Question

For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Noise Levels in Hospitals In a hospital study, it was found that the standard deviation of the sound levels from 20 randomly selected areas designated as "casualty doors" "was 4.1 dB A and the standard deviation of 24 randomly selected areas designated as operating theaters was 7.5 dB A. At \(\alpha=0.05,\) can you substantiate the claim that there is a difference in the standard deviations?

Short answer

There is a significant difference in standard deviations at \(\alpha = 0.05\).

Step-by-step solution

State the Hypotheses and Identify the Claim

We need to determine if there is a difference in the standard deviations of sound levels in two areas: casualty doors and operating theaters. The hypotheses are: Null Hypothesis (\(H_0\)): \( \sigma_1 = \sigma_2 \) (There is no difference in standard deviations).Alternative Hypothesis (\(H_1\)): \( \sigma_1 eq \sigma_2 \) (There is a difference in standard deviations).The claim is that there is a difference in the standard deviations, which is represented by the alternative hypothesis.

Find the Critical Value

Since we are testing for a difference in standard deviations, we'll use an F-test.The formula for the F-test statistic is:\[F = \frac{s_1^2}{s_2^2}\]Given \(s_1 = 7.5\) for operating theaters, \(s_2 = 4.1\) for casualty doors, and the sample sizes \(n_1 = 24\), \(n_2 = 20\), the degrees of freedom are: \(df_1 = n_1 - 1 = 23\) and \(df_2 = n_2 - 1 = 19\).Using an F-distribution table and the significance level \(\alpha = 0.05\), look up the critical value for \(df_1 = 23\) and \(df_2 = 19\). The two-tailed critical value of F at \(\alpha=0.05\) is approximately 2.54 and 0.39.

Compute the Test Value

Calculate the F statistic using the formula:\[F = \frac{s_1^2}{s_2^2} = \frac{(7.5)^2}{(4.1)^2} = \frac{56.25}{16.81} \approx 3.35\]This computed F value will be compared with the critical values to determine whether we reject the null hypothesis.

Make the Decision

Compare the computed F value (3.35) with the critical values from the F-distribution table. Since 3.35 exceeds the upper critical value of 2.54, we reject the null hypothesis \(H_0\). This indicates that there is a significant difference in the standard deviations between the two areas.

Summarize the Results

At the \(\alpha = 0.05\) significance level, there is enough evidence to support the claim that there is a difference in the standard deviations of sound levels between casualty doors and operating theaters.

Key concepts

F-test

An F-test is a statistical test that compares the variances of two populations to determine if they are significantly different. It derives its name from the F-distribution, which this test employs for analyzing variances. The F-test is particularly useful when conducting hypothesis tests that involve comparing two sample variances, such as in the case of comparing variations in noise levels across different environments in a hospital study.
The test statistic for an F-test is calculated using the ratio of the variances of the two groups. For example, if group one has a variance of 56.25 and group two has a variance of 16.81, the F statistic can be represented as:\[F = \frac{s_1^2}{s_2^2} = \frac{56.25}{16.81} \approx 3.35\]The goal of performing an F-test is to determine if the observed F-test statistic is significantly different based on a critical value obtained from the F-distribution table. This is done by defining the null and alternative hypotheses and calculating if the difference in variance is beyond what would be expected by random chance.

critical value

In hypothesis testing, the critical value is a threshold that tells us whether to reject the null hypothesis or not. In the context of an F-test, it refers to a value derived from the F-distribution table which corresponds to a specific level of significance, denoted as \(\alpha\).
For example, if you conduct an F-test at a significance level of \(\alpha = 0.05\), you will look for the critical value in the F-distribution table that corresponds to your sample sizes and degrees of freedom. This value serves as a cutoff or boundary. If your computed F statistic exceeds this value, it indicates a significant difference in variances.
In our scenario, the critical value for the test was approximately 2.54 for an upper tail, with 23 and 19 degrees of freedom. Since the computed F value of 3.35 exceeded this critical value, it led to the rejection of the null hypothesis. This means there is a statistically significant difference between the variances of sound levels in casualty doors and operating theaters.

test statistic

The test statistic is a crucial component in hypothesis testing that allows statisticians to make informed decisions based on data. In the context of an F-test, the test statistic is a calculated value that compares the variances of two groups.
Essentially, the test statistic determines how far the observed data deviates from the null hypothesis. For example, the formula used to find the F statistic is:\[F = \frac{s_1^2}{s_2^2}\]Where \(s_1\) and \(s_2\) are the standard deviations of two different samples. As demonstrated in the hospital noise level exercise, a calculated F statistic of 3.35 was obtained using the sample variances.
This computed F value was then compared against the critical values found in the F-distribution table, to decide whether to accept or reject the null hypothesis. The test statistic serves as a quantitative measure that signals if there is enough information to draw a conclusion about the population parameters.

null hypothesis

The null hypothesis (\(H_0\)) represents a statement of no effect or no difference, commonly used as a starting point for statistical hypothesis testing. It essentially proposes that any observed difference in data is due to sampling or experimental error.
In the exercise regarding hospital noise levels, the null hypothesis claimed that there is no difference in the standard deviations of sound levels between two distinct areas: casualty doors and operating theaters:\[H_0: \sigma_1 = \sigma_2\]The null hypothesis serves as the default assumption until evidence from a statistical test, like an F-test, suggests otherwise. If the computed statistic falls outside the range defined by critical values, statisticians reject the null hypothesis in favor of the alternative hypothesis.
In our case, the test led to a rejection of the null hypothesis, offering substantial evidence to believe that the standard deviations in these two hospital areas are indeed different, underlining potential real-world distinctions in their noise levels.