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Using the result of the preceding problem find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) when \(\mathrm{n}=10, \mathrm{~m}=7, \underline{\mathrm{X}}=4.2\) \(\mathrm{Y}=3.4, \mathrm{~S}_{1}^{2}=49, \mathrm{~S}_{2}^{2}=32\)

Short Answer

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\(= 0.8\) #tag_title#Step 2: Calculate the standard error for the difference in means#tag_content# Next, compute the standard error for the difference in means using the formula: $$SE(\underline{X} - \underline{Y}) = \sqrt{\frac{S_{1}^{2}}{n} + \frac{S_{2}^{2}}{m}} = \sqrt{\frac{49}{10} + \frac{32}{7}}\) \(= \sqrt{7.142}\) #tag_title#Step 3: Find the t-value for the given confidence level#tag_content# Now, find the t-value corresponding to a 90% confidence level using the degrees of freedom: $$df = min(n - 1, m - 1) = min(9, 6) = 6$$ Consulting a t-table or using calculator yields \(t = 1.943\). #tag_title#Step 4: Calculate the margin of error#tag_content# Multiply the standard error by the t-value to find the margin of error: $$ME = t \times SE = 1.943 \times \sqrt{7.142} = 4.16\) #tag_title#Step 5: Construct the confidence interval#tag_content# Finally, construct the 90% confidence interval for the difference in means: $$(\underline{X} - \underline{Y}) \pm ME = 0.8 \pm 4.16$$ The 90% confidence interval for \(\mu_{1} - \mu_{2}\) is \((-3.36, 4.96)\).

Step by step solution

01

Compute the point estimate for the difference in means

First, we compute the point estimate of the difference in means, which is the difference of the sample means: $$\underline{X} - \underline{Y} = 4.2 - 3.4$$

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Key Concepts

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

Difference in Means
The difference in means is a concept that we use when we are comparing two different groups to see how their average values, or means, differ. In this context, we're examining the difference in means between two sets of data, often referred to as two samples.
We calculate it by subtracting one sample's mean from the other.
When we talk about a "difference in means," we're typically trying to understand if the difference we observe is significant or just due to random chance. In practice, it helps us put a number on how much one group tends to differ from another.
  • If the difference is large, it can indicate important underlying differences between the groups.
  • If it is small or zero, both groups might be quite similar in the characteristics being measured.
In the given exercise, the difference is captured by the expression \(\underline{X} - \underline{Y}\), which, in our specific problem, computes to 4.2 - 3.4. This value gives us the numerical difference in means between the two samples, serving as a foundational step in constructing a confidence interval for the difference.
Sample Means
Sample means are averages that we calculate from a sample of data. A sample is essentially a set or a subset of data taken from a larger population for the purpose of analysis.
The mean of that sample is a key statistic that serves as an estimate of the population mean. In many statistical analyses, like this one, calculating the sample mean is a critical step to understanding the central tendency of that sample. This mean is represented often as \(\underline{X}\) or \(\underline{Y}\) when we have multiple groups or samples under investigation.
  • \(\underline{X}\) represents the mean of the first sample, for example, which in this exercise is given as 4.2.
  • \(\underline{Y}\) represents the mean of the second sample, and here it is 3.4.
Sample means form the basis for calculating other statistics like the variance, standard deviation, and forming the point estimate for the difference between two means.
Point Estimate
A point estimate is a single value that is our best guess for a particular parameter, such as a mean, in a population based on a sample statistic. It's the starting point for making further statistical conclusions, such as constructing confidence intervals.
In the context of the problem given, the point estimate is crucial for determining the difference between the two sample means, \(\underline{X} - \underline{Y}\). This result can then be used to predict the actual difference between the population means \(\mu_{1} - \mu_{2}\).
  • The point estimate simplifies our view by providing a singular number like 0.8 for the difference, which we derived from the sample means (4.2 for \(\underline{X}\) and 3.4 for \(\underline{Y}\)).
  • It provides a basis for the confidence interval, which helps determine the reliability and range of our estimate.
Without the point estimate, it would be challenging to translate sample information into meaningful predictions about the general population.

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Most popular questions from this chapter

Neighboring fanners, Mr. Solinger and Mr. Rottner, were planting different varieties of corn. Each planted 100 acres under similar conditions. Mr. Solinger's yield was 84 bushels per acre with a standard deviation, \(\sigma_{\mathrm{S}}\), of 5 bushels. Mr. Rottner had a yield of 80 bushels per acre with a standard deviation, \(\sigma_{R}\), of \(6 .\) Assume all sampling normal, (a) What is the maximum likelihood estimate of the difference in means? (b) Make a \(90 \%\) confidence interval estimate for the mean difference in yield between the two farms.

Tom and Joe like to throw darts. Tom throws 100 times and hits the target 54 times; Joe throws 100 times and hits the target 49 times. Find a 95 per cent confidence interval for \(\mathrm{p}_{1}-\mathrm{p}_{2}\) where \(\mathrm{p}_{1}\) represents the true proportion of hit in Tom's tosses, and \(\mathrm{p}_{2}\) represents the true proportion of hits in Joe's tosses.

The last \(21 \mathrm{Mr}\). Americas' had a variance of \(6 \mathrm{in}^{2}\) in their chest size while the last \(16 \mathrm{Mr}\). U.S.A.s' had a variance of 4 in \(^{2}\) in the same category. Assuming normality, find a \(.98\) confidence interval estimate of [(Variance Mr. America) / (Variance Mr. U.S.A.)] .

During the \(1976-77\) season Coach Jerry Tarkanian outfitted his University of Nevada at Las Vegas basketball team with new sneakers. The 16 member team had an average size of \(14.5\) and a standard deviation of \(5 .\) Find a 90 percent confidence interval for the mean sneaker size of all collegiate basketball players. Assume the population is normal and the variance is not known.

In Grand Central Station, there is a coffee machine which is regulated so that the amount of coffee dispensed is normally distributed with a standard deviation of \(.5\) ounces per cup. A random sample of 50 cups had an average of 5 ounces per cup. (a) Find the maximum likelihood estimate for the average amount of coffee in each cup dispensed by the machine. (b) Find a \(95 \%\) confidence interval for the mean of all cups dispensed.

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