Question

To compare the effect of stress in the form of noise on the ability to perform a simple task, 70 subjects were divided into two groups. The first group of 30 subjects acted as a control, while the second group of 40 were the experimental group. Although each subject performed the task in the same control room, each of the experimental group subjects had to perform the task while loud rock music was played. The time to finish the task was recorded for each subject and the following summary was obtained: $$\begin{array}{lll} & \text { Control } & \text { Experimental } \\\\\hline n & 30 & 40 \\\\\bar{x} & 15 \text { minutes } & 23 \text { minutes } \\ s & 4 \text { minutes } & 10 \text { minutes }\end{array}$$ a. Find a \(99 \%\) confidence interval for the difference in mean completion times for these two groups. b. Based on the confidence interval in part a, is there sufficient evidence to indicate a difference in the average time to completion for the two groups? Explain.

Short answer

#tag_title#Question#tag_content#Based on the provided information, calculate the 99% confidence interval for the difference in mean completion times between control and experimental groups, and determine if there's sufficient evidence to conclude that there's a difference in the average completion times for each group.

Step-by-step solution

Calculate the Standard Error of the Difference

First, we need to calculate the standard error of the difference in means. This is done using the formula: $$SE_{diff} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$$ where \(s_1\) and \(s_2\) are the sample standard deviations for the control and experimental groups, respectively, and \(n_1\) and \(n_2\) are the corresponding sample sizes. Plugging in the given values, we get: $$SE_{diff} = \sqrt{\frac{4^2}{30} + \frac{10^2}{40}}$$

Calculate the Critical Value

Next, we need to determine the critical value for the t-distribution. Since we have a 99% confidence interval, our alpha (\(α\)) is \(1 - 0.99 = 0.01\). We'll use the t-distribution because the population standard deviation is unknown. We must calculate the degrees of freedom for the estimation: $$v = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1 - 1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2 - 1}}$$ After calculating the degrees of freedom, we look up the t-score for the 99% confidence level. We'll call this t-score \(t_{\alpha/2}\).

Calculate the Confidence Interval

Now, we can calculate the confidence interval: $$CI = (\bar{x_1} - \bar{x_2}) \pm t_{\alpha/2} \times SE_{diff}$$ where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means for the control and experimental groups, respectively. Plugging in the values will give us the 99% confidence interval for the difference in mean completion times between the control and the experimental groups.

Determine if there is Sufficient Evidence for a Difference in Means

To determine if there's sufficient evidence to indicate a difference in mean completion times between the two groups, we need to analyze the confidence interval calculated in Step 3. If the interval does not contain 0, then we can conclude that there's a significant difference (at a 99% confidence level) between the two groups. Otherwise, we cannot conclude that there's a significant difference in mean completion times.

Key concepts

Understanding Standard Error

Standard error (SE) is one of the most fundamental concepts in statistics, particularly when it comes to understanding the precision of an estimated mean. Standard error measures the dispersion of sample mean estimates from the actual population mean. If we were to repeatedly draw samples from a population and calculate the mean for each sample, the standard error would tell us how much those sample means would typically vary from the true population mean.

In the context of our exercise, we computed the standard error of the difference between two sample means to gauge the expected variability in this difference, should new samples be taken. By including each group's sample standard deviation and size, we establish how much uncertainty or 'error' exists when we estimate the mean difference between the control and experimental groups from our sample data. A smaller standard error suggests that our sample mean is a more precise estimate of the true population mean.

The Role of t-Distribution

In statistics, there's a special relevance to the t-distribution when dealing with small sample sizes or when the population standard deviation is unknown. The shape of the t-distribution is similar to the normal distribution but with thicker tails, meaning it predicts a higher probability of values further from the mean, accommodating for the increased uncertainty with smaller samples.

For hypothesis testing and constructing confidence intervals, the t-distribution comes into play as it allows us to make inferences about the population mean, even when some population parameters are unknown. The critical t-values are found using degrees of freedom, which in simpler terms, refers to the number of independent values in a calculation. The more the degrees of freedom, the closer the t-distribution resembles a normal distribution.

Sample Mean: A Crucial Estimator

The sample mean serves as an estimator for the population mean. In our scenario, we have two sample means: one for the control group and one for the experimental group. These means provide an average measure of the time it takes to complete a task under two different conditions: with and without loud rock music. When we compare the sample means from our two groups, we essentially compare these conditions' effects on task performance.

The difference between these sample means is pivotal in our analysis. If the means are significantly different, it suggests that the presence of loud rock music (our experimental variable) may have an appreciable effect on completion times. However, we must also account for the random variability in our samples, which brings us back to the importance of the standard error in giving context to the sample mean.

Hypothesis Testing: Making Inferences

Hypothesis testing is a statistical method for making decisions about population parameters based on sample data. The core idea is to assume a null hypothesis (often a hypothesis of no effect or no difference) and then determine whether the observed data provide enough evidence to reject this null hypothesis.

In our textbook problem, we test the hypothesis that loud rock music does not affect task completion time. By constructing a confidence interval for the difference in mean completion times and checking whether this interval contains the value zero (indicating no difference), we can either support the null hypothesis or suggest there is a significant difference that warrants rejecting it. If our confidence interval does not contain zero, we infer that the effect of rock music is statistically significant, providing evidence against the null hypothesis at the stated confidence level.