91Ó°ÊÓ

Researchers from the University of British Columbia conducted a study to investigate the effects of color on cognitive tasks. Words were displayed on a computer screen with background colors of red and blue. Results from scores on a test of word recall are given below. Higher scores correspond to greater word recall. a. Use a \(0.05\) significance level to test the claim that the samples are from populations with the same mean. b. Construct a confidence interval appropriate for the hypothesis test in part (a). What is it about the confidence interval that causes us to reach the same conclusion from part (a)? c. Does the background color appear to have an effect on word recall scores? If so, which color appears to be associated with higher word memory recall scores? $$ \begin{array}{|l|l|} \hline \text { Red Background } & n=35, \bar{x}=15.89, s=5.90 \\ \hline \text { Blue Background } & n=36, \bar{x}=12.31, s=5.48 \\ \hline \end{array} $$

Step by step solution

01

State the Null and Alternative Hypotheses

For the hypothesis test, we need to state the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)): \(H_0\): The means of the populations are equal (\(\mu_1 = \mu_2\)).\(H_1\): The means of the populations are not equal (\(\mu_1 eq \mu_2\)).

02

Determine the Significance Level and Test Statistic

The significance level is given as \(\alpha = 0.05\). The test statistic for comparing two means when population variances are unknown but assumed to be equal is calculated using the independent two-sample t-test formula:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2(\frac{1}{n_1} + \frac{1}{n_2})}}\]where \(s_p^2\) is the pooled variance given by:\[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}.\]

03

Calculate the Pooled Variance

Using the given data:\(n_1 = 35, \bar{x}_1 = 15.89, s_1 = 5.90$$n_2 = 36, \bar{x}_2 = 12.31, s_2 = 5.48\)We first compute the pooled variance,\[ s_p^2 = \frac{(35 - 1)5.90^2 + (36 - 1)5.48^2}{35 + 36 - 2} = \frac{34\cdot34.81 + 35\cdot30.03}{69} = \frac{1183.54 + 1051.05}{69} \approx 32.34\]

04

Calculate the Test Statistic

We use the pooled variance to find the test statistic,\[ t = \frac{15.89 - 12.31}{\sqrt{32.34\left(\frac{1}{35} + \frac{1}{36}\right)}} = \frac{3.58}{\sqrt{32.34 \cdot 0.056}} = \frac{3.58}{\sqrt{1.811}} \approx \frac{3.58}{1.346} \approx 2.66 \]

05

Determine the Critical Value and Compare

For a two-tailed test with \(\alpha = 0.05\) and \(df = n_1 + n_2 - 2 = 69\), the critical t-values can be found from the t-distribution table. The critical t-value for \(df = 69\) is approximately \(\pm 2.00\). Since the absolute value of our test statistic \(|t| = 2.66\) is greater than \(2.00\), we reject the null hypothesis.

06

Construct the Confidence Interval

Construct a 95% confidence interval for the difference in means:\[(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \cdot \sqrt{s_p^2\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}\]Using \(t_{\alpha/2} \approx 2.00\) and our earlier calculated quantities:\[3.58 \pm 2.00 \cdot 1.346 = 3.58 \pm 2.692 = [0.888, 6.272]\]

07

Interpret the Confidence Interval

The confidence interval [0.888, 6.272] does not contain 0, indicating a statistically significant difference between the means. This supports our decision to reject the null hypothesis.

08

Answer the Research Question

Considering the results, the background color does affect word recall scores. Specifically, the red background is associated with higher word recall scores since the mean score (15.89) is higher than that for the blue background (12.31).

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.

Independent Two-Sample T-Test

An Independent Two-Sample T-Test is a statistical method used to compare the means of two independent groups. In our case, we have word recall scores for two different background colors—red and blue. We want to see if the background color affects word recall ability.
Here's how it works:

• First, gather the sample data from two groups (in this case, red background and blue background).
• Calculate the sample means and standard deviations for each group.
• Use the formula for the t-statistic to compare these two sample means while accounting for variability within the groups.

We then compare this t-statistic to a critical value to decide if there's a significant difference between the means.

Pooled Variance

When the variances of two populations are assumed to be equal, we calculate something called Pooled Variance. It combines the variances of the two independent samples into a single estimate.
This is achieved using the formula:
\[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}\]
In our example:

• \(n_1=35\), \(s_1=5.90\) for the red background group.
• \(n_2=36\), \(s_2=5.48\) for the blue background group.

Plug these values into the formula to get the pooled variance \(s_p^2\). This gives us a more accurate measure of the variability within the data.

Significance Level

The Significance Level, denoted by \(\alpha\), is the probability of rejecting the null hypothesis when it is actually true. It's a threshold we set before conducting our test. For our example, the significance level is set at \(\alpha = 0.05\). This means we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.

• If the p-value we calculate from our test is less than \(\alpha\), we reject the null hypothesis.
• If the p-value is greater than \(\alpha\), we do not reject the null hypothesis.

Setting the right significance level is crucial because it balances the risk of Type I (false positive) and Type II (false negative) errors.

Confidence Interval

A Confidence Interval gives us a range of values within which we expect the true population parameter (such as the difference in means) to fall. In this case, we are constructing a 95% confidence interval to estimate the difference between the means of the two groups.
Here's the formula we use:
\[ (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \cdot \sqrt{s_p^2\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \]
For our study:

• The sample mean difference is \(3.58\).
• \(t_{\alpha/2}\) is approximately \(2.00\) for a 95% confidence level and our degrees of freedom.

So, our interval is approximately \[ [0.888, 6.272] \]. This range does not include zero, supporting our hypothesis that there is a significant difference.

Null Hypothesis

The Null Hypothesis, denoted as \(H_0\), assumes there is no effect or no difference between groups. In our study, the null hypothesis is that the mean word recall scores are the same for both background colors.
It's typically the hypothesis we aim to test against.

• If we fail to reject \(H_0\), it means we did not find enough evidence to show a difference.
• If we reject \(H_0\), it suggests there is a significant difference.

In our example, \(H_0\) states that the means of word recall for red and blue backgrounds are equal.

Alternative Hypothesis

The Alternative Hypothesis, denoted as \(H_1\), is what we take on if we reject the null hypothesis. In this case, \(H_1\) states that the means of word recall for the red and blue backgrounds are not equal.
This is a two-sided test because we are open to there being either a positive or negative difference between the groups.

• We hope to find sufficient evidence to support \(H_1\), implying that background color does affect word recall.

In our study, the higher word recall scores with a red background vs. a blue background provided evidence to reject \(H_0\) and support \(H_1\).