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(a) Suppose a \(95 \%\) confidence interval for the difference of means contains both positive and negative numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain both positive and negative numbers? Explain. What about a \(90 \%\) confidence interval? Explain. (b) Suppose a \(95 \%\) confidence interval for the difference of proportions contains all positive numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain all positive numbers as well? Explain. What about a \(90 \%\) confidence interval? Explain.

Short Answer

Expert verified
(a) A 99% interval may include zero, the 90% interval might not. (b) A 99% interval may include zero, the 90% interval likely remains positive.

Step by step solution

01

Understanding Confidence Intervals

Confidence intervals provide a range of values within which we expect the true parameter to lie, with a certain probability. A wider interval corresponds to a higher confidence level, while a narrower interval corresponds to a lower confidence level.
02

Analyzing 95% Confidence Interval with Positive and Negative Values

If a 95% confidence interval contains both positive and negative numbers, it indicates that zero is within the range of plausible values for the parameter. Thus, there is no significant evidence to suggest whether the parameter is positive or negative.
03

Exploring 99% Confidence Interval for Difference of Means

A 99% confidence interval will be wider than a 95% interval because increasing the confidence level requires accounting for more variability. The interval is likely to contain more values, and if the 95% interval includes both positive and negative numbers, the 99% interval will likely also include zero and, hence, both positive and negative numbers.
04

Exploring 90% Confidence Interval for Difference of Means

A 90% confidence interval will be narrower than a 95% interval. Therefore, if the 95% interval contains both positive and negative values, the 90% interval might not. It may exclude zero if the narrower interval falls entirely to one side of zero.
05

Analyzing 95% Confidence Interval with All Positive Values for Proportions

If a 95% confidence interval for the difference of proportions contains all positive numbers, it shows evidence that the difference is likely positive.
06

Exploring 99% Confidence Interval for Difference of Proportions

A 99% confidence interval will be wider than a 95% interval, so there's a possibility it might include zero even if the 95% interval showed only positive values. This larger range accounts for more variability, potentially including both sides of zero.
07

Exploring 90% Confidence Interval for Difference of Proportions

A 90% confidence interval will be narrower, and if the 95% interval showed all positive values, the 90% is also likely to be positive, as the narrowed range would still lie within the 95% interval.

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Key Concepts

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

Difference of Means
The difference of means is a fundamental concept used when comparing two distinct groups to understand how their averages differ. In many research scenarios, you're often interested in finding out whether the average score, measurement, or outcome in one group varies significantly from another.

To determine this, statisticians calculate the means of each group and then subtract one mean from the other, giving a statistical "difference of means." Understanding the nature of this difference helps researchers make informed decisions about whether an observed effect is meaningful or likely due to chance.

If the confidence interval for the difference of means includes zero, it suggests that there is no significant difference between the group means at the chosen confidence level. This is because zero within the interval implies that the means could be equal.

It's crucial to consider both the confidence level and the width of the confidence interval to interpret whether the observed difference is statistically significant. More variability in the data yields a wider interval, especially as the confidence level increases.
Difference of Proportions
The difference of proportions is similar to the difference of means but is used for binary data or percentages. It's aimed at understanding whether two different groups differ in terms of proportions, such as the proportion of success rates, preference rates, or any other binary outcome.

Just like with means, you calculate separate proportions for each group and then subtract one from the other. This difference can help determine whether the discrepancy between groups is statistically significant.

When evaluating confidence intervals for proportions, a key focus is whether the interval includes zero. If a confidence interval for the difference of proportions only contains positive values, it indicates a significant difference favoring one group over the other at that confidence level.

However, as the confidence level increases, such as moving from a 95% to a 99% confidence interval, the range of plausible values widens. This could potentially introduce zero into the interval if the difference was initially small, raising questions about the statistical significance of the observed difference.
Confidence Level
Confidence level is a critical concept that defines the probability that a confidence interval contains the true parameter. Commonly used levels are 90%, 95%, and 99%, each indicating a different degree of certainty. A higher confidence level means we want to be more certain about our estimate containing the true parameter, so we widen the interval to account for more variability.

If you're using a 90% confidence level, you're saying there's a 90% probability that the true parameter value lies within your calculated interval. Conversely, for a 99% confidence level, you claim a 99% probability.

However, a higher confidence level leads to a wider interval. This is because you're allowing for more variability, thus sacrificing some precision in your estimate's specificity to gain confidence in covering the true value. Conversely, a lower confidence level yields a narrower interval, which might provide a more precise estimate at the cost of certainty.
  • Higher confidence level = Wider interval
  • Lower confidence level = Narrower interval
Choosing the right confidence level involves balancing the need for precision and certainty, often depending on the context and stakes of the research at hand.

