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Data from two samples gave the following results: $$ \begin{array}{|lcc|}\hline & \text { Sample 1 } & \text { Sample 2 } \\\\\hline \bar{y} & 96.2 & 87.3 \\\\\mathrm{SE} & 3.7 & 4.6 \\\\\hline\end{array}$$ $$ \text { Compute the standard error of }\left(\bar{Y}_{1}-\bar{Y}_{2}\right) $$.

Short Answer

Expert verified
The standard error of the difference between the two sample means is approximately 5.9.

Step by step solution

01

Understand the Concept of Standard Error for Difference in Means

The standard error (SE) of the difference between two sample means is used to measure the spread of the sampling distribution of that difference. The formula to calculate the standard error of the difference between two independent sample means is \( SE(\bar{Y}_1 - \bar{Y}_2) = \sqrt{SE_1^2 + SE_2^2} \), where \( SE_1 \) and \( SE_2 \) are the standard errors of the two samples, respectively.
02

Plug in the Given Values

Use the standard errors given for Sample 1 and Sample 2 to calculate the standard error of the difference between the two sample means. Given \( SE_1 = 3.7 \) and \( SE_2 = 4.6 \), plug these values into the formula: \( SE(\bar{Y}_1 - \bar{Y}_2) = \sqrt{3.7^2 + 4.6^2} \).
03

Calculate the Standard Error of the Difference

Perform the calculation: \( SE(\bar{Y}_1 - \bar{Y}_2) = \sqrt{3.7^2 + 4.6^2} = \sqrt{13.69 + 21.16} = \sqrt{34.85} \approx 5.9 \). Round to an appropriate number of significant figures based on the context.

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Key Concepts

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

Sampling Distribution
When conducting experiments or research, understanding the concept of a sampling distribution is crucial. Imagine you draw a multitude of samples, each one the same size, from a particular population. Each of these samples has its own mean, and if we were to plot all these sample means, we’d get what is called the sampling distribution of the mean. In essence, it is a probability distribution of all possible means that could be obtained from a given sample size.

Importantly, the central limit theorem tells us that, regardless of the population's shape, the sampling distribution of the mean will tend to be normally distributed if the sample size is large enough. This holds true especially if the samples are independent of each other and are drawn from a population with a finite level of variance. By knowing this, we can move on to understanding why the standard error is so important in statistical analysis.
Sample Means
The sample mean, often denoted as \( \bar{y} \), is simply the average of all the measurements in a sample. It serves as an estimator of the population mean. For example, if you take a random sample of people and measure their heights, the sample mean height would give you a good idea of the average height in the entire population if your sample is truly representative.

However, we have to consider that different samples will yield different means. This variability among sample means can tell us a lot about the population's characteristics and the precision of our sample mean as an estimate of the population mean. The more samples we take, the more we can be confident about our estimates; that's where the concept of the standard error comes into play, which measures the variation of the sample means.
Standard Error Calculation
The standard error is a statistical term that measures the accuracy with which a sample distribution represents a population by using standard deviation. In simpler terms, it tells us how far off we can expect our sample mean to be from the true population mean. The calculation for the standard error of a single sample mean is \( \frac{\sigma}{\sqrt{n}} \) where \( \sigma \) is the population standard deviation and \( n \) is the sample size.

For the case involving two sample means, the standard error of the difference between these means helps us understand how much variation exists between the two. It's important when comparing two groups. You calculate it using the formula provided in the solution you mentioned: \( SE(\bar{Y}_1 - \bar{Y}_2) = \sqrt{SE_1^2 + SE_2^2} \). This calculation is essential when we're looking to gauge whether a difference between two sample means is statistically significant or if it might just be due to random chance.

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Most popular questions from this chapter

The diameter of the stem of a wheat plant is an important trait because of its relationship to breakage of the stem, which interferes with harvesting the crop. An agronomist measured stem diameter in eight plants of the Tetrastichon cultivar of soft red winter wheat. All obser-- \(-\) wheat vations were made 3 weeks after flowering of the plant. The stem diameters (mm) were as follows \(^{50}\). $$ \begin{array}{llllllll} 2.3 & 2.6 & 2.4 & 2.2 & 2.3 & 2.5 & 1.9 & 2.0 The mean of these data is 2.275 and the standard deviation is \(0.238 .\) (a) Calculate the standard error of the mean. (b) Construct a \(95 \%\) confidence interval for the population mean. (c) Define in words the population mean that you estimated in part (b). (See Example 6.1.1.) \end{array} $$

For each of the following, decide whether the description fits the SD or the SE. (a) This quantity is a measure of the accuracy of the sample mean as an estimate of the population mean. (b) This quantity tends to stay the same as the sample size goes up. (c) This quantity tends to go down as the sample size goes up.

In evaluating a forage crop, it is important to measure the concentration of various constituents in the plant tissue. In a study of the reliability of such measurements, a batch of alfalfa was dried, ground, and passed through a fine screen. Five small \((0.3 \mathrm{gm})\) aliquots of the alfalfa were then analyzed for their content of insoluble ash. The results \((\mathrm{gm} / \mathrm{kg})\) were as follows: $$ \begin{array}{lllll} 10.0 & 8.9 & 9.1 & 11.7 & 7.9 \end{array} $$ For these data, calculate the mean, the standard deviation, and the standard error of the mean.

In an experiment on soybean varieties, individually potted soybean plants were grown in a greenhouse, with 10 plants of each variety used in the experiment. From the harvest of each plant, five seeds were chosen at random and individually analyzed for their percentage of oil. This gave a total of 50 measurements for each variety. To calculate the standard error of the mean for a variety, the experimenter calculated the standard deviation of the 50 observations and divided by \(\sqrt{50}\). Why would this calculation be of doubtful validity?

For the 28 lamb birthweights of Example \(6.2 .2,\) the mean is \(5.1679 \mathrm{~kg},\) the \(\mathrm{SD}\) is \(0.6544 \mathrm{~kg},\) and the \(\mathrm{SE}\) is \(0.1237 \mathrm{~kg}\) (a) Construct a \(95 \%\) confidence interval for the population mean. (b) Construct a \(99 \%\) confidence interval for the population mean. (c) Interpret the confidence interval you found in part (a). That is, explain what the numbers in the interval mean. (Hint: See Examples 6.3 .4 and \(6.3 .5 .)\) (d) Often researchers will summarize their data in reports and articles by writing \(\bar{y} \pm \mathrm{SD}(5.17 \pm 0.65)\) or \(\bar{y} \pm \mathrm{SE}(5.17 \pm 0.12) .\) If the researcher of this study is planning to compare the mean birthweight of these Rambouillet lambs to another breed, Booroolas, which style of presentation should she use?

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