Problem 1 Most recent tests and calculatio... [FREE SOLUTION] (2024)

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Chapter 7: Problem 1

Most recent tests and calculations estimate at the \(95 \%\) confidence levelthat mitochondrial Eve, the maternal ancestor to all living humans, lived\(138,000 \pm 18,\) ooo years ago. What is meant by "95\% confidence" in thiscontext? (A) A confidence interval of the true age of mitochondrial Eve has beencalculated using \(z\) -scores of ±1.96 . (B) A confidence interval of the true age of mitochondrial Eve has beencalculated using \(t\) -scores consistent with \(d f\) \(=n-1\) and tailprobabilities of \(\pm 0.025 .\) (C) There is a 0.95 probability that mitochondrial Eve lived between 120,000and 156,000 years ago. (D) If 20 random samples of data are obtained by this method and a \(95 \%\)confidence interval is calculated from each, the true age of mitochondrial Evewill be in 19 of these intervals. (E) Of all random samples of data obtained by this method, \(95 \%\) will yieldintervals that capture the true age of mitochondrial Eve.

Short Answer

Expert verified

D and E

Step by step solution

01

Understand the concept of confidence interval

A confidence interval gives an estimated range of values which is likely to include an unknown population parameter. The '95% confidence' means we are 95% certain that the true parameter lies within our calculated confidence interval.

02

Identify key characteristics of a 95% confidence interval

For a 95% confidence interval, typically, we use a z-score of ±1.96 for large samples. However, the t-score is used for small sample sizes and depending on degrees of freedom.

03

Evaluation of each option

Let's evaluate each option to see which one correctly describes the concept of a 95% confidence interval:

04

Option A

Option A suggests a confidence interval calculated using z-scores of ±1.96, which is generally correct for large sample sizes.

05

Option B

Option B talks about using t-scores consistent with degrees of freedom, which is suitable for smaller sample sizes.

06

Option C

Option C incorrectly interprets the confidence interval as stating there is a 0.95 probability that mitochondrial Eve lived between specific years. This misunderstanding is common but incorrect.

07

Option D

Option D correctly states that if we were to take 20 different samples and construct a 95% confidence interval for each, we'd expect about 19 of these intervals to contain the true age. This reflects the idea behind how confidence intervals are interpreted.

08

Option E

Option E is another correct interpretation, stating that of all random samples obtained by this method, 95% will yield intervals capturing the true age of mitochondrial Eve.

09

Conclusion

Options D and E are valid interpretations of what a 95% confidence interval means. Both options emphasize that in repeated sampling, 95% of the constructed intervals will contain the true parameter.

Key Concepts

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

confidence level

The confidence level is a statistical measure that quantifies the level of certainty in a confidence interval. For instance, a 95% confidence level means that if we repeatedly took random samples and constructed confidence intervals, 95% of those intervals would capture the true parameter value. This does not mean there's a 95% chance the true value is within a specific interval once it's calculated—rather, it's about long-run frequency. The higher the confidence level, the wider the confidence interval, as it aims to ensure the true parameter is captured more often.

z-scores

Z-scores are standardized scores that indicate how many standard deviations an element is from the mean. They are used in statistics for calculating confidence intervals, especially when the sample size is large (typically n > 30). For a 95% confidence interval, we use a z-score of ±1.96. This value comes from the standard normal distribution and suggests that 95% of the distribution is within 1.96 standard deviations from the mean. Hence, when constructing a confidence interval with a large sample, we use these z-scores to determine the margin of error.

t-scores

T-scores are used instead of z-scores when the sample size is small (usually n < 30). Unlike z-scores, t-scores account for additional uncertainty from having fewer data points. T-scores are derived from the t-distribution, which has thicker tails than the normal distribution, reflecting the greater spread in smaller samples. The t-score needed to calculate a confidence interval depends on both the desired confidence level and the degrees of freedom (df), which is usually the sample size minus one (n-1). As the sample size increases, the t-distribution approaches the standard normal distribution, making t-scores and z-scores more similar.

population parameter estimation

Population parameter estimation is the process of using sample data to estimate a parameter of the overall population, such as the mean or proportion. Confidence intervals are a key tool in this process, providing a range of values within which we expect the true population parameter to lie, with a certain level of confidence. Estimations help in making inferences about the population without needing to collect data from every individual, saving time and resources. Accurate estimation relies on representative sampling methods and appropriate use of statistical formulas and distributions.

degrees of freedom

Degrees of freedom (df) refer to the number of values in a calculation that are free to vary without breaking a constraint. In the context of confidence intervals and t-scores, the degrees of freedom are typically the sample size minus one (n-1). This concept is crucial because it affects the shape of the t-distribution used to calculate the confidence interval. More degrees of freedom mean the t-distribution becomes more like the normal distribution, reducing the impact of sample size on the interval. Thus, understanding and correctly calculating degrees of freedom ensure accurate and reliable statistical inferences.

