimens adhered to steel appeared in the article: \(\begin{array}{lllll}107.1 & 109.5 & 107.4 & 106.8 & 108.1\end{array}\) a. Is it plausible that the given sample observations were selected from a normal distribution? b. Calculate a two-sided \(95 \%\) confidence interval for the true average work of adhesion for UHPC adhered to steel. Does the interval suggest that 107 is a plausible value for the true average work of adhesion for UHPC adhered to steel? What about 110 ? c. Predict the resulting work of adhesion value resulting from a single future replication of the experiment by calculating a \(95 \%\) prediction interval, and compare the width of this interval to the width of the CI from (b). d. Calculate an interval for which you can have a high degree of confidence that at least \(95 \%\) of all UHPC specimens adhered to steel will have work of adhesion values between the limits of the interval.

A normal probability plot of the \(n=26\) observations on escape time given in Exercise 36 of Chapter I shows a substantial linear pattern; the sample mean and sample standard deviation are \(370.69\) and \(24.36\), respectively. a. Calculate an upper confidence bound for population mean escape time using a confidence level of \(95 \% .\) b. Calculate an upper prediction bound for the escape time of a single additional worker using a prediction level of \(95 \%\). How does this bound compare with the confidence bound of part (a)? c. Suppose that two additional workers will be chosen to participate in the simulated escape exercise. Denote their escape times by \(X_{27}\) and \(X_{28}\), and let \(\bar{X}_{\text {aew }}\) denote the average of these two values. Modify the formula for a PI for a single \(x\) value to obtain a PI for \(\bar{X}_{\text {mew }}\), and calculate a \(95 \%\) two-sided interval based on the given escape data.

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