For Exercises 2 through 12, perform each of these steps....

For Exercises 2 through $12,$ 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. $$ \begin{array}{l|ccccccc}{\text { Child }} & {1} & {2} & {3} & {4} & {5} & {6} & {7} \\ \hline \text { Trial } 1 & {100} & {150} & {150} & {110} & {130} & {120} & {118} \\ \hline \text { Trial 2} & {90} & {130} & {150} & {90} & {105} & {110} & {120}\end{array} $$

For Exercises 2 through $12,$ 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.
$$
\begin{array}{l|ccccccc}{\text { Child }} & {1} & {2} & {3} & {4} & {5} & {6} & {7} \\ \hline \text { Trial } 1 & {100} & {150} & {150} & {110} & {130} & {120} & {118} \\ \hline \text { Trial 2} & {90} & {130} & {150} & {90} & {105} & {110} & {120}\end{array}
$$

Key Concepts

Assumptions of Normality

Many parametric tests, including the paired samples t-test, assume that the data or the differences between paired observations are normally or approximately normally distributed. This assumption is crucial for the validity of the test results.

Confidence Interval

A confidence interval is a range of values derived from sample data that is believed, with a specified level of confidence, to contain the true value of the population parameter. It provides an estimate of the uncertainty associated with the sample statistic.

P-Value Method

The p-value method involves calculating the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value under the assumption that the null hypothesis is true. A small p-value indicates strong evidence against the null hypothesis, leading to its rejection if it is below the predetermined significance level.

Paired Samples t-Test

The paired samples t-test is used to compare the means of two related groups, typically by analyzing the differences between paired observations. It is particularly useful when measurements are taken before and after an intervention on the same subjects.

Hypothesis Testing

Hypothesis testing involves formulating a null hypothesis and an alternative hypothesis about a population parameter, then using sample data to determine whether there is enough evidence to reject the null hypothesis. This process forms the foundation of inferential statistics.

Significance Level and Critical Values

The significance level, denoted by alpha (?), defines the threshold for rejecting the null hypothesis and represents the probability of committing a Type I error. Critical values corresponding to this level are used to determine the rejection region for the test statistic.

Transcript

00:01 We have a sample of six students who assembled a toy once and was timed, and then they assembled a toy a second time and were timed again.

00:10 And we want to test if at a significance level of 0 .01, if there's evidence to support the case that it took them less time to assemble it the second time.

00:23 We're asked to do this test using the p -value method, and then we're asked to construct a 99 % confidence interval for the difference in the means.

00:31 So we're doing this test at a significance level of 0 .01.

00:39 So first step is to identify our hypotheses.

00:42 So the no hypothesis is that the mean of the difference in times is zero, or that there was no change.

00:53 And the alternative hypothesis is that the mean of the differences is not zero.

01:01 In fact, it should be greater than zero, because we want to test whether the times were reduced on the second construction of the toy.

01:11 And from the problem we have n is equal to 7.

01:15 So since we're using the p -value method to test this, we can go straight to calculating our test value that's given by this formula.

01:32 And so we have n from the question.

01:36 And we need to find the average difference in times and the sample standard deviation of the difference in times.

01:44 So i've calculated these parameters in excel.

01:47 So i copied and pasted the sample data from the question here.

01:51 And in this column, i've calculated the differences in times, or trial 1 minus trial 2.

01:57 And here i've calculated the average of the sample differences.

02:01 This is the average for column j.

02:04 And here i've calculated the sample standard deviation for the differences.

02:13 So now we have 11 .86 divided by 10 .37 over the square root of 7.

02:30 And this comes out to 3 .026.

02:36 So now we can go find our p value, knowing that the degrees of freedom is n -1, and the significance level is 0 .01...

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