/*! 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 28 Noise and Stress In Exercise \(8... [FREE SOLUTION] | 91Ó°ÊÓ

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Noise and Stress In Exercise \(8.52,\) you compared the effect of stress in the form of noise on the ability to perform a simple task. Seventy subjects were divided into two groups; the first group of 30 subjects acted as a control, while the second group of 40 was the experimental group. Although each subject performed the task, the experimental group subjects had to perform the task while loud rock music was played. The time to finish the task was recorded for each subject and the following summary was obtained: $$\begin{array}{lll} & \text { Control } & \text { Experimental } \\\\\hline n & 30 & 40 \\\\\bar{x} & 15 \text { minutes } & 23 \text { minutes } \\\s & 4 \text {minutes } & 10 \text { minutes }\end{array}$$

Short Answer

Expert verified
Answer: Yes, there is a significant difference in completion times between the control group and experimental group when noise is added.

Step by step solution

01

State the null and alternative hypotheses

Our null hypothesis (H0) states that there is no significant difference in the completion times between the control and experimental groups. The alternative hypothesis (H1) states that there is a significant difference between the two groups. H0: µ1 = µ2 H1: µ1 ≠ µ2 where µ1 is the mean completion time for the control group and µ2 is the mean completion time for the experimental group.
02

Calculate the t-statistic

Use the following formula to calculate the t-statistic: t = \frac{(\bar{X}_1 - \bar{X}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} Plug in the values: - \(\bar{X}_1 = 15\) (control group mean completion time) - \(\bar{X}_2 = 23\) (experimental group mean completion time) - \(s_1 = 4\) (control group standard deviation) - \(s_2 = 10\) (experimental group standard deviation) - \(n_1 = 30\) (control group sample size) - \(n_2 = 40\) (experimental group sample size) t = \(\frac{(15 - 23)}{\sqrt{\frac{4^2}{30} + \frac{10^2}{40}}}\) = \(\frac{-8}{\sqrt{\frac{16}{30} + \frac{100}{40}}}\) Now, calculate the denominator.
03

Calculate the denominator

Evaluate the expression inside the square root in the denominator: \(\sqrt{\frac{16}{30} + \frac{100}{40}} = \sqrt{\frac{8}{15} + \frac{5}{4}} = \sqrt{1.5333...}\)
04

Compute the t-statistic

Now, divide the numerator by the calculated denominator to get the t-statistic: t = \(\frac{-8}{\sqrt{1.5333...}} = -6.55\)
05

Calculate the degrees of freedom

Use the following formula to calculate the degrees of freedom: df = \(\frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1 - 1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2 - 1}}\) Plug in the values: - \(s_1 = 4\) (control group standard deviation) - \(s_2 = 10\) (experimental group standard deviation) - \(n_1 = 30\) (control group sample size) - \(n_2 = 40\) (experimental group sample size) df = \(\frac{(\frac{4^2}{30} + \frac{10^2}{40})^2}{\frac{(\frac{4^2}{30})^2}{30 - 1} + \frac{(\frac{10^2}{40})^2}{40 - 1}}\)
06

Compute the degrees of freedom

Now, calculate the degrees of freedom: df = \(\frac{((\frac{16}{30}) + (\frac{100}{40}))^2}{\frac{(\frac{16}{30})^2}{29} + \frac{(\frac{100}{40})^2}{39}} \approx 56.46\) The degrees of freedom are approximately 56.
07

Interpret the result

Using a t-table with 56 degrees of freedom and a two-tailed test with an alpha level of 0.05, we find the critical value to be ±2.004. However, our calculated t-statistic is -6.55, which is beyond the critical value. Therefore, we can reject the null hypothesis and conclude that there is a significant difference in completion times between the control group and experimental group when noise is added.

