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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.

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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 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.

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Most popular questions from this chapter

Suppose it is required that the mean operating life of size "D" batteries be 22 hours. Suppose also that the operating life of the batteries is normally distributed. It is known that the standard deviation of the operating life of all such batteries produced is \(3.0\) hours. If a sample of 9 batteries has a mean operating life of 20 hours, can we conclude that the mean operating life of size "D" batteries is not 22 hours? Then suppose the standard deviation of the operating life of all such batteries is not known but that for the sample of 9 batteries the standard deviation is \(3.0 .\) What conclusion would we then reach?

A certain printing press is known to turn out an average of 45 copies a minute. In an attempt to increase its output, an alteration is made to the machine, and then in 3 short test runs it turns out 46,47, and 48 copies a minute. Is this increase statistically significant, or is it likely to be simply the result of chance variation? Use a significance level of \(.05\).

A manufacturer of car batteries guarantees that his batteries will last, on the average, 3 years with a standard deviation of 1 year. If 5 of these batteries have lifetimes of \(1.9,2.4,3.0\), \(3.5\), and \(4.2\) years, is the manufacturer still convinced that his batteries have a standard deviation of 1 year?

For a shipment of cable, suppose that the specifications call for a mean breaking strength of 2,000 pounds. A sampling of the breaking strength of a number of segments of the cable has a mean breaking strength of 1955 pounds with an associated standard error of the mean of 25 pounds. Using the 5 percent level, test the significance of the difference found.

A sports magazine reports that the people who watch Monday night football games on television are evenly divided between men and women. Out of a random sample of 400 people who regularly watch the Monday night game, 220 are men. Using a \(.10\) level of significance, can be conclude that the report is false?

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