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Find a \((1-\alpha) 100 \%\) confidence interval for a population mean \(\mu\) for these values: a. \(\alpha=.01, n=38, \bar{x}=34, s^{2}=12\) b. \(\alpha=.10, n=65, \bar{x}=1049, s^{2}=51\) c. \(\alpha=.05, n=89, \bar{x}=66.3, s^{2}=2.48\)

Short Answer

Expert verified
a. (32.47, 35.53) b. (1046.35, 1051.65) c. (65.83, 66.77)

Step by step solution

01

Find the t-score

Find the t-score \(t_{\frac{\alpha}{2}, n-1}\) for the given significance level \(\alpha\) and sample size \(n\), using a t-distribution table, calculator, or statistical software. a. For \(\alpha=0.01\) and \(n=38\), the degrees of freedom are \(df=38-1=37\). Using a t-distribution table, the t-score is approximately \(2.701\). b. For \(\alpha=0.10\) and \(n=65\), the degrees of freedom are \(df=65-1=64\). Using a t-distribution table, the t-score is approximately \(1.669\). c. For \(\alpha=0.05\) and \(n=89\), the degrees of freedom are \(df=89-1=88\). Using a t-distribution table, the t-score is approximately \(1.984\).
02

Calculate the sample standard deviation

Calculate the sample standard deviation \(s\) by taking the square root of the given sample variance, \(s^2\). a. \(s = \sqrt{12} \approx 3.46\) b. \(s = \sqrt{51} \approx 7.14\) c. \(s = \sqrt{2.48} \approx 1.57\)
03

Calculate the Confidence Interval

Now, we will use the formula to calculate the confidence interval: \(\bar{x} \pm t_{\frac{\alpha}{2}, n-1} \cdot \frac{s}{\sqrt{n}}\) a. The confidence interval for case a is \(34 \pm 2.701 \cdot \frac{3.46}{\sqrt{38}}\), which is approximately \((32.47, 35.53)\). b. The confidence interval for case b is \(1049 \pm 1.669 \cdot \frac{7.14}{\sqrt{65}}\), which is approximately \((1046.35, 1051.65)\). c. The confidence interval for case c is \(66.3 \pm 1.984 \cdot \frac{1.57}{\sqrt{89}}\), which is approximately \((65.83, 66.77)\).
04

Final answer

The confidence intervals are as follows: a. \((32.47, 35.53)\) b. \((1046.35, 1051.65)\) c. \((65.83, 66.77)\)

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Key Concepts

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

t-distribution
The t-distribution is an essential concept when working with small sample sizes. Unlike the normal distribution, which is ideal for large samples, the t-distribution takes into account the additional variability inherent in smaller datasets.
This distribution is symmetric and bell-shaped, similar to the normal distribution, but with thicker tails. This means it is more prone to producing values that fall far from its mean, reflecting more uncertainty.
  • The thicker tails account for small sample sizes and an unknown population standard deviation.
  • As the sample size increases, the t-distribution approaches a normal distribution.
You use the t-distribution when constructing confidence intervals for a population mean, especially when the population standard deviation is unknown. This distribution allows you to determine the t-score, which is crucial in calculating the interval range. The t-score is simply a factor derived from your desired confidence level and your degrees of freedom.
sample variance
Sample variance is a measure of the spread or dispersion within a sample data set. It quantifies how much the individual data points deviate from the sample mean.
The formula for sample variance, denoted as \(s^2\), involves the average of the squared differences from the mean.
It gives an idea of how much variability there is in the sample.
  • Calculated as \(s^2 = \frac{1}{n-1} \sum_{i=1}^{n}(x_i - \bar{x})^2\)
  • Useful in calculating the sample standard deviation, \(s\), which is the square root of the sample variance.
  • A smaller variance indicates that the data points tend to be closer to the mean.
In the context of confidence intervals, the sample variance helps determine the width of the interval by affecting the standard error. A larger variance typically results in a wider confidence interval, reflecting increased uncertainty about the population mean.
population mean
The population mean, denoted by \(\mu\), is a key parameter in statistics. It represents the average of all measurements in a population. In practice, it's often impossible to measure every single data point, so we estimate \(\mu\) using a sample.
  • The sample mean, \(\bar{x}\), is calculated as the sum of all observations divided by the number of observations.
  • The sample mean serves as an unbiased estimator of the population mean \(\mu\).
When constructing a confidence interval, the goal is to estimate the range within which the true population mean lies, with a certain level of confidence.
This is especially valuable when making predictions or deciding on policy, as the mean provides a central measure of the data set, helping to understand general trends.
degrees of freedom
Degrees of freedom pertain to the number of independent values in a calculation. In statistical analysis, especially when estimating parameters, it's crucial to understand and correctly apply degrees of freedom.
For a confidence interval of the population mean, degrees of freedom are calculated as the sample size minus one, \(n - 1\).
  • They adjust for the estimation of the sample mean, which reduces the number of available independent values.
  • The degrees of freedom impact the critical t-score used in the t-distribution.
With more degrees of freedom, the t-distribution resembles the normal distribution more closely. Consequently, with larger sample sizes, the degrees of freedom increase, making the confidence interval calculations more precise.
Understanding the concept of degrees of freedom ensures you apply the correct t-score when determining the confidence interval.

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Most popular questions from this chapter

Multimedia Kids Do our children spend as much time enjoying the outdoors and playing with family and friends as previous generations did? Or are our children spending more and more time glued to the television, computer, and other multimedia equipment? A random sample of 250 children between the ages of 8 and 18 showed that 170 children had a TV in their bedroom and that 120 of them had a video game player in their bedroom. a. Estimate the proportion of all 8 - to 18 -year-olds who have a TV in their bedroom, and calculate the margin of error for your estimate. b. Estimate the proportion of all 8 - to 18 -year-olds who have a video game player in their bedroom, and calculate the margin of error for your estimate.

If 36 measurements of the specific gravity of aluminum had a mean of 2.705 and a standard deviation of .028 , construct a \(98 \%\) confidence interval for the actual specific gravity of aluminum.

You want to estimate the mean hourly yield for a process that manufactures an antibiotic. You observe the process for 100 hourly periods chosen at random, with the results \(\bar{x}=34\) ounces per hour and \(s=3\). Estimate the mean hourly yield for the process using a \(95 \%\) confidence interval.

Samples of 400 printed circuit boards were selected from each of two production lines \(A\) and \(B\). Line A produced 40 defectives, and line B produced 80 defectives. Estimate the difference in the actual fractions of defectives for the two lines with a confidence coefficient of \(.90 .\)

How likely are you to vote in the next national election? In a survey by \(P e w\) Research, \(^{10}\) fully \(77 \%\) of the registered Republican voters are absolutely going to vote this year while only \(65 \%\) of Democrats are absolutely going to vote in the next election. The sample consisted of 469 registered Republicans, 490 registered Democrats, and 480 registered Independents. a. Construct a \(98 \%\) confidence interval for the proportion of registered Republicans who say they are absolutely going to vote in the next election. If a Republican senator predicts that at least \(85 \%\) of registered Republicans will absolutely vote in the next election, is this figure realistic? b. Construct a \(99 \%\) confidence interval for the proportion of registered Democrats who say they are absolutely going to vote in the next election.

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