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From appropriately selected samples, two sets of IQ scores are obtained. For group \(1, \underline{\mathrm{X}}=104, \mathrm{~S}=10\), and \(\mathrm{n}=16\); for group \(2, \underline{\mathrm{X}}=112, \mathrm{~S}=8\), and \(\mathrm{n}=14\). At the \(5 \%\) significance level is there a significant difference between the 2 groups?

Short Answer

Expert verified
We conducted a two-sample t-test to determine if there is a significant difference between the IQ scores of two groups. Our calculated t-value was approximately -2.18, and the critical t-value at the 5% significance level with 13 degrees of freedom was approximately 2.16. Since the calculated t-value is more extreme than the critical t-value, we reject the null hypothesis and conclude that there is a significant difference between the two groups at the 5% significance level.

Step by step solution

01

Identify the hypothesis

First, we need to set up our null and alternative hypotheses. The null hypothesis states that there is no significant difference between the two groups, or that the difference in means is equal to 0. The alternative hypothesis states that the difference in means is not zero. \(H_0: \mu_1 - \mu_2 = 0\) \(H_A: \mu_1 - \mu_2 \neq 0\)
02

Find the t-value

We need the following information to perform the t-test: 1. Sample means: \(\underline{X}_1 = 104\), \(\underline{X}_2 = 112\) 2. Sample standard deviations: \(S_1 = 10\), \(S_2 = 8\) 3. Sample sizes: \(n_1 = 16\), \(n_2 = 14\) The formula for the t-value in a two-sample t-test is: \(t = \frac{(\underline{X}_1 - \underline{X}_2) - (\mu_1 - \mu_2)} {\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}}\) We can plug in our values with \(\mu_1 - \mu_2 = 0\) since we are testing against the null hypothesis. \(t = \frac{(104 - 112) - 0} {\sqrt{\frac{10^2}{16} + \frac{8^2}{14}}}\)
03

Calculate the t-value

Now, let's compute t-value: \(t = \frac{-8}{\sqrt{\frac{100}{16} + \frac{64}{14}}}\) \(t \approx -2.18\)
04

Find the critical value

To determine whether the t-value is significant at the \(5\%\) significance level, we need to find the critical t-value. Since this is a two-tailed t-test, we will refer to the t-distribution table at the \(97.5\%\) level or the \(2.5\%\) level in the tails. First, we need to find the degrees of freedom for the t-distribution. The formula for degrees of freedom in a two-sample t-test is: \(df = min(n_1-1, n_2-1) = min(15, 13) = 13\) Now, we can refer to the \(97.5\%\) level of the t-distribution table with 13 degrees of freedom to get the critical t-value: \(t_{critical} \approx 2.16\)
05

Determine the significance

Since we have our calculated t-value and critical t-value, we can now compare them: \(t \approx -2.18\) \(t_{critical} \approx 2.16\) The calculated t-value is more extreme than the critical t-value (in the negative direction). This means that we reject the null hypothesis and conclude that there is a significant difference between the two groups at the \(5\%\) significance level.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often denoted by \( H_0 \), is a statistical statement asserting there is no effect or no difference between two or more groups or factors. In the context of a two-sample t-test, the null hypothesis presumes that the means of two independent samples are equal. For instance, if we're comparing the average IQ scores between two groups, the null hypothesis would state \( H_0: \mu_1 - \mu_2 = 0 \), signifying no real difference in intelligence between the two groups. The goal of the t-test is to determine whether the observed data provides enough evidence to reject this null hypothesis.
Alternative Hypothesis
Contrary to the null hypothesis, the alternative hypothesis \( H_A \) or \( H_1 \) anticipates some form of effect or difference that contradicts \( H_0 \). In the two-sample t-test, the alternative hypothesis posits that there is a significant difference between the sample means. It can take a non-directional form (two-tailed) stating \( H_A: \mu_1 - \mu_2 \eq 0 \), which means the means are not equal, without specifying which one is greater, as seen in our IQ comparison example. The test’s result might lead to either retaining the null hypothesis or rejecting it in favor of the alternative.
T-Value Calculation
The t-value calculation is a crucial step in a t-test, as it assesses the extent of deviation between the group means when measured against the variability in the data. The formula encapsulates this as \( t = \frac{(\underline{X}_1 - \underline{X}_2) - (\mu_1 - \mu_2)} {\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}} \). In this equation, \( \underline{X}_1 \) and \( \underline{X}_2 \) represent the sample means, \( S_1 \) and \( S_2 \) are the standard deviations, and \( n_1 \) and \( n_2 \) are the sample sizes. The higher the absolute value of the t-value, the stronger the evidence against the null hypothesis, indicating a larger difference between the groups.
Degrees of Freedom
Degrees of freedom, abbreviated as df, refers to the number of independent values or quantities that can vary in an analysis without violating any given restrictions. In a two-sample t-test, the degrees of freedom can be calculated using the formula \( df = min(n_1-1, n_2-1) \), where \( n_1 \) and \( n_2 \) denote the sample sizes of each group. This calculation assists in determining the exact shape of the t-distribution, which is pivotal for establishing the critical t-value to gauge the significance of the test's results.
Significance Level
The significance level, often denoted as alpha \( (\alpha) \), is the threshold chosen by the researcher to gauge the strength of evidence required to reject the null hypothesis. Commonly set at 0.05 (or 5%), it represents a 5% risk of concluding that there is a difference when there is no actual difference. This threshold helps to mitigate the rate of false positives, ensuring that results are not due to mere random chance. To interpret the t-test results, the calculated t-value is compared against the critical t-value at the chosen significance level. If the absolute calculated t-value is greater than the critical value, the null hypothesis is rejected, implying a statistically significant difference between the group means.

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Most popular questions from this chapter

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