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compute the standard error of \(\left(\bar{Y}_{1}-\bar{Y}_{2}\right)\) for the following data: $$ \begin{array}{|lcc|} \hline & \text { Sample 1 } & \text { Sample 2 } \\ \hline n & 10 & 10 \\ \bar{y} & 125 & 217 \\ s & 44.2 & 28.7 \\ \hline \end{array} $$

Short Answer

Expert verified
The standard error of the difference between the means \(\bar{Y}_{1} - \bar{Y}_{2}\) is approximately 16.66.

Step by step solution

01

Understand the Standard Error of the Difference in Sample Means

The standard error of the difference between two sample means, \(\bar{Y}_{1} - \bar{Y}_{2}\), is calculated using the standard errors of both samples. The formula is \(SE_{\bar{Y}_{1} - \bar{Y}_{2}} = \sqrt{{SE_{\bar{Y}_{1}}^2 + SE_{\bar{Y}_{2}}^2}}\), where \(SE_{\bar{Y}_{i}} = \frac{s_i}{\sqrt{n_i}}\) for \(i = 1, 2\).
02

Calculate the Standard Error of Each Sample Mean

The standard error for each sample mean is given by the formula \(SE_{\bar{Y}_{i}} = \frac{s_i}{\sqrt{n_i}}\). For Sample 1, the standard error is \(SE_{\bar{Y}_{1}} = \frac{44.2}{\sqrt{10}}\). For Sample 2, it is \(SE_{\bar{Y}_{2}} = \frac{28.7}{\sqrt{10}}\).
03

Compute the Standard Errors for Both Samples

Compute \(SE_{\bar{Y}_{1}}\) and \(SE_{\bar{Y}_{2}}\) using the values from Step 2. For Sample 1: \(SE_{\bar{Y}_{1}} = \frac{44.2}{3.162} \approx 13.97\). For Sample 2: \(SE_{\bar{Y}_{2}} = \frac{28.7}{3.162} \approx 9.07\).
04

Calculate the Standard Error of the Difference Between Sample Means

Use the computed standard errors of the individual sample means to find the standard error of the difference: \(SE_{\bar{Y}_{1} - \bar{Y}_{2}} = \sqrt{SE_{\bar{Y}_{1}}^2 + SE_{\bar{Y}_{2}}^2}\). Substitute the values to get \(SE_{\bar{Y}_{1} - \bar{Y}_{2}} = \sqrt{(13.97)^2 + (9.07)^2}\).
05

Final Calculation

Carry out the arithmetic to find the standard error of the difference: \(SE_{\bar{Y}_{1} - \bar{Y}_{2}} = \sqrt{(13.97)^2 + (9.07)^2} \approx \sqrt{195.2 + 82.3} \approx \sqrt{277.5} \approx 16.66\).

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Key Concepts

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

Sample Mean
The sample mean is a fundamental statistic that represents the average of a set of individual data points from a sample of a larger population. It is denoted by \( \bar{Y} \), where \( Y \) signifies the data points and the bar symbol indicates an average. The sample mean is calculated by summing all the observations in the sample and then dividing by the number of observations, expressed as \( \bar{Y} = \frac{1}{n}\sum_{i=1}^{n} Y_i \). In the given exercise, we have two sample means, \( \bar{Y}_{1} \) and \( \bar{Y}_{2} \), with respective values of 125 and 217.

Understanding the concept of the sample mean is essential because it provides a measure of the central tendency of the data and serves as a point of comparison for further statistical analysis. It is particularly important in estimating the true mean of the population from which the sample is drawn. When comparing two sample means, as in the exercise, we can gain insights into the differences that may exist between two populations or groups.
Standard Error Calculation
The standard error measures the precision of the sample mean as an estimate of the population mean. It is the standard deviation of the sampling distribution of the sample mean. The calculation of standard error helps in understanding the variability within the samples, taking into account the sample size. Larger samples tend to have smaller standard errors, as they provide more reliable estimates of the population mean.

The standard error of the sample mean (SE) is computed using the formula \( SE_{\bar{Y}} = \frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation and \( n \) is the sample size. In our exercise, the standard errors for Sample 1 and Sample 2 are calculated using their respective standard deviations (44.2 and 28.7) and sample sizes (10 for both).

The resulting standard errors, 13.97 for Sample 1 and 9.07 for Sample 2, indicate the variability of each sample mean estimate. When we investigate the difference between two sample means, we are essentially exploring whether this observed difference could be due to random variation alone or if it reflects a true difference in the underlying populations.
Statistical Inference
Statistical inference involves using data from a sample to make estimates or test hypotheses about the characteristics of a larger population. A key part of statistical inference is determining the likelihood that any observed differences in sample statistics, such as sample means, are not due to chance alone.

The standard error of the difference between two sample means plays a crucial role in this process. By calculating the standard error of the difference, as detailed in the exercise, which turns out to be approximately 16.66, we can assess the precision of the estimate of the difference in population means. This value allows us to construct confidence intervals or perform hypothesis tests, such as a t-test, to evaluate whether the difference between the sample means is statistically significant.

Statistical inference, therefore, extends beyond simple comparisons and provides a methodological framework for making informed decisions about the populations from which the samples are drawn, based on sample data. Interpreting standard errors and using them to infer population characteristics is a key skill in many fields, including economics, psychology, and medicine.

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Most popular questions from this chapter

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 \(100 ?\)

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?

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.

A zoologist measured tail length in 86 individuals, all in the 1-year age group, of the deermouse Peromyscus. The mean length was \(60.43 \mathrm{~mm}\) and the standard deviation was \(3.06 \mathrm{~mm} .\) A \(95 \%\) confidence interval for the mean is (59.77,61.09) (a) True or false (and say why): We are \(95 \%\) confident that the average tail length of the 86 individuals in the sample is between \(59.77 \mathrm{~mm}\) and \(61.09 \mathrm{~mm} .\) (b) True or false (and say why): We are \(95 \%\) confident that the average tail length of all the individuals in the population is between \(59.77 \mathrm{~mm}\) and \(61.09 \mathrm{~mm}\).

An agronomist measured the heights of \(n\) corn plants. \({ }^{5}\) The mean height was \(220 \mathrm{~cm}\) and the standard deviation was \(15 \mathrm{~cm}\). Calculate the standard error of the mean if (a) \(n=25\) (b) \(n=100\)

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