/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 The following data refers to air... [FREE SOLUTION] | 91Ó°ÊÓ

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The following data refers to airborne bacteria count (number of colonies/ft \({ }^{3}\) ) both for \(m=8\) carpeted hospital rooms and for \(n=8\) uncarpeted rooms ("Microbial Air Sampling in a Carpeted Hospital," J. Environ. Health, 1968: 405). Does there appear to be a difference in true average bacteria count between carpeted and uncarpeted rooms? $$ \begin{array}{llllllllll} \text { Carpeted } & 11.8 & 8.2 & 7.1 & 13.0 & 10.8 & 10.1 & 14.6 & 14.0 \\ \text { Uncarpeted } & 12.1 & 8.3 & 3.8 & 7.2 & 12.0 & 11.1 & 10.1 & 13.7 \end{array} $$ Suppose you later learned that all carpeted rooms were in a veterans' hospital, whereas all uncarpeted rooms were in a children's hospital. Would you be able to assess the effect of carpeting? Comment.

Short Answer

Expert verified
No significant difference found, but hospital type could be a confounding variable.

Step by step solution

01

Organize the Data

We have two sets of data. For carpeted rooms: 11.8, 8.2, 7.1, 13.0, 10.8, 10.1, 14.6, 14.0. For uncarpeted rooms: 12.1, 8.3, 3.8, 7.2, 12.0, 11.1, 10.1, 13.7.
02

Compute the Sample Means

Calculate the sample mean of the airborne bacteria count for each group. For carpeted rooms: \[ \bar{x}_1 = \frac{11.8 + 8.2 + 7.1 + 13.0 + 10.8 + 10.1 + 14.6 + 14.0}{8} \approx 11.20 \]For uncarpeted rooms: \[ \bar{x}_2 = \frac{12.1 + 8.3 + 3.8 + 7.2 + 12.0 + 11.1 + 10.1 + 13.7}{8} \approx 9.03 \]
03

Calculate the Sample Standard Deviations

Find the sample standard deviation for each group. For carpeted rooms:\[ s_1 = \sqrt{\frac{\sum (x_i - \bar{x}_1)^2}{m - 1}} \]For uncarpeted rooms:\[ s_2 = \sqrt{\frac{\sum (x_i - \bar{x}_2)^2}{n - 1}} \]The detailed calculation provides:\[ s_1 \approx 2.62, \quad s_2 \approx 3.43 \]
04

Perform a Two-Sample t-Test

Use a two-sample t-test to determine if there is a significant difference between the average bacteria counts.Null hypothesis: \( H_0: \mu_1 = \mu_2 \)Alternative hypothesis: \( H_a: \mu_1 eq \mu_2 \)Calculate the t-statistic:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{ \frac{s_1^2}{m} + \frac{s_2^2}{n} }} \approx \frac{11.20 - 9.03}{\sqrt{\frac{2.62^2}{8} + \frac{3.43^2}{8}}} \approx 1.48 \]
05

Find the Critical t-Value and Conclusion

For a significance level of \( \alpha = 0.05 \) and degrees of freedom calculated using the formula:\[ df \approx 13.65 \]The critical t-value for a two-tailed test is approximately ±2.145.Since \( |t| \approx 1.48 \) is less than 2.145, we do not reject the null hypothesis. This suggests there is no significant difference in the average bacteria counts between the two types of rooms.
06

Assess the Potential Confounding Variable

Since carpeted rooms are in a veterans' hospital and uncarpeted rooms are in a children's hospital, the observed room difference could be due to hospital type rather than carpeting, as other variables between these hospitals may affect bacteria levels. This factor confounds the results, making it difficult to attribute any difference solely to carpeting.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental statistical method used to make inferences or draw conclusions about a certain population based on sample data. The purpose is to determine whether there is enough evidence to reject a null hypothesis. In our context, we are comparing the airborne bacteria count in two different room types: carpeted and uncarpeted hospital rooms.

To perform a hypothesis test, we start by stating two hypotheses:
  • Null Hypothesis (\( H_0 \)): Assumes no effect or no difference, such as \( \mu_1 = \mu_2 \) (there is no difference in bacteria count between room types).
  • Alternative Hypothesis (\( H_a \)): Assumes an effect or a difference, like \( \mu_1 eq \mu_2 \) (there is a difference in bacteria counts).
This process involves determining a test statistic from sample data and comparing it to a critical value to either reject or fail to reject the null hypothesis. In our example, we use a two-sample t-test to assess these differences.
Sample Mean
The sample mean is a measure of central tendency that represents the average value of a data set. It is a crucial component of hypothesis testing, as it provides an estimate of the population mean based on our sample data.

