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The formula used to compute a confidence interval for the mean of a normal population when \(n\) is small is $$\bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}}$$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=17\) b. \(90 \%\) confidence, \(n=12\) c. \(99 \%\) confidence, \(n=24\) d. \(90 \%\) confidence, \(n=25\) e. \(90 \%\) confidence, \(n=13\) f. \(95 \%\) confidence, \(n=10\)

Short Answer

Expert verified
The t critical values will vary based on the specific t table used or software. They should be close to the ones provided in any standard t table for respective degrees of freedom and confidence levels for scenarios a to f given.

Step by step solution

01

Understanding the t critical value

The t critical value would depend on two things: the desired level of confidence and the degree of freedom. The level of confidence is specified in the question. The degrees of freedom is calculated as \(n - 1\), where \(n\) is the sample size.
02

Use a t table or software

With the level of confidence and degrees of freedom, we can then lookup the t critical value in a t table or use statistical software or online calculator that would provide the desired values.
03

Calculate t critical values

Compute the t critical values for each of the scenarios given: a. 95% confidence, n=17: Degrees of freedom=16. b. 90% confidence, n=12: Degrees of freedom=11. c. 99% confidence, n=24: Degrees of freedom=23. d. 90% confidence, n=25: Degrees of freedom=24. e. 90% confidence, n=13: Degrees of freedom=12. f. 95% confidence, n=10: Degrees of freedom=9.

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Key Concepts

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

t Critical Value
The t critical value is a crucial component in constructing confidence intervals, especially when dealing with small sample sizes. It determines how many standard deviations away from the mean the confidence interval should extend in a t-distribution.
To find the t critical value, two main components are needed: the level of confidence and the degrees of freedom. With these, you can look up the t critical value in a t distribution table or use statistical software.
Unlike the standard normal distribution (z distribution), which is used for larger sample sizes (n > 30), the t-distribution adjusts for smaller samples, providing more reliable intervals. This makes it important because, in smaller samples, data variability can make it harder to precisely estimate the population mean.
Degrees of Freedom
Degrees of freedom are a statistical concept used to approximate the deviation of sample data from the actual population mean. For the t-distribution, the degrees of freedom (df) are calculated as the sample size minus one, expressed as: \[ df = n - 1 \] Where \(n\) is the sample size.
This adjustment accounts for the variability and uncertainty introduced by estimating the population mean using the sample data.
In practice:
  • For a sample size of 17, the degrees of freedom is 16 (17 - 1).
  • For a sample size of 12, the degrees of freedom is 11 (12 - 1).
  • The concept ensures each data point in the sample influences the calculation of variances, leading to more accurate estimates.
Level of Confidence
The level of confidence indicates the probability that the confidence interval will contain the true population parameter. It is usually expressed as a percentage, such as 90%, 95%, or 99%.
A higher confidence level means a wider interval, which increases the certainty that the interval covers the true mean, but at the cost of precision. Conversely, a lower confidence level offers more precision with less certainty.
In constructing confidence intervals:
  • A 95% confidence level implies that if we were to take 100 different samples and compute a confidence interval from each, we expect about 95 of them to contain the true population mean.
  • Statistical software or tables help associate these confidence levels with t critical values, adjusting intervals appropriately based on sample size and distribution.
Normal Distribution
The normal distribution is a fundamental concept in statistics, characterized by its symmetrical, bell-shaped curve. It describes how data is distributed around the mean and is defined by its mean and standard deviation.
For small sample sizes, the t-distribution is used instead of the normal distribution because it accounts for the additional uncertainty inherent in smaller samples (more breadth in the tails of the curve).
Both normal and t distributions are continuous probability distributions, but the t distribution has thicker tails. This feature provides more robustness in estimating population parameters under conditions where the normal distribution might not fully apply (i.e., small sample sizes or unknown population variances).
Understanding the relationship between these distributions helps ensure the appropriate statistical tools are applied correctly, maximizing the reliability of inferential statistics.

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Most popular questions from this chapter

Given a variable that has a \(t\) distribution with the specified degrees of freedom, what percentage of the time will its value fall in the indicated region? a. \(10 \mathrm{df}\), between \(-1.81\) and \(1.81\) b. \(10 \mathrm{df}\), between \(-2.23\) and \(2.23\) c. 24 df, between \(-2.06\) and \(2.06\) d. \(24 \mathrm{df}\), between \(-2.80\) and \(2.80\) f. \(24 \mathrm{df}\), to the right of \(2.80\) g. \(10 \mathrm{df}\), to the left of \(-1.81\)

The article "Most Canadians Plan to Buy Treats. Many Will Buy Pumpkins, Decorations and/or Costumes" (Ipsos-Reid, October 24, 2005) summarized results from a survey of 1000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween during \(2005 .\) The resulting sample mean and standard deviation were \(\$ 46.65\) and \(\$ 83.70\), respectively. a. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be larger than the mean anticipated expense. b. Is it reasonable to think that the distribution of the variable anticipated Halloween expense is approximately normal? Explain why or why not. c. Is it appropriate to use the \(t\) confidence interval to estimate the mean anticipated Halloween expense for Canadian residents? Explain why or why not. d. If appropriate, construct and interpret a \(99 \%\) confidence interval for the mean anticipated Halloween expense for Canadian residents.

Recent high-profile legal cases have many people reevaluating the jury system. Many believe that juries in criminal trials should be able to convict on less than a unanimous vote. To assess support for this idea, investigators asked each individual in a random sample of Californians whether they favored allowing conviction by a \(10-2\) verdict in criminal cases not involving the death penalty. The Associated Press (San Luis Obispo Telegram-Tribune, September 13,1995 ) reported that \(71 \%\) supported the \(10-2\) verdict. Suppose that the sample size for this survey was \(n=900\). Compute and interpret a \(99 \%\) confidence interval for the proportion of Californians who favor the \(10-2\) verdict.

The paper "The Curious Promiscuity of Queen Honey Bees (Apis mellifera): Evolutionary and Behavioral Mechanisms" (Annals of Zoology Fennici [2001]:255- 265) describes a study of the mating behavior of queen honeybees. The following quote is from the paper: Queens flew for an average of \(24.2 \pm 9.21\) minutes on their mating flights, which is consistent with previous findings. On those flights, queens effectively mated with \(4.6 \pm 3.47\) males (mean \(\pm \mathrm{SD})\). a. The intervals reported in the quote from the paper were based on data from the mating flights of \(n=\) 30 queen honeybees. One of the two intervals reported is stated to be a confidence interval for a population mean. Which interval is this? Justify your choice. b. Use the given information to construct a \(95 \%\) confidence interval for the mean number of partners on a mating flight for queen honeybees. For purposes of this exercise, assume that it is reasonable to consider these 30 queen honeybees as representative of the population of queen honeybees.

In the article "Fluoridation Brushed Off by Utah" (Associated Press, August 24, 1998 ), it was reported that a small but vocal minority in Utah has been successful in keeping fluoride out of Utah water supplies despite evidence that fluoridation reduces tooth decay and despite the fact that a clear majority of Utah residents favor fluoridation. To support this statement, the article included the result of a survey of Utah residents that found \(65 \%\) to be in favor of fluoridation. Suppose that this result was based on a random sample of 150 Utah residents. Construct and interpret a \(90 \%\) confidence interval for \(p\), the true proportion of Utah residents who favor fluoridation. Is this interval consistent with the statement that fluoridation is favored by a clear majority of residents?

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