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Chapter 7: Problem 2

Suppose (25,30) is a \(90 \%\) confidence interval estimate for a population mean \(\mu\). 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 of the resulting intervals. (E) If \(90 \%\) confidence intervals are calculated from all possible samples of the given size, \(\mu\) will be in \(90 \%\) of these intervals.

Short Answer

Expert verified
Option (E) is a true statement.

Step by step solution

01

Understanding Confidence Interval

A confidence interval is a range of values that's likely to contain a population parameter with a certain level of confidence. Here, the interval (25, 30) is a 90% confidence interval.
02

Statement Evaluation - Option A

Option (A) states: 'There is a 0.90 probability that \(\bar{x}\) is between 25 and 30.' This is incorrect because the confidence interval pertains to the population mean \(\mu\), not the sample mean \(\bar{x}\).
03

Statement Evaluation - Option B

Option (B) states: 'Of the sample values, 90% are between 25 and 30.' This is incorrect because the confidence interval does not indicate where the sample values fall, but rather where the population mean might lie based on the sample data.
04

Statement Evaluation - Option C

Option (C) states: 'There is a 0.90 probability that \(\mu\) is between 25 and 30.' This may seem true, but it is not a correct interpretation of a confidence interval. The confidence interval means if we were to take many samples and compute many confidence intervals, 90% of them would contain \(\mu\); it does not give the probability for a single interval.
05

Statement Evaluation - Option D

Option (D) states: '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 of the resulting intervals.' This is a correct interpretation of a confidence interval. It means that the method used to construct the interval will capture the true population mean in 90 out of 100 samples.
06

Statement Evaluation - Option E

Option (E) states: 'If 90% confidence intervals are calculated from all possible samples of the given size, \(\mu\) will be in 90% of these intervals.' This is also correct and aligns with the definition of the confidence interval.
07

Conclusion

Both options (D) and (E) are correct. They correctly explain the interpretation of a 90% confidence interval. However, according to the structure of the problem, usually only one option is accepted, typically (E) since it universally explains the confidence level concept.

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Key Concepts

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

Population Mean
The population mean, denoted by \(\mu\), is the average value of a characteristic in a population. It is a fixed value and not subject to change. In the case of our exercise, when we refer to the population mean, we are talking about the true average that exists in the entire group of interest, usually too large to measure completely.

For example, if you want to understand the average height of all students in a school, the population mean would be the true average height of every single student in that school. Because it is often impractical to measure everyone in the population, we take a sample from the population and use it to estimate the population mean.
Sample Data
Sample data are the values collected from a subset of the population. The mean of these sample values is known as the sample mean and is denoted by \(\bar{x}\). Sample data are easier to gather compared to measuring the entire population.

Consider the previous example of a school. Instead of measuring the height of all students, you might measure the height of a random group of 30 students. These measurements form your sample data. Your sample mean, \(\bar{x}\), would be the average height of those 30 students. We use sample data to make inferences about the population mean because it is much more manageable to collect and analyze.
Confidence Level
The confidence level represents the degree of certainty we have that the population parameter lies within the calculated confidence interval. It is typically expressed as a percentage, such as 90%, 95%, or 99%.

In our example, a 90% confidence level means that if we were to repeat the sampling process and calculate a confidence interval 100 times, about 90 of those intervals would be expected to contain the true population mean \(\mu\). It is important to note that this does not mean there is a 90% probability that the population mean is within a single given interval; rather, it is about the reliability of the method used to compute the interval.
Interval Estimation
Interval estimation involves calculating a range of values, derived from sample data, that is likely to include the population parameter. The range is known as the confidence interval.

For our exercise, the interval \(25, 30\) is a 90% confidence interval for the population mean \(\mu\). This means we are 90% confident the true population mean falls within this interval.

If we sampled 100 different groups and calculated the confidence interval for each, we would expect around 90 of those intervals to contain the true population mean. Interval estimation provides a way of expressing uncertainty about a population parameter and is a crucial concept for statistical inference.

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Most popular questions from this chapter

One gallon of gasoline is put into each of an SRS of 30 test autos, and the resulting mileage figures are tabulated with \(\bar{x}\) \(=28.5\) and \(s=1.2 .\) Determine a \(95 \%\) confidence interval estimate of the mean mileage of all comparable autos. (A) \(28.5 \pm 2.045(1.2)\) (B) \(28.5 \pm 2.045\left(\frac{1.2}{\sqrt{29}}\right)\) (C) \(28.5 \pm 2.045\left(\frac{1.2}{\sqrt{29}}\right)\) (D) \(28.5 \pm 1.96\left(\frac{1.2}{\sqrt{29}}\right)\) (E) \(28.5 \pm 1.96\left(\frac{1.2}{\sqrt{29}}\right)\)

A pharmaceutical company claims that a medicine will produce a desired effect for a mean time of 58.4 minutes. A government researcher runs a hypothesis test of 40 patients and calculates a mean of \(\bar{x}=59.5\) with a standard deviation 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 for the mean mpg for all cars on the road in 2019. Suppose the mpg measurements for the population of interest is actually sharply skewed right. For which of the 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 distributions can be approximately normal. (E) Because of the central limit theorem, all sampling distributions with \(n \geq 30\) are equally approximately normal.

Do high school girls apply to more colleges than high school boys? A two- sample \(t\) -test of the hypotheses \(H_{0}: \mu_{\text {girls }}=\mu_{\text {boys }}\) versus \(H_{a}: \mu_{\text {girls }}>\mu_{\text {boys }}\) results in a \(P\) -value of 0.02 . Which of the following statements must be true? I. A \(90 \%\) confidence interval for the difference in means contains o. II. A \(95 \%\) confidence interval for the difference in means contains o. III. A 99\% confidence interval for the difference in means contains o. (A) I only (B) III only (C) I and II only (D) II and III only (E) I, II, and III

Thirty students volunteer to test which of two strategies for taking multiple- choice exams leads to higher average results. Each student flips a coin, and if heads, uses Strategy A on the first exam and Strategy \(\mathrm{B}\) on the second, while if tails, uses Strategy B on the first exam and Strategy A on the second. The average of all 30 Strategy A results is then compared to the average of all 30 Strategy B results. What is the conclusion at the \(5 \%\) significance level if a two- sample hypothesis test, \(H_{0}: \mu_{1}=\mu_{2}, H_{\mathrm{a}}: \mu_{1} \neq \mu_{2},\) results in a \(P\) -value of \(0.18 ?\) (A) The observed difference in average scores is significant. (B) The observed difference in average scores is not significant. (C) A conclusion is not possible without knowing the average scores resulting from using each strategy. (D) A conclusion is not possible without knowing the average scores and the standard deviations resulting from using each strategy. (E) A two-sample hypothesis test should not be used here.
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