/*! 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 594 Two separate groups of subjects ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two separate groups of subjects were tested. The experimental group (Group E) had 10 subjects; the control group (Group C) had 9 subjects. The data are given below; the scores are assumed to be normally distributed. Group E: \(12,13,16,14,15,12,15,14,13\), and 16 . Group C: \(10,13,14,12,15,16,12,14\), and 11 . Determine whether the means of the two groups differ significantly at the \(.05\) level of sionificance.

Short Answer

Expert verified
We fail to reject the null hypothesis, so at the \(0.05\) significance level, there is not enough evidence to conclude that the means of Group E and Group C differ significantly.

Step by step solution

01

Identify the data sets and their sizes

From the provided information, we have the following data sets: Group E: \(12,13,16,14,15,12,15,14,13\), and \(16\) Group C: \(10,13,14,12,15,16,12,14\), and \(11\) The sample size for Group E, \(n_E\), is \(10\), and the sample size for Group C, \(n_C\), is \(9\).
02

Calculate the mean and standard deviation for each group

Calculate the mean and the standard deviation for each group. For Group E: Mean, \(\bar{x}_E = \frac{12+13+16+14+15+12+15+14+13+16}{10} = 14\) Standard deviation, \(s_E = \sqrt{\frac{(\Sigma (x_i-\bar{x}_E)^2)}{n_E-1}} = \sqrt{\frac{(4+1+2+0+1+4+1+0+1+2)}{10-1}} = \sqrt{\frac{20}{9}} = 1.491\) For Group C: Mean, \(\bar{x}_C = \frac{10+13+14+12+15+16+12+14+11}{9} = 13\) Standard deviation, \(s_C = \sqrt{\frac{(\Sigma (x_i-\bar{x}_C)^2)}{n_C-1}} = \sqrt{\frac{(9+0+1+1+4+9+1+1+4)}{9-1}} = \sqrt{\frac{24}{8}} =1.633\)
03

Set up the null hypothesis and alternative hypothesis

The null hypothesis, \(H_0\), is that the means of both groups do not differ significantly. The alternative hypothesis, \(H_1\), is that the means of both groups differ significantly. \(H_0: \mu_E = \mu_C\) \(H_1: \mu_E \neq \mu_C\)
04

Conduct a two-sample t-test

Perform a two-sample t-test using the sample means, standard deviations, and sample sizes. The t-value is calculated using the formula: \(t = \frac{(\bar{x}_E - \bar{x}_C)}{\sqrt{\frac{(s_E^2)}{n_E} + \frac{(s_C^2)}{n_C}}}\) \(t = \frac{(14 - 13)}{\sqrt{\frac{(1.491^2)}{10} + \frac{(1.633^2)}{9}}}= \frac{1}{\sqrt{\frac{2.22}{10} + \frac{2.667}{9}}} = 1.326 \)
05

Calculate the critical t-value and compare

We have a two-tailed t-test with a significance level (\(\alpha\)) of \(0.05\), and degrees of freedom (\(df\)) equal to the smaller of \(n_E-1\) and \(n_C-1\); in our case, \(df = 9 - 1 = 8\). Using a t-table or calculator, we find the critical t-value for \(\alpha = 0.05\) and \(df = 8\): \(t_{critical} = \pm 2.306\) Since our calculated t-value, \(1.326\), is NOT in the rejection region of \(t < -2.306\) or \(t > 2.306\), we fail to reject the null hypothesis.
06

