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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

Four treatments were compared for their effect on the growth of spinach cells in cell culture flasks. The experimenter randomly allocated two flasks to each treatment. After a certain time on treatment, he randomly drew three aliquots ( 1 cc each) from each flask and measured the cell density in each aliquot; thus, he had six cell density measurements for each treatment. In calculating the standard error of a treatment mean, the experimenter calculated the standard deviation of the six measurements and divided by \(\sqrt{6} .\) On what grounds might an objection be raised to this method of calculating the SE?

In a study of the effect of aluminum intake on the mental development of infants, a group of 92 infants who had been born prematurely were given a special aluminum-depleted intravenous-feeding solution. \({ }^{18}\) At age 18 months the neurologic development of the infants was measured using the Bayley Mental Development Index. (The Bayley Mental Development Index is similar to an IQ score, with 100 being the average in the general population.) A \(95 \%\) confidence interval for the mean is (93.8,102.1) (a) Interpret this interval. That is, what does the interval tell us about neurologic development in the population of prematurely born infants who receive intravenousfeeding solutions? (b) Does this interval indicate that the mean IQ of the sampled population is below the general population average of \(100 ?\)

In a study of larval development in the tufted apple budmoth (Platynota idaeusalis), an entomologist measured the head widths of 50 larvae. All 50 larvae had been reared under identical conditions and had moulted six times. The mean head width was \(1.20 \mathrm{~mm}\) and the standard deviation was \(0.14 \mathrm{~mm}\). Construct a \(90 \%\) confidence interval for the population mean. \(^{17}\)

Ferulic acid is a compound that may play a role in disease resistance in corn. A botanist measured the con- \(-\) centration of soluble ferulic acid in corn seedlings grown in the dark or in a light/dark photoperiod. The results (nmol acid per gm tissue) were as shown in the table. \({ }^{41}\) $$ \begin{array}{|ccc|} \hline & \text { Dark } & \text { Photoperiod } \\ \hline n & 4 & 4 \\ \bar{y} & 92 & 115 \\ s & 13 & 13 \\ \hline \end{array} $$ (a) Construct a \(95 \%\) confidence interval for the difference in ferulic acid concentration under the two lighting conditions. (Assume that the two populations from which the data came are normally distributed.) [Note: Formula (6.7.1) yields 6 degrees of freedom for these data. (b) Repeat part (a) for a \(90 \%\) level of confidence.

An experiment is being planned to compare the effects of several diets on the weight gain of beef cattle, measured over a 140 -day test period. \(^{20}\) In order to have enough precision to compare the diets, it is desired that the standard error of the mean for each diet should not exceed \(5 \mathrm{~kg}\). (a) If the population standard deviation of weight gain is guessed to be about \(20 \mathrm{~kg}\) on any of the diets, how many cattle should be put on each diet in order to achieve a sufficiently small standard error? (b) If the guess of the standard deviation is doubled, to \(40 \mathrm{~kg}\), does the required number of cattle double? Explain.

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