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Key Concepts

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

t-statistic
A t-statistic is a handy way to check if there's a significant difference between the means of two groups. In our exercise, it's the number that tells us whether noise affects task completion time. By calculating the t-statistic, we compare the effect of noise on people in the experimental group with those in the control group.
To compute the t-statistic, we use the formula:\[t = \frac{(\bar{X}_1 - \bar{X}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where:
  • \(\bar{X}_1\) and \(\bar{X}_2\) are the mean times.
  • \(s_1\) and \(s_2\) denote the standard deviations.
  • \(n_1\) and \(n_2\) signify the sample sizes.
In our case, the control group mean was 15 minutes, and the experimental group mean was 23 minutes. Plugging in these values helps you see if the noise made a real difference in times altogether.
degrees of freedom
Degrees of freedom (df) is like assigning flexibility to the data. It shows how many values can vary within the data set, while keeping certain restrictions.
In hypothesis testing, degrees of freedom influence how precise our t-statistic and other statistical measures will be.
To calculate degrees of freedom for comparing two means, we'd use the following formula:\[df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}\]This value guides us to the appropriate critical value from the t-table. Having a generous number of subjects in both groups means that in our task completion example, we calculate the degrees of freedom to be around 56.
null and alternative hypotheses
In any experiment, we begin with a statement about what we're testing. The null hypothesis (H0) predicts no difference or effect, while the alternative hypothesis (H1) suggests there is a noticeable change or effect. In statistical testing, these are crucial to determining whether our results are just by chance.
For the noise and stress exercise:
  • **Null Hypothesis (H0):** The noise doesn't affect the time it takes to complete the task, meaning the completion times for both groups are the same. Mathematically, \(\mu_1 = \mu_2\).
  • **Alternative Hypothesis (H1):** The noise does impact completion time, indicating a difference in the means for each group. Symbolically, \(\mu_1 eq \mu_2\).
By testing these hypotheses, we can make informed decisions about the influence of noise.
control group vs. experimental group
In studies like ours, clear distinctions between the control group and the experimental group help uncover the effects of what's being tested. The control group acts as our baseline, experiencing no changes or additions. Here, they complete the task without loud music.
The experimental group, on the other hand, undergoes a specific treatment—in this scenario, they're subjected to loud rock music while finishing their task. By comparing outcomes from both groups, we assess whether the added noise significantly impacts task performance.
This setup ensures we have a fairway to analyze whether external factors like noise genuinely affect performance by isolating its influence from other variables.
significant difference
A significant difference indicates an essential alteration or effect brought by a certain factor, like during our exercise. When we say there's a significant difference, we mean the change isn't by random chance—it really matters!
In hypothesis testing, once we calculate our t-statistic, we compare it to a critical value found for our degrees of freedom. If our t-value exceeds this critical threshold, it confirms a significant difference, leading us to reject the null hypothesis.
In the noise and task example, our calculated t-statistic of -6.55 was way beyond the critical value (±2.004). This reinforced the idea that noise did, in fact, significantly alter task completion times for subjects.

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Most popular questions from this chapter

Define the power of a statistical test. As the alternative value of \(\mu\) gets farther from \(\mu_{0}\), how is the power affected?

Suppose you wish to detect a difference between \(\mu_{1}\) and \(\mu_{2}\) (either \(\mu_{1}>\mu_{2}\) or \(\mu_{1}<\mu_{2}\) ) and, instead of running a two-tailed test using \(\alpha=.05,\) you use the following test procedure. You wait until you have collected the sample data and have calculated \(\bar{x}_{1}\) and \(\bar{x}_{2}\). If \(\bar{x}_{1}\) is larger than \(\bar{x}_{2},\) you choose the alternative hypothesis \(H_{\mathrm{a}}: \mu_{1}>\mu_{2}\) and run a one-tailed test placing \(\alpha_{1}=.05\) in the upper tail of the \(z\) distribution. If, on the other hand, \(\bar{x}_{2}\) is larger than \(\bar{x}_{1},\) you reverse the procedure and run a one-tailed test, placing \(\alpha_{2}=.05\) in the lower tail of the \(z\) distribution. If you use this procedure and if \(\mu_{1}\) actually equals \(\mu_{2},\) what is the probability \(\alpha\) that you will conclude that \(\mu_{1}\) is not equal to \(\mu_{2}\) (i.e., what is the probability \(\alpha\) that you will incorrectly reject \(H_{0}\) when \(H_{0}\) is true)? This exercise demonstrates why statistical tests should be formulated prior to observing the data.

Find the appropriate rejection regions for the large-sample test statistic \(z\) in these cases: a. A left-tailed test at the \(1 \%\) significance level. b. A two-tailed test with \(\alpha=.01\). c. Suppose that the observed value of the test statistic was \(z=-2.41 .\) For the rejection regions constructed in parts a and b, draw the appropriate conclusion for the tests. If appropriate, give a measure of the reliability of your conclusion.

Potency of an Antibiotic A drug manufacturer claimed that the mean potency of one of its antibiotics was \(80 \%\). A random sample of \(n=100\) capsules were tested and produced a sample mean of \(\bar{x}=79.7 \%\) with a standard deviation of \(s=.8 \% .\) Do the data present sufficient evidence to refute the manufacturer's claim? Let \(\alpha=.05 .\) a. State the null hypothesis to be tested. b. State the alternative hypothesis. c. Conduct a statistical test of the null hypothesis and state your conclusion.

A random sample of \(n=35\) observations from a quantitative population produced a mean \(\bar{x}=2.4\) and a standard deviation \(s=.29 .\) Suppose your research objective is to show that the population mean \(\mu\) exceeds 2.3 a. Give the null and alternative hypotheses for the tes b. Locate the rejection region for the test using a \(5 \%\) significance level. c. Find the standard error of the mean. d. Before you conduct the test, use your intuition to decide whether the sample mean \(\bar{x}=2.4\) is likely or unlikely, assuming that \(\mu=2.3 .\) Now conduct the test. Do the data provide sufficient evidence to indicate that \(\mu>2.3 ?\)

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