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Problem 1 Most recent tests and calculatio... [FREE SOLUTION] (3)

Most popular questions from this chapter

A pharmaceutical company claims that a medicine will produce a desired effectfor a mean time of 58.4 minutes. A government researcher runs a hypothesistest of 40 patients and calculates a mean of \(\bar{x}=59.5\) with a standarddeviation of \(s=8.3 .\) What is the \(P\) -value? (A) \(P\left(t>\frac{59.5-58.4}{8.3 / \sqrt{40}}\right)\) with \(d f=39\) (B) \(P\left(t>\frac{59.5-58.4}{8.3 / \sqrt{40}}\right)\) with \(d f=40\) (C) \(2 P\left(t>\frac{59.5-58.4}{8.3 / \sqrt{40}}\right)\) with \(d f=39\) (D) \(2 P\left(t>\frac{59.5-58.4}{8.3 / \sqrt{40}}\right)\) with \(d f=40\) (E) \(2 P\left(z>\frac{59.5-58.4}{8.3 / \sqrt{40}}\right)\)A consumer testing agency plans to calculate a \(99 \%\) confidence interval forthe mean mpg for all cars on the road in 2019. Suppose the mpg measurementsfor the population of interest is actually sharply skewed right. For which ofthe sample sizes, \(n=30,50,\) or \(70,\) would the sampling distribution of\(\bar{x}\) be closest to normal? (A) 30 (B) 50 (C) 70 (D) Because of skewness of the population, none of the sampling distributionscan be approximately normal. (E) Because of the central limit theorem, all sampling distributions with \(n\geq 30\) are equally approximately normal.A company manufactures a synthetic rubber bungee cord with a braided coveringof natural rubber and a minimum breaking strength of \(450 \mathrm{~kg}\). Ifthe mean breaking strength of a sample drops below a specified level, theproduction process is halted and the machinery inspected. Which of thefollowing would result from a Type I error? (A) Halting the production process when too many cords break (B) Halting the production process when the breaking strength is below thespecified level (C) Halting the production process when the breaking strength is withinspecifications (D) Allowing the production process to continue when the breaking strength isbelow specifications (E) Allowing the production process to continue when the breaking strength iswithin specificationsIn a study aimed at reducing developmental problems in lowbirth-weight babies(under 2500 grams), 347 infants were exposed to a special educationalcurriculum while 561 did not receive any special help. After 3 years, thechildren exposed to the special curriculum showed a mean IQ of 93.5 with a standard deviationof 19.1; the other children had a mean IQ of 84.5 with a standard deviation of \(19.9 .\) Find a \(95 \%\) confidenceinterval estimate for the difference in mean IQs of all low-birth-weightbabies who receive special intervention and those who do not. (A) \((93.5-84.5) \pm 1.97\sqrt{\frac{(19.1)^{2}}{347}+\frac{(19.9)^{2}}{561}}\) (B) \((93.5-84.5) \pm1.97\left(\frac{19.1}{\sqrt{307}}+\frac{19.9}{\sqrt{561}}\right)\) (C) \((93.5-84.5) \pm 1.97\sqrt{\frac{(19.1)^{2}}{347}+\frac{(19.9)^{2}}{561}}\) (D) \((93.5-84.5) \pm1.65\left(\frac{19.1}{\sqrt{37}}+\frac{19.9}{\sqrt{561}}\right)\) (E) \((93.5-84.5) \pm 1.65 \sqrt{\frac{(19.1)^{2}+(19.9)^{2}}{347+561}}\)Suppose (25,30) is a \(90 \%\) confidence interval estimate for a populationmean \(\mu\). Which of the following is a true statement? (A) There is a 0.90 probability that \(\bar{x}\) is between 25 and \(30 .\) (B) Of the sample values, \(90 \%\) are between 25 and 30 . (C) There is a o.9o probability that \(\mu\) is between 25 and 30 . (D) If 100 random samples of the given size are picked and a \(90 \%\)confidence interval estimate is calculated from each, \(\mu\) will be in 90 ofthe resulting intervals. (E) If \(90 \%\) confidence intervals are calculated from all possible samplesof the given size, \(\mu\) will be in \(90 \%\) of these intervals.
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Problem 1 Most recent tests and calculatio... [FREE SOLUTION] (2024)
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