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Most popular questions from this chapter

Finance: \(\mathrm{P} / \mathrm{E}\) Ratio The price of a share of stock divided by a company's estimated future earnings per share is called the P/E ratio. High P/E ratios usually indicate "growth" stocks, or maybe stocks that are simply overpriced. Low \(\mathrm{P} / \mathrm{E}\) ratios indicate "value" stocks or bargain stocks. A random sample of 51 of the largest companies in the United States gave the following P/E ratios (Reference: Forbes). $$ \begin{array}{rrrrrrrrrrrr} 11 & 35 & 19 & 13 & 15 & 21 & 40 & 18 & 60 & 72 & 9 & 20 \\ 29 & 53 & 16 & 26 & 21 & 14 & 21 & 27 & 10 & 12 & 47 & 14 \\ 33 & 14 & 18 & 17 & 20 & 19 & 13 & 25 & 23 & 27 & 5 & 16 \\ 8 & 49 & 44 & 20 & 27 & 8 & 19 & 12 & 31 & 67 & 51 & 26 \\ 19 & 18 & 32 & & & & & & & & & \end{array} $$ (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 25.2\) and \(s \approx 15.5\). (b) Find a \(90 \%\) confidence interval for the \(\mathrm{P} / \mathrm{E}\) population mean \(\mu\) of all large U.S. companies. (c) Find a \(99 \%\) confidence interval for the \(\mathrm{P} / \mathrm{E}\) population mean \(\mu\) of all large U.S. companies. (d) Interpretation Bank One (now merged with J.P. Morgan) had a P/E of 12 , AT\&T Wireless had a \(\mathrm{P} / \mathrm{E}\) of 72 , and Disney had a \(\mathrm{P} / \mathrm{E}\) of 24 . Examine the confidence intervals in parts (b) and (c). How would you describe these stocks at the time the sample was taken? (e) Check Requirements In previous problems, we assumed the \(x\) distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: See the central limit theorem in Section \(6.5\).

22\. | Opinion Poll: Crime and Violence A New York Times/CBS poll asked the question, "What do you think is the most important problem facing this country today?" Nineteen percent of the respondents answered, "Crime and violence." The margin of sampling error was plus or minus 3 percentage points. Following the convention that the margin of error is based on a \(95 \%\) confidence interval, find a \(95 \%\) confidence interval for the percentage of the population that would respond, "Crime and violence" to the question asked by the pollsters.

Medical: Plasma Compress At Community Hospital, the burn center is experimenting with a new plasma compress treatment. A random sample of \(n_{1}=316\) patients with minor burns received the plasma compress treatment. Of these patients, it was found that 259 had no visible scars after treatment. Another random sample of \(n_{2}=419\) patients with minor burns received no plasma compress treatment. For this group, it was found that 94 had no visible scars after treatment. Let \(p_{1}\) be the population proportion of all patients with minor burns receiving the plasma compress treatment who have no visible scars. Let \(p_{2}\) be the population proportion of all patients with minor burns not receiving the plasma compress treatment who have no visible scars. (a) Check Requirements Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation Explain the meaning of the confidence interval found in part (b) in the context of the problem. Does the interval contain numbers that are all positive? all negative? both positive and negative? At the \(95 \%\) level of confidence, does treatment with plasma compresses seem to make a difference in the proportion of patients with visible scars from minor burns?

Pro Football and Basketball: Heights of Players Independent random samples of professional football and basketball players gave the following information (References: Sports Encyclopedia of Pro Football and Official NBA Basketball Encyclopedia). Note: These data are also available for download at the Online Study Center. $$ \begin{aligned} &\text { Heights (in } \mathrm{ft} \text { ) of pro football players: } x_{1} ; n_{1}=45\\\ &\begin{array}{llllllllll} 6.33 & 6.50 & 6.50 & 6.25 & 6.50 & 6.33 & 6.25 & 6.17 & 6.42 & 6.33 \\ 6.42 & 6.58 & 6.08 & 6.58 & 6.50 & 6.42 & 6.25 & 6.67 & 5.91 & 6.00 \\ 5.83 & 6.00 & 5.83 & 5.08 & 6.75 & 5.83 & 6.17 & 5.75 & 6.00 & 5.75 \\ 6.50 & 5.83 & 5.91 & 5.67 & 6.00 & 6.08 & 6.17 & 6.58 & 6.50 & 6.25 \\ 6.33 & 5.25 & 6.67 & 6.50 & 5.83 & & & & & \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Heights (in } \mathrm{ft} \text { ) of pro basketball players: } x_{2} ; n_{2}=40\\\ &\begin{array}{llllllllll} 6.08 & 6.58 & 6.25 & 6.58 & 6.25 & 5.92 & 7.00 & 6.41 & 6.75 & 6.25 \\ 6.00 & 6.92 & 6.83 & 6.58 & 6.41 & 6.67 & 6.67 & 5.75 & 6.25 & 6.25 \\ 6.50 & 6.00 & 6.92 & 6.25 & 6.42 & 6.58 & 6.58 & 6.08 & 6.75 & 6.50 \\ 6.83 & 6.08 & 6.92 & 6.00 & 6.33 & 6.50 & 6.58 & 6.83 & 6.50 & 6.58 \end{array} \end{aligned} $$ (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 6.179, s_{1} \approx 0.366, \bar{x}_{2} \approx 6.453\), and \(s_{2} \approx 0.314 .\) (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2}\). Find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Interpretation Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(90 \%\) level of confidence, do professional football players tend to have a higher population mean height than professional basketball players? (d) Check Requirements Which distribution (standard normal or Student's \(t)\) did you use? Why? Do you need information about the height distributions? Explain.

FBI Report: Larceny Thirty small communities in Connecticut (population near 10,000 each) gave an average of \(\bar{x}=138.5\) reported cases of larceny per year. Assume that \(\sigma\) is known to be \(42.6\) cases per year (Reference: Crime in the United States, Federal Bureau of Investigation). (a) Find a \(90 \%\) confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (b) Find a \(95 \%\) confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (c) Find a \(99 \%\) confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (d) Compare the margins of error for parts (a) through (c). As the confidence levels increase, do the margins of error increase? (e) Critical Thinking: Compare the lengths of the confidence intervals for parts (a) through (c). As the confidence levels increase, do the confidence intervals increase in length?

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