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Determine the \(t\) critical value that will capture the desired \(t\) curve area in each of the following cases: a. Central area \(=.95\), df \(=10\) b. Central area \(=.95, \mathrm{df}=20\) c. Central area \(=99\), df \(=20\) d. Central area \(=.99, \mathrm{df}=50\) e. Upper-tail area \(=.01, \mathrm{df}=25\) f. Lower-tail area \(=.025\), df \(=5\)

Short Answer

Expert verified
a: ±2.228, b: ±2.086, c: ±2.845, d: ±2.678, e: 2.485, f: -2.571.

Step by step solution

01

Understanding the Problem

The problem requires us to find the critical values of the t-distribution for different cases of central and tail areas given different degrees of freedom (df). For instances with central areas, we'll divide the remaining area equally between both tails.
02

Case a - Find Critical t for Central Area = 0.95, df = 10

For a central area of 0.95 with 10 degrees of freedom, the area in each tail is (1 - 0.95) / 2 = 0.025. We look for the t-distribution value with df = 10 and upper-tail probability of 0.025. Using a t-table or calculator, the critical value is approximately t = ±2.228.
03

Case b - Find Critical t for Central Area = 0.95, df = 20

With a central area of 0.95 and df = 20, the tail probability is again 0.025 on each side. Using a t-table or calculator, the critical value is approximately t = ±2.086.
04

Case c - Find Critical t for Central Area = 0.99, df = 20

Here the central area is 0.99. Thus, the tail probability is (1 - 0.99) / 2 = 0.005 per tail. Using a t-table or calculator, the critical value for df = 20 and upper-tail probability 0.005 is approximately t = ±2.845.
05

Case d - Find Critical t for Central Area = 0.99, df = 50

With a central area of 0.99 and df = 50, each tail then has an area of 0.005. Using a t-table or calculator, the critical value is approximately t = ±2.678.
06

Case e - Find Critical t for Upper-Tail Area = 0.01, df = 25

For an upper-tail area of 0.01 with df = 25, the critical value can be directly looked up for the upper-tail probability of 0.01. Using a t-table or calculator, the t-value is approximately t = 2.485.
07

Case f - Find Critical t for Lower-Tail Area = 0.025, df = 5

This case specifies a lower-tail area of 0.025 with df = 5. We need the negative t-value corresponding to a lower tail probability of 0.025. Using a t-table or calculator, the critical value is approximately t = -2.571.

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Key Concepts

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

Degrees of Freedom (df)
The concept of 'degrees of freedom' (df) is fundamental in statistics, especially when dealing with the t-distribution. Simply put, degrees of freedom represent the number of independent values or quantities which can be assigned to a statistical distribution. In most cases, it is calculated as the sample size minus one \[ df = n - 1 \] where \( n \) is the sample size. It reflects the amount of variation or freedom we allow for change within our sample data. In the context of the t-distribution, knowing the degrees of freedom is crucial because it affects the shape of the t-distribution curve. A smaller df value results in a curve with heavier tails, which means there's more probability at the extremes. Conversely, a larger df results in a curve more closely resembling a standard normal distribution.
Understanding this allows us to determine how much uncertainty or variability we have when making estimations from our sample data to the broader population.
Central Area
The 'central area' of a t-distribution curve refers to the probability mass found between two critical values within the tails of the distribution. It represents a confidence interval that accounts for the bulk of the variability within a data set, usually implying how confident we can be that a parameter falls within the specified range.
For example, if we have a central area of 0.95, it means that 95% of our data is covered within the range, leaving 5% in the two tails of the distribution. To find critical values corresponding to such a central area, you equally distribute the leftover probability across both tails. Here is the formula to calculate it:
\[ \text{Tail Probability} = \frac{1 - \text{Central Area}}{2} \] Each tail would thus hold a probability of 0.025 when the central area is 0.95. Knowing the central area allows us to calculate these tail probabilities, which are essential when drafting confidence intervals or performing hypothesis tests.
Tail Probability
'Tail probability' is the probability that accounts for the values that fall in the tails of a distribution, beyond a certain point in the distribution's overall range. It plays a key role in statistical inference, especially for determining critical values using the t-distribution.
When we talk about tails in a t-distribution, we refer to the outer edges of the curve, where extremely low or high values lie. The tail probability tells us how likely it is, given our distribution, that a value will fall beyond our specified critical points. In practice, once a central area is specified, the remaining probability is split between the two tails, each receiving half. For an upper-tail area or a lower-tail area solely focused on one side, no splitting is needed, and the specified area directly corresponds to the tail probability. For example, a lower-tail probability of 0.025 means there's a 2.5% chance that any given value from our data set will fall below a certain cutoff point. Understanding these probabilities is vital for interpreting data in hypothesis tests and assessing statistical significance.
T-Table or Calculator
Using a t-table or a calculator is indispensable when it comes to determining critical values in the t-distribution, which are necessary for confidence intervals and hypothesis testing. A t-table provides a tabular form of critical values for different degrees of freedom and probabilities, typically arranged by one-tailed and two-tailed tests. To determine a critical value using a t-table, you need to know your degrees of freedom and the probability corresponding to your analysis, whether it be for tail or central area applications.
For example, look up the df value in one column and then move to the row that corresponds with your required tail probability to find the critical t-value. In instances where precision is key or manual lookup is cumbersome, statistical calculators or statistical software can be utilized. These tools allow you to input the degrees of freedom and desired tail probabilities or the central area directly to generate accurate critical values quickly. Utilizing these resources can significantly improve accuracy and efficiency in statistical analysis.