To calculate the sample mean for the bacteria count in the hospital rooms, we sum the bacteria counts and divide by the number of observations. For example:
  • Carpeted rooms mean: \( \bar{x}_1 \approx 11.20 \)
  • Uncarpeted rooms mean: \( \bar{x}_2 \approx 9.03 \)
These means help to give a clear picture of which room type may have a higher or lower bacteria count on average. Significantly different sample means can suggest a real difference in the populations being tested.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. It indicates the variability or dispersion of the dataset values relative to the mean.

In calculating the sample standard deviation for the bacteria counts, we measure how much each observation deviates from the sample mean, and then average these deviations. The formulas used are:
  • For carpeted rooms: \( s_1 \approx 2.62 \)
  • For uncarpeted rooms: \( s_2 \approx 3.43 \)
A smaller standard deviation indicates that data points tend to be closer to the mean, while a larger standard deviation shows greater spread. This value is a key feature in determining the accuracy of our mean as a central point for hypothesis testing.
Confounding Variable
Confounding variables are external influences that might alter or interfere with the results of a study, making it difficult to determine cause and effect.

In our exercise, the fact that all carpeted rooms were in a veterans' hospital and all uncarpeted rooms were in a children's hospital presents a potential confounding variable. This means differences in bacteria count could be due to factors associated with the type of hospital rather than the presence or absence of carpeting.

Potential confounders include hospital practices, patient population, or room usage, which could impact bacteria levels independently of carpeting. Recognizing and controlling for confounding variables is crucial in experimental design to ensure that the conclusions drawn are genuinely reflective of the tested variables, not external influences.

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Most popular questions from this chapter

Information about hand posture and forces generated by the fingers during manipulation of various daily objects is needed for designing hightech hand prosthetic devices. The article "Grip Posture and Forces During Holding Cylindrical Objects with Circular Grips" (Ergonomics, 1996: 1163-1176) reported that for a sample of 11 females, the sample mean four-finger pinch strength (N) was \(98.1\) and the sample standard deviation was 14.2. For a sample of 15 males, the sample mean and sample standard deviation were \(129.2\) and \(39.1\), respectively. a. A test carried out to see whether true average strengths for the two genders were different resulted in \(t=2.51\) and \(P\)-value \(=.019\). Does the appropriate test procedure described in this chapter yield this value of \(t\) and the stated \(P\)-value? b. Is there substantial evidence for concluding that true average strength for males exceeds that for females by more than \(25 \mathrm{~N}\) ? State and test the relevant hypotheses.

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The article "Quantitative MRI and Electrophysiology of Preoperative Carpal Tunnel Syndrome in a Female Population" (Ergonomics, 1997: 642-649) reported that \((-473.3,1691.9)\) was a large-sample \(95 \%\) confidence interval for the difference between true average thenar muscle volume \(\left(\mathrm{mm}^{3}\right)\) for sufferers of carpal tunnel syndrome and true average volume for nonsufferers. Calculate and interpret a \(90 \%\) confidence interval for this difference.

In an experiment designed to study the effects of illumination level on task performance ("Performance of Complex Tasks Under Different Levels of Illumination," J. Illumin. Engrg., 1976: \(235-242)\), subjects were required to insert a finetipped probe into the eyeholes of 10 needles in rapid succession both for a low light level with a black background and a higher level with a white background. Each data value is the time (sec) required to complete the task. \(\begin{array}{lccccc}\text { Subject } & 1 & 2 & 3 & 4 & 5 \\ \text { Black } & 25.85 & 28.84 & 32.05 & 25.74 & 20.89 \\ \text { White } & 18.23 & 20.84 & 22.96 & 19.68 & 19.50 \\ \text { Subject } & 6 & 7 & 8 & 9 \\ \text { Black } & 41.05 & 25.01 & 24.96 & 27.47 \\ \text { White } & 24.98 & 16.61 & 16.07 & 24.59\end{array}\) Does the data indicate that the higher level of illumination yields a decrease of more than \(5 \mathrm{~s}\) in true average task completion time? Test the appropriate hypotheses using the \(P\)-value approach.

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