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Chapter 9: Problem 3

Noise Levels in Hospitals The mean noise level of 20 randomly selected areas designated as "casualty doors", was \(63.1 \mathrm{dBA}\), and the sample standard deviation is \(4.1 \mathrm{dBA}\). The mean noise level for 24 randomly selected areas designated as operating theaters was \(56.3 \mathrm{dBA}\), and the sample standard deviation was \(7.5 \mathrm{dBA}\). At \(\alpha=0.05,\) can it be concluded that there is a difference in the means?

Short Answer

Expert verified
Reject the null hypothesis; there is a difference in mean noise levels.

Step by step solution

01

Identify the Hypotheses

First, we identify the null and alternative hypotheses. The null hypothesis \(H_0\) claims there is no difference in the means of the noise levels, while the alternative hypothesis \(H_1\) claims there is a difference. Thus, the hypotheses are:\[ H_0: \mu_1 = \mu_2 \]\[ H_1: \mu_1 eq \mu_2 \]
02

Calculate the Test Statistic

We need to calculate the test statistic for a two-sample t-test. The formula for the test statistic \(t\) is:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Where \(\bar{x}_1 = 63.1, \bar{x}_2 = 56.3\), \(s_1 = 4.1, s_2 = 7.5\), and \(n_1 = 20, n_2 = 24\). Plugging in these values, we get:\[ t = \frac{63.1 - 56.3}{\sqrt{\frac{4.1^2}{20} + \frac{7.5^2}{24}}} \approx \frac{6.8}{1.975} \approx 3.44 \]
03

Determine the Critical Value and Decision Rule

We refer to the t-distribution table with degrees of freedom approximated via the formula for unequal variances:\[ df \approx \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}} \approx 32.36 \]Approximately \(df = 32\), and at \(\alpha = 0.05\) for a two-tailed test, the critical value \(t_c\) is approximately 2.037. The decision rule is: if the absolute value of the test statistic \(|t|\) is greater than 2.037, reject the null hypothesis.
04

Make a Conclusion

Since the calculated test statistic \(t = 3.44\) is greater than the critical value \(t_c = 2.037\), we reject the null hypothesis \(H_0\). Therefore, there is sufficient evidence at the 5% significance level to conclude that there is a difference in the means of the noise levels between casualty doors and operating theaters.

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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 statistical method used to make decisions or infer conclusions about a population based on sample data. It starts by establishing two hypotheses:
  • The null hypothesis (\(H_0\)), which is a statement of no effect or no difference. It often posits that any kind of observed effect or difference is due to sampling or experimental error.
  • The alternative hypothesis (\(H_1\)), which is the statement you want to test. It suggests there is an effect or a difference.
The goal of hypothesis testing is to determine whether the sample data provides enough evidence to reject the null hypothesis in favor of the alternative hypothesis. This involves computing a test statistic and comparing it to a critical value determined by your chosen significance level (\(\alpha\)). When conducting hypothesis testing, it's essential to define your hypotheses clearly and ensure they reflect what you are trying to prove or disprove.

In our exercise, the hypotheses were \(H_0: \mu_1 = \mu_2\) and \(H_1: \mu_1 eq \mu_2\), indicating that we were testing for a difference in noise levels between two different hospital areas.
critical value
The critical value is a key part of hypothesis testing. It serves as a threshold that determines the cutoff point where the test statistic either rejects or fails to reject the null hypothesis. This value is derived from the probability distribution associated with the test statistic under the null hypothesis.
  • For our t-test, the critical value comes from the t-distribution, which is suitable for situations where the sample size is small and the population standard deviation is unknown.
  • The critical value depends on the chosen significance level (\(\alpha\)), typically set at 0.05 for a 5% level of significance, and the degrees of freedom of the test.
In a two-tailed t-test, as seen in our exercise, we look for both ends of the t-distribution. If the test statistic exceeds the critical value range, we reject the null hypothesis. For our case, the critical value was approximately 2.037, which means if the test statistic calculated was greater than this value, it would indicate a statistically significant difference between the means.
degrees of freedom
Degrees of freedom ( $df$ ) are an integral concept in statistical tests like the t-test. They refer to the number of independent values or quantities that can vary in an analysis without breaking any constraints. In the context of a t-test, they affect the shape of the t-distribution used to calculate the critical value.
  • For a two-sample t-test, the degrees of freedom can be calculated using a specific formula, especially when the variances are assumed to be unequal.
  • This complex calculation involves the sample sizes and standard deviations of the two groups.
  • In our exercise, the degrees of freedom were approximately 32.36, which we rounded to 32 for practical purposes when determining the critical value from the t-distribution table.
Degrees of freedom account for the size of the data set and enhance the accuracy of the statistical analysis. They ensure that the conclusions drawn are reliable, particularly when estimating parameters or testing hypotheses.
significance level
The significance level, denoted by \(\alpha\), is the probability of rejecting the null hypothesis when it is actually true. Essentially, it measures the risk of making a Type I error鈥攆alsely declaring a difference when none exists.
  • Common significance levels are 0.05, 0.01, and 0.10, with 0.05 being the most frequently used in social sciences and many experimental studies.
  • A 0.05 significance level implies a 5% risk of concluding that there is an effect or a difference when there is none, offering a balance between sensitivity and specificity.
  • The \(\alpha\) level determines the critical value: if the test statistic exceeds this critical value, the null hypothesis is rejected.
In the exercise, the significance level was set at 0.05, indicating a moderate level of risk in making a Type I error. This level guided our decision to determine the critical value (\(t_c = 2.037\)) and conclude whether a significant difference was present based on the calculated test statistic.

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Most popular questions from this chapter

What are the two different degrees of freedom associated with the \(F\) distribution?

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Improving Study Habits As an aid for improving students' study habits, nine students were randomly selected to attend a seminar on the importance of education in life. The table shows the number of hours each student studied per week before and after the seminar. At \(\alpha=0.10\), did attending the seminar increase the number of hours the students studied per week? $$ \begin{array}{l|ccccccccc}{\text { Before }} & {9} & {12} & {6} & {15} & {3} & {18} & {10} & {13} & {7} \\ \hline \text { After } & {9} & {17} & {9} & {20} & {2} & {21} & {15} & {22} & {6}\end{array} $$

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