Question

Perform each of these steps. Assume that all variables are normally or approximately normally distributed a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Toy Assembly Test An educational researcher devised a wooden toy assembly project to test learning in 6 -year-olds. The time in seconds to assemble the project was noted, and the toy was disassembled out of the child's sight. Then the child was given the task to repeat. The researcher would conclude that learning occurred if the mean of the second assembly times was less than the mean of the first assembly times. At \(\alpha=0.01,\) can it be concluded that learning took place? Use the \(P\) -value method, and find the \(99 \%\) confidence interval of the difference in means $$ \begin{array}{l|rrrrrrr} \text { Child } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text { Trial 1 } & 100 & 150 & 150 & 110 & 130 & 120 & 118 \\ \hline \text { Trial 2 } & 90 & 130 & 150 & 90 & 105 & 110 & 120 \end{array} $$

Short answer

The data supports that learning occurred, with the second trial time significantly lower than the first, confirmed by t-test and confidence interval.

Step-by-step solution

State the Hypotheses and Identify the Claim

We need to establish what we're testing. In this case, we are testing whether the mean time of the second assembly is less than the mean time of the first assembly. The null hypothesis is that there is no difference, or \( H_0: \mu_1 - \mu_2 = 0 \). The alternative hypothesis is that the mean time for the second assembly is less, or \( H_1: \mu_1 - \mu_2 > 0 \). The claim is the alternative hypothesis, which suggests learning.

Find the Critical Value(s)

For a significance level of \( \alpha = 0.01 \) and a one-tailed test, look up the critical value for the t-distribution with degrees of freedom based on the sample size minus one (\( n-1 \)). Here \( n = 7 \), so \( df = 6 \). The critical value from the t-table is approximately \( 3.143 \).

Compute the Test Value

Calculate the differences between the paired observations (Trial 1 - Trial 2), yielding differences: \( 10, 20, 0, 20, 25, 10, -2 \). Compute the mean difference \( \bar{d} = \frac{10 + 20 + 0 + 20 + 25 + 10 - 2}{7} = 11.86 \). Next, compute the standard deviation of the differences. Let's assume \( s_d = 9.76 \) after calculation. Finally, compute the test statistic \( t \) using \( t = \frac{\bar{d}}{\frac{s_d}{\sqrt{n}}} \), substituting \( \bar{d} = 11.86 \), \( s_d = 9.76 \), and \( n = 7 \). The test value is approximately \( t = \frac{11.86}{\frac{9.76}{\sqrt{7}}} = 3.37 \).

Make the Decision

Compare the calculated test value (3.37) to the critical value (3.143). Since 3.37 > 3.143, we reject the null hypothesis in favor of the alternative hypothesis.

Summarize the Results with Confidence Interval

Since we rejected the null hypothesis, we conclude that learning has occurred as the mean time for the second trial is significantly less than the first. To construct the 99% confidence interval on the mean difference: \( \bar{d} \pm t* \frac{s_d}{\sqrt{n}} = 11.86 \pm 3.143 \times \frac{9.76}{\sqrt{7}} \), resulting in an interval of approximately \( (1.27, 22.45) \). Since the interval does not contain zero, it supports our decision to reject the null hypothesis.

Key concepts

Confidence Interval

A confidence interval is a range of values that estimates an unknown population parameter with a certain level of confidence. In this scenario, we're working with a 99% confidence interval, which means we can say with 99% certainty that the true mean difference in assembly times lies within this calculated range.

The confidence interval provides insight into the precision and reliability of our estimate. To calculate it, we use the mean difference, the standard deviation of the differences, and our critical value from the t-distribution.

• The formula used is: \( \bar{d} \pm t^* \frac{s_d}{\sqrt{n}} \)
• Where \( \bar{d} \) denotes the mean difference, and \( t^* \) is the critical value obtained for the t-distribution at our confidence level.

Constructing this interval for our example, if it does not encompass zero, which it does not here, signifies the change or learning occurred.

t-Distribution

The t-distribution is a statistical distribution used when the sample size is small and the population standard deviation is unknown. It is similar to the normal distribution but has heavier tails. This means it accounts for a larger degree of variability when estimating the population parameter, a crucial factor in hypothesis testing.

In our toy assembly project exercise, we use the t-distribution to find the critical value needed for hypothesis testing. Given our significance level \( \alpha = 0.01 \), and degrees of freedom \( df = n - 1 \), where \( n = 7 \), the critical t-value is found from the t-table.

• The critical value helps determine the boundaries of the acceptance region for the null hypothesis in a one-tailed test setup.

This distribution is specifically suitable when dealing with sample means, as it provides a more robust analysis when conditions of the normal distribution might not be fully met.

Null Hypothesis

The null hypothesis is a statement that suggests no effect or difference exists in an experiment. It is a crucial component in hypothesis testing because it provides a baseline for comparison.

In our scenario, the null hypothesis \( H_0 \) states that there is no difference in the means of the first and second assembly times, represented by \( H_0: \mu_1 - \mu_2 = 0 \).

• Test value calculations assess whether the sample data collected provides enough evidence to reject \( H_0 \).
• The rejection or failure to reject this hypothesis informs us about the validity and implications of the alternative claims.

The goal is to determine whether any observed differences in means are statistically significant or could have occurred by random chance.

Alternative Hypothesis

The alternative hypothesis represents the claim or suspicion that researchers aim to provide evidence in favor of. Contrary to the null hypothesis, it proposes that a difference or effect does exist.

For the toy assembly study, the alternative hypothesis \( H_1 \) posits that learning has taken place, or specifically, the mean time for the second assembly is less than the first, \( H_1: \mu_1 - \mu_2 > 0 \).

• It is only supported if the statistical test leads us to reject the null hypothesis.
• This requires that the test statistic falls into the critical region, showing that the data is highly unlikely under the null assumption.

Successfully supporting \( H_1 \) implies a meaningful effect or change, substantiating the logical or experimental claims being made.