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Most popular questions from this chapter

Even as traditional markets for sweetgum lumber have declined, large section solid timbers traditionally used for construction bridges and mats have become increasingly scarce. The article "Development of Novel Industrial Laminated Planks from Sweetgum Lumber" (J. of Bridge Engr., 2008: 64-66) described the manufacturing and testing of composite beams designed to add value to low- grade sweetgum lumber. Here is data on the modulus of rupture (psi; the article contained summary data expressed in MPa): \(\begin{array}{llllll}6807.99 & 7637.06 & 6663.28 & 6165.03 & 6991.41 & 6992.23 \\ 6981.46 & 7569.75 & 7437.88 & 6872.39 & 7663.18 & 6032.28 \\\ 6906.04 & 6617.17 & 6984.12 & 7093.71 & 7659.50 & 7378.61 \\ 7295.54 & 6702.76 & 7440.17 & 8053.26 & 8284.75 & 7347.95 \\ 7422.69 & 7886.87 & 6316.67 & 7713.65 & 7503.33 & 7674.99\end{array}\) a. Verify the plausibility of assuming a normal population distribution. b. Estimate the true average modulus of rupture in a way that conveys information about precision and reliability. c. Predict the modulus for a single beam in a way that conveys information about precision and reliability. How does the resulting prediction compare to the estimate in (b).

Each of the following is a confidence interval for \(\mu=\) true average (i.c., population mean) resonance frequency \((\mathrm{Hz})\) for all tennis rackets of a certain type: \((114.4,115.6) \quad(114.1,115.9)\) a. What is the value of the sample mean resonance frequency? b. Both intervals were calculated from the same sample data. The confidence level for one of these intervals is \(90 \%\) and for the other is \(99 \%\). Which of the intervals has the \(90 \%\) confidence level, and why?

A sample of 66 obese adults was put on a lowcarbohydrate diet for a year. The average weight loss was \(11 \mathrm{lb}\) and the standard deviation was \(19 \mathrm{lb}\). Calculate a \(99 \%\) lower confidence bound for the true average weight loss. What does the bound say about confidence that the mean weight loss is positive?

Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting \(95 \%\) confidence interval is \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. b. Consider the following statement There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: We can be highly confident that \(95 \%\) of all bottles of this type of cough syrup have an alcohol content that is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? d. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

Here are the lengths (in minutes) of the 63 nineinning games from the first week of the 2001 major league baseball season: \(\begin{array}{llllllllll}194 & 160 & 176 & 203 & 187 & 163 & 162 & 183 & 152 & 177 \\ 177 & 151 & 173 & 188 & 179 & 194 & 149 & 165 & 186 & 187 \\ 187 & 177 & 187 & 186 & 187 & 173 & 136 & 150 & 173 & 173 \\ 136 & 153 & 152 & 149 & 152 & 180 & 186 & 166 & 174 & 176 \\ 198 & 193 & 218 & 173 & 144 & 148 & 174 & 163 & 184 & 155 \\ 151 & 172 & 216 & 149 & 207 & 212 & 216 & 166 & 190 & 165 \\ 176 & 158 & 198 & & & & & & & \end{array}\) Assume that this is a random sample of nineinning games (the mean differs by \(12 \mathrm{~s}\) from the mean for the whole season). a. Give a \(95 \%\) confidence interval for the population mean. b. Give a \(95 \%\) prediction interval for the length of the next nine-inning game. On the first day of the next week, Boston beat Tampa Bay 3-0 in a nine- inning game of \(152 \mathrm{~min}\). Is this within the prediction interval? c. Compare the two intervals and explain why one is much wider than the other. d. Explore the issue of nomality for the data and explain how this is relevant to parts (a) and (b).

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