91Ó°ÊÓ

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9 - 1 along with "Table" answers based on Table \(A\) - 3 with df equal to the smaller of \(\boldsymbol{n}_{I}-\boldsymbol{I}\) and \(\boldsymbol{n}_{2}-\boldsymbol{I} .\) ) Color and Creativity Researchers from the University of British Columbia conducted trials to investigate the effects of color on creativity. Subjects with a red background were asked to think of creative uses for a brick; other subjects with a blue background were given the same task. Responses were scored by a panel of judges and results from scores of creativity are given below. Higher scores correspond to more creativity. The researchers make the claim that "blue enhances performance on a creative task." a. Use a 0.01 significance level to test the claim that blue enhances performance on a creative task. b. Construct the 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)? $$\begin{array}{l|l} \hline \text { Red Background: } & n=35, \bar{x}=3.39, s=0.97 \\ \hline \text { Blue Background: } & n=36, \bar{x}=3.97, s=0.63 \\ \hline \end{array}$$

Step by step solution

01

- State the hypotheses

Formulate the null and alternative hypotheses. Let \(\bar{x}_1\) and \(\bar{x}_2\) be the sample means, and \(s_1\) and \(s_2\) be the sample standard deviations for Red and Blue backgrounds respectively.\[ H_0: \bar{x}_1 = \bar{x}_2 \]\[ H_1: \bar{x}_1 < \bar{x}_2 \]Here, we are testing if the mean score for Blue background is significantly greater than the mean score for Red background.

02

- Calculate the test statistic

For two independent samples, we use the t-test for unequal variances (Welch’s t-test). The test statistic is given by:\[ t = \frac{ \bar{x}_2 - \bar{x}_1 }{ \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} } \]Substitute the given values:\[ \bar{x}_1 = 3.39, \bar{x}_2 = 3.97, s_1 = 0.97, s_2 = 0.63, n_1 = 35, n_2 = 36 \]Calculate the pooled standard error:\[ SE = \sqrt{\frac{0.97^2}{35} + \frac{0.63^2}{36}} \]\[ SE = \sqrt{\frac{0.9409}{35} + \frac{0.3969}{36}} \]\[ SE = \sqrt{0.02688 + 0.01102} \]\[ SE = \sqrt{0.0379} = 0.1947 \]Now calculate the t statistic:\[ t = \frac{3.97 - 3.39}{0.1947} \]\[ t = \frac{0.58}{0.1947} \]\[ t = 2.98 \]

03

- Determine degrees of freedom

Since we are using the unequal variances t-test, we approximate the degrees of freedom (df) using the following formula:\[ df \approx \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]Substituting the values, we get:\[ df \approx \frac{(0.02688 + 0.01102)^2}{\frac{(0.02688)^2}{34} + \frac{(0.01102)^2}{35}} \]\[ df \approx \frac{0.0379^2}{\frac{0.000722}{34} + \frac{0.000121}{35}} \]\[ df \approx \frac{0.001437}{0.000021 + 0.000003} \]\[ df \approx \frac{0.001437}{0.000024} \]\[ df \approx 59.875 \]The degree of freedom is approximately 60.

04

- Determine the critical value and make a decision

For a one-tailed test at significance level \( \alpha = 0.01 \) and df approximately 60, the critical value (t-critical) from the t-distribution table is approximately 2.39.Since our calculated t value (2.98) is greater than the critical value (2.39), we reject the null hypothesis. Therefore, there is significant evidence at the \(0.01\) level to support the claim that blue enhances performance on a creative task.

05

- Construct a confidence interval

Construct the confidence interval for the difference in means using the formula:\[ (\bar{x}_2 - \bar{x}_1) \pm t_{critical} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Using the previously calculated standard error and the critical value for a 98% confidence interval (since it corresponds to a 0.01 significance level in a one-tailed test):\[ 0.58 \pm 2.39 \cdot 0.1947 \]Calculate the margin of error (MOE):\[ MOE = 2.39 \cdot 0.1947 = 0.4653 \]Therefore, the 98% confidence interval for the difference in means is:\[ 0.58 \pm 0.4653 \]This simplifies to:\[ (0.1147, 1.0453) \]Since the entire confidence interval is above zero, it supports the conclusion that the mean score for the Blue background is significantly higher than the mean score for the Red background.

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Key Concepts

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

Independent Samples

When conducting a study or experiment, it's essential to understand the concept of independent samples. Independent samples consist of two groups that are not related to each other in any way. For instance, in the provided exercise, subjects were divided into two groups: one with a red background and one with a blue background. Each group's performance was measured independently, with no overlap or connection between the individuals in each sample.
Independent samples help ensure that the results are not influenced by external factors or relationships among participants. This provides a clearer picture of the effect being studied, in this case, the effect of color on creativity.
A practical benefit of using independent samples is that it reduces biases and ensures the validity of the findings. For students and researchers, it's crucial to understand this concept to design robust experiments that provide reliable results.

T-Test

The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In our example, we use the t-test to compare the creativity scores of subjects with red and blue backgrounds. The t-test for independent samples helps us determine if any observed differences in sample means are statistically significant or just due to random chance.
An important variant is Welch's t-test, which we used here. It is particularly useful when the variances of the two samples are unequal. The formula for Welch’s t-test is:
\[ t = \frac{ \bar{x}_2 - \bar{x}_1 }{ \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} } \]
Using this formula allows us to calculate a t-value which can then be compared against a critical value from the t-distribution table. If our calculated t-value exceeds the critical value, we reject the null hypothesis and conclude that there is a significant difference between the two sample means.
Knowing how to perform and interpret a t-test is fundamental for anyone involved in data analysis or research.

Confidence Interval

A confidence interval provides a range of values within which we expect the true population parameter to lie, based on our sample data. For our hypothesis test, a confidence interval helps determine the reliability of the difference in means between two groups.
We calculate the confidence interval using the results from the t-test and the standard error. The formula for constructing the confidence interval for the difference in means is:
\[ (\bar{x}_2 - \bar{x}_1) \pm t_{critical} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]
In our example, we found that the 98% confidence interval for the difference in creativity scores was (0.1147, 1.0453). This interval does not cross zero, which indicates a significant difference between the two groups. This supports the hypothesis that the blue background enhances creative performance.
A confidence interval provides more information than a simple hypothesis test, as it shows the range of plausible values for the parameter being studied. It is a crucial concept for interpreting data in research.

Significance Level

The significance level, denoted by \( \alpha \), is the probability threshold for rejecting the null hypothesis. It represents the risk of making a Type I error, which is rejecting a true null hypothesis. A common significance level used is 0.05, but in our problem, a more stringent level of 0.01 is utilized.
At a significance level of 0.01, there is only a 1% risk of concluding that a difference exists when there is none. This stricter criterion is often used in fields where the cost of a Type I error is high. By setting a low \( \alpha \) level, we require stronger evidence to reject the null hypothesis.
In our exercise, since the calculated t-value (2.98) was greater than the critical value (2.39), we rejected the null hypothesis at the 0.01 significance level. This indicates strong evidence that the blue background indeed enhances creative performance.
Understanding the significance level is crucial as it informs the robustness and reliability of your conclusions, helping you make informed decisions based on your data.