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In a study published in the journal Teaching of Psychology, the article "Fudging the Numbers: Distributing Chocolate Influences Student Evaluations of an Undergraduate Course" states that distributing chocolate to students prior to teacher evaluations increases results. The authors randomly divided three sections of a course taught by the same instructor into two groups. Fifty of the students were given chocolate by an individual not associated with the course and 50 of the students were not given chocolate. The mean score from students who received chocolate was \(4.2,\) while the mean score for the nonchocolate groups was 3.9. Suppose that the sample standard deviation of both the chocolate and nonchocolate groups was 0.8 . Does chocolate appear to improve teacher evaluations? Use the \(\alpha=0.1\) level of significance.

Step by step solution

01

State the hypotheses

Define the null hypothesis ( H_0 ) and the alternative hypothesis ( H_a ). The null hypothesis states that there is no difference in teacher evaluations between students who receive chocolate and those who do not. The alternative hypothesis states that students who receive chocolate give higher evaluations. H_0: = H_a: >

02

Choose the appropriate test

Since the data involves comparing the means of two independent groups, use a two-sample t-test.

03

Set the significance level

Set the level of significance ( = 0.1 ) Consider - values.

04

Calculate the test statistic

Use the formula for the two-sample t-test statistic: t = - / + / where = 4.2 , = 3.9 , = 0.8 , n_1 = 0.8 ,n = 50 t = ( 4.2 - 3.9 ) / ( / sqrt( 1 = ) / sqrt / = ( = )+ .137 / sqrt(5/2) = 0.3 + t Using

05

Find the critical value

From the t-distribution table, find the critical value for a one-tailed test with ( d.f. = 98 and = 1 - .1 = 0.9 .

06

Compare the test statistic to the critical value

Compare the calculated test statistic to the critical value. If the test statistic is greater than the critical value, reject the null hypothesis.

07

Make a decision

Since ( t_calculated > t_critical) The null hypothesis is rejected.

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Key Concepts

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

Two-Sample T-Test

A two-sample t-test is used to determine if there is a significant difference between the means of two independent groups. In our study, we have two groups: one that received chocolate and one that didn't. Our aim is to see if the chocolate influences teacher evaluation scores.
We start by calculating the mean difference between the two groups. We also need to know the sample sizes and standard deviations of both groups. The t-test formula helps us determine if the observed difference in means is statistically significant or if it happened by chance.
Key things to remember about the two-sample t-test:

• It assumes that the data in both groups come from normal distributions.
• The variances of the two populations are equal.
• The samples are independent of each other.

If these assumptions hold true, we can trust the results of our t-test.

Level of Significance

The level of significance, denoted as \(\alpha\), is the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it's the risk we are willing to take to say there is an effect when there is none.
In our problem, we set \(\alpha = 0.1\). This means we have a 10% risk of concluding that chocolate improves teacher evaluations when, in fact, it does not. Higher alpha levels increase the risk of a false positive, while lower levels reduce this risk but might miss a real effect.
Always define your level of significance before conducting your test to avoid bias.

Critical Value

The critical value is a point (or points) on the test distribution that is compared to the test statistic to decide whether to reject the null hypothesis. For the two-sample t-test, we obtain this value from the t-distribution table, considering our \(\alpha\) level and degrees of freedom (d.f.).
In our example, the degrees of freedom are calculated as: \(d.f. = n_1 + n_2 - 2\). With \(n_1 = n_2 = 50\), we have \(d.f. = 98\).
We would look up the critical value in the t-distribution table for a one-tailed test with \(d.f. = 98\) and \(\alpha = 0.1\). Comparing our test statistic to this critical value helps us determine the statistical significance.

Null Hypothesis

The null hypothesis (\(H_0\)) is a statement that there is no effect or no difference. It serves as the default assumption that we seek to test against. For our study, the null hypothesis is:
\(H_0: \mu_{chocolate} = \mu_{no\ chocolate}\)
This hypothesis posits that there is no difference in teacher evaluation scores between the chocolate and non-chocolate groups.
If our test produces a statistic that falls beyond the critical value, we reject the null hypothesis, indicating there’s sufficient evidence supporting an effect.

Alternative Hypothesis

The alternative hypothesis (\(H_a\)) is what you might believe to be true or hope to prove. It is a statement that there is a meaningful effect or a difference. In our case, the alternative hypothesis is:
\(H_a: \mu_{chocolate} > \mu_{no\ chocolate}\)
This hypothesis suggests that students who received chocolate give higher teacher evaluation scores than those who did not receive chocolate. By rejecting the null hypothesis in favor of the alternative, we provide evidence supporting the positive impact of chocolate on evaluations.