Make a conclusion

We fail to reject the null hypothesis, so at the \(0.05\) significance level, there is not enough evidence to conclude that the means of Group E and Group C differ significantly.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics that helps determine if there is enough evidence to support a specific hypothesis about a population parameter.
In this exercise, hypothesis testing is being used to compare the means of two independent groups (Group E and Group C).
The process starts with setting two opposing hypotheses:
  • The null hypothesis (\(H_0\)) suggests no difference between the group means, e.g., \(\mu_E = \mu_C\).
  • The alternative hypothesis (\(H_1\)) suggests a significant difference, e.g., \(\mu_E eq \mu_C\).
Next, a statistical test, such as a two-sample t-test, is conducted to evaluate the likelihood of the data under the assumption that \(H_0\) is true.
If the test statistic falls into a region of rejection, the null hypothesis is rejected in favor of the alternative hypothesis, indicating a significant difference between the groups.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold for determining the statistical significance of a hypothesis test. In this exercise, a significance level of \(0.05\) is used, which means there is a 5% risk of rejecting the null hypothesis if it is true.
Choosing the right significance level depends on the context of the study. Commonly used levels include \(0.01\), \(0.05\), and \(0.10\). A smaller \(\alpha\) value indicates a stricter criterion for rejecting the null hypothesis, reducing the chances of a Type I error (false positive).
For this problem, if the t-value falls beyond the critical values (\(-2.306\) and \(2.306\) for 8 degrees of freedom), the null hypothesis would be rejected. However, the computed t-value (\(1.326\)) does not meet this criterion, leading to a decision to not reject \(H_0\).
Mean and Standard Deviation Calculation
Calculating the mean and standard deviation is crucial in understanding the central tendency and variability of a data set. In this exercise, each group’s scores need their mean and standard deviation calculated to perform the two-sample t-test.
  • The mean (\(\bar{x}\)) provides a measure of the average, for example, Group E has a mean of 14.
  • The standard deviation (\(s\)) indicates the variation in the data set. A larger standard deviation means the data points are more spread out. For instance, Group E's standard deviation is approximately 1.491.
To calculate the mean, sum the data values and divide by the number of observations.
For the standard deviation, compute the square root of the variance. The variance is the average of the squared differences from the Mean (\(\bar{x}\)). These calculations prepare the data for subsequent t-test analysis.
Normal Distribution
The assumption of a normal distribution is fundamental to many statistical procedures, particularly parametric tests such as the two-sample t-test.
In this context, it implies that the data follows a symmetric, bell-shaped curve, where most observations cluster around the central peak and probabilities for values taper off as they move away from the mean.
Normal distribution underpins the validity of the t-test results, ensuring that sample means approximate normality according to the Central Limit Theorem, especially when sample sizes are reasonably large.
For small sample sizes, normality ensures accurate test results. Hence, the assumption that the group scores in this problem are normally distributed allows the t-test to proceed with confidence that the test statistics yield reliable and valid comparisons.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A pharmaceutical company claims that its new vaccine is \(90 \%\) effective, but the federal drug agency suspects that it is only \(40 \%\) effective. Devise a procedure to test its effectiveness and use it to find the probability that (a) the government will incorrectly grant the company claim (when the effectiveness is in fact \(40 \%\), and (b) the company claim will be incorrectly denied (when the effectiveness is in fact \(90 \% .)\)

\(\mathrm{H}_{0}: \mu=\mu_{0}\) \(\mathrm{H}_{1}: \mu=\mu_{1}\) and \(\alpha\) and \(\beta\) are probabilities of making type I and type II errors respectively, show that for a two-tailed test the required sample size \(\mathrm{n}\) is given approximately by $$ \mathrm{n}=\left[\left\\{\left(\mathrm{Z}_{\alpha / 2}+\mathrm{Z}_{\mathrm{B}}\right)^{2} \sigma^{2}\right\\} /\left(\mu_{1}-\mu_{0}\right)^{2}\right] \text { , } $$ provided that $$ \operatorname{Pr}\left[Z<-Z_{\alpha / 2}-\left\\{\left(\sqrt{n}\left|\mu_{1}-\mu_{0}\right|\right) / \sigma\right\\}\right] $$ is small when \(\mu=\mu_{1}\).

Two independent reports on the value of a tincture for treating a disease in camels were available. The first report made on a small pilot series showed the new tincture to be probably superior to the old treatment with a Yates' \(\mathrm{X}^{2}\) of \(3.84, \mathrm{df}=1, \alpha=.05\). The second report with a larger trial gave a "not significant" result with a Yates \(\mathrm{X}^{2}=2.71, \mathrm{df}=1\), \(\alpha=.10 .\) Can the results of the 2 reports be combined to form a new conclusion?

Suppose you are a buyer of large supplies of light bulbs. You want to test, at the \(5 \%\) significance level, the manufacturer's claim that his bulbs last more than 800 hours. You test 36 bulbs and find that the sample mean, \(\underline{X}\), is 816 hours and the sample standard deviation \(\mathrm{s}=70\) hours. Should you accept the claim?

A sample of Democrats and a sample of Republicans were polled on an issue. Of 200 Republicans, 90 would vote yes on the issue; of 100 Democrats, 58 would vote yes. Can we say that more Democrats than Republicans favor the issue at the \(1 \%\) level of significance?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.