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A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=5.2, s_{1}=2.7, n_{1}=10\) and \(\bar{x}_{2}=4.9, s_{2}=2.8, n_{2}=8\)

Short Answer

Expert verified
The \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) is (-2.261, 2.861).

Step by step solution

01

Identify the given statistics.

\(\bar{x}_{1}=5.2, s_{1}=2.7, n_{1}=10, \bar{x}_{2}=4.9, s_{2}=2.8, n_{2}=8\)
02

Insert the given values into the formula.

Substitute the given values into the confidence interval formula: \(5.2 - 4.9 ± 1.96 \sqrt{\frac{2.7^{2}}{10} + \frac{2.8^{2}}{8}}\).
03

Perform the calculations inside the square root.

Calculate the values inside the square root: \(5.2 - 4.9 ± 1.96 \sqrt{\frac{7.29}{10} + \frac{7.84}{8}}\).
04

Simplify the expression inside the square root.

Calculate the values inside the square root to get: \(5.2 - 4.9 ± 1.96 \sqrt{0.729 + 0.98}\).
05

Perform the calculation inside the square root again.

By adding the values together inside the square root we get: \(5.2 - 4.9 ± 1.96 \sqrt{1.709}\).
06

Calculate the final confidence interval.

Solve the rest of the above expression to finally obtain: \(0.3 ± 1.96(1.307)\). Calculate the values: \(0.3 ± 2.561 = (-2.261, 2.861)\)
07

Interpreting the final confidence interval.

This interval represents the range of values that the difference in population means, \(\mu_{1}-\mu_{2}\), is likely to fall within, with a 95% confidence.

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Key Concepts

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

Population Means
The concept of population means refers to the average of a set of values drawn from the entire population. When we are interested in comparing two groups, like in the given problem, we denote these averages with symbols such as \(\mu_1\) and \(\mu_2\). In practical terms, it's often impossible or impractical to measure every individual in a population. Thus, we use samples to estimate these true population means.

For example, we might be comparing average heights of men and women in a particular city. Rather than measuring every individual, we take a sample from each group. The averages of these samples are our best estimates of \(\mu_1\) and \(\mu_2\), the true population means. However, it's important to remember that these sample means are just estimates, and the true values might differ slightly. This uncertainty is why concepts like confidence intervals are important.
Sample Statistics
Sample statistics are values computed from a sample that help us estimate population parameters. The key sample statistics used to calculate a confidence interval are:

  • The sample mean (\(\bar{x}\)), which provides an estimate of the population mean \(\mu\).
  • The sample standard deviation (\(s\)), which is used to measure the spread or dispersion of sample values.
  • The sample size (\(n\)), which indicates how many observations are included in the sample.

In the exercise, we're given multiple sample statistics for two populations. These include \(\bar{x}_1 = 5.2\), \(s_1 = 2.7\), and \(n_1 = 10\) for the first group, and \(\bar{x}_2 = 4.9\), \(s_2 = 2.8\), and \(n_2 = 8\) for the second group. By utilizing these values, we can proceed to compute the confidence interval which estimates the difference between the population means \(\mu_1 - \mu_2\). The precision of our estimates largely depends on the sample size; larger samples tend to give more reliable results.
Confidence Level
The confidence level is a measure of certainty or reliability associated with a confidence interval. It's usually expressed as a percentage, such as \(95\%\), which indicates the degree of confidence that the interval contains the true population parameter.

Choosing a confidence level is critical because it affects the width of the confidence interval:
  • A higher confidence level (like \(99\%\)) leads to a wider interval, reflecting higher certainty.
  • A lower confidence level (like \(90\%\)) results in a narrower interval, but with less certainty.

In our example, we are working with a \(95\%\) confidence level, which is a commonly accepted standard in statistics. This means that if we were to take 100 different samples and compute a confidence interval for each, we could expect about 95 of those intervals to contain the true difference between the population means \((\mu_1 - \mu_2)\). Therefore, the confidence interval provides a range that likely includes the true parameter, offering a balance between certainty and specificity. This helps to make informed conclusions based on statistical evidence.

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Most popular questions from this chapter

How big is the home field advantage in the National Football League (NFL)? In Exercise 6.240 on page 419 , we examine a difference in means between home and away teams using two separate samples of 80 games from each group. However, many factors impact individual games, such as weather conditions and the scoring of the opponent. It makes more sense to investigate this question using a matched pairs design, using scores for home and away teams matched for the same game. The data in NFLScores2011 include the points scored by the home and away team in 256 regular season games in \(2011 .\) We will treat these games as a sample of all NFL games. Estimate average home field scoring advantage and find a \(90 \%\) confidence interval for the mean difference.

Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=-2.6, s_{d}=4.1\) \(n_{d}=18\)

Involve scores from the high school graduating class of 2010 on the SAT (Scholastic Aptitude Test). The average score on the Mathematics part of the SAT exam for males is 534 with a standard deviation of 118 , while the average score for females is 500 with a standard deviation of 112 . (a) If random samples are taken with 40 males and 60 females, find the mean and standard deviation of the distribution of differences in sample means, \(\bar{x}_{m}-\bar{x}_{f},\) where \(\bar{x}_{m}\) represents the sample mean for the males and \(\bar{x}_{f}\) represents the sample mean for the females. (b) Repeat part (a) if the random samples contain 400 males and 600 females. (c) What effect do the different sample sizes have on center and spread of the distribution?

Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllllll} \hline \text { Situation } 1 & 125 & 156 & 132 & 175 & 153 & 148 & 180 & 135 & 168 & 157 \\ \text { Situation } 2 & 120 & 145 & 142 & 150 & 160 & 148 & 160 & 142 & 162 & 150 \\ \hline \end{array} $$

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 50 from a population with proportion 0.25

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