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91Ó°ÊÓ

To compare two elementary schools regarding teaching of reading skills, 12 sets of identical twins were used. In each case, one child was selected at random and sent to school A, and his or her twin was sent to school B. Near the end of fifth grade, an achievement test was given to each child. The results follow: $$ \begin{array}{l|cccccc} \hline \text { Twin Pair } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { School A } & 177 & 150 & 112 & 95 & 120 & 117 \\ \hline \text { School B } & 86 & 135 & 115 & 110 & 116 & 84 \\ \hline \end{array} $$ $$ \begin{array}{l|cccccc} \hline \text { Twin Pair } & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text { School A } & 86 & 111 & 110 & 142 & 125 & 89 \\ \hline \text { School B } & 93 & 77 & 96 & 130 & 147 & 101 \\ \hline \end{array} $$ Use a \(0.05\) level of significance to test the hypothesis that the two schools have the same effectiveness in teaching reading skills against the alternate hypothesis that the schools are not equally effective.

Short Answer

Expert verified
There is no significant difference in reading skills effectiveness between the schools.

Step by step solution

01

Formulate Hypotheses

We start by formulating our null and alternative hypotheses. The null hypothesis is that both schools have the same mean effectiveness: \[ H_0: \mu_A = \mu_B \]The alternative hypothesis is that the schools do not have the same effectiveness:\[ H_1: \mu_A eq \mu_B \]
02

Calculate Differences

For each twin pair, calculate the difference in test scores: \( D_i = \text{Score from School A} - \text{Score from School B} \). These differences are:\[ \begin{array}{cccccc} 91 & 15 & -3 & -15 & 4 & 33 & -7 & 34 & 14 & 12 & -22 & -12 \end{array} \]
03

Compute Sample Mean and Standard Deviation of Differences

Calculate the sample mean \( \bar{D} \) and sample standard deviation \( s_D \) of the differences. The mean is:\[ \bar{D} = \frac{\sum D_i}{n} = \frac{145}{12} \approx 12.08 \]The standard deviation \( s_D \) is computed as:\[ s_D = \sqrt{\frac{\sum (D_i - \bar{D})^2}{n-1}} \approx 27.65 \]
04

Conduct the Paired t-test

Use the paired t-test formula:\[ t = \frac{\bar{D}}{s_D / \sqrt{n}} = \frac{12.08}{27.65 / \sqrt{12}} \approx 1.253 \]with degrees of freedom \( n-1 = 11 \).
05

Determine Critical Value

At a \(0.05\) significance level with \(11\) degrees of freedom, the critical value for a two-tailed test is approximately \(2.201\).
06

Compare t-statistic to Critical Value

Since the calculated \(t\)-value \(1.253\) is less than the critical value \(2.201\), we fail to reject the null hypothesis.
07

Conclusion

There is not enough evidence to conclude that there is a significant difference in the effectiveness of teaching reading skills between schools A and B at the \(0.05\) significance level.

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Key Concepts

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

Paired t-test
A paired t-test is a statistical method used to compare two related groups. It is particularly useful when the two groups are matched or related in some way. In this exercise, we are comparing the reading skill effectiveness of two schools using pairs of twins. Each twin pair serves as a natural match where one twin attended School A and the other attended School B. This makes the observations dependent and simplifies the process of comparing the two groups.

Instead of looking at just the raw scores, the paired t-test focuses on the differences between each pair. By computing the difference in achievement test scores for each pair of twins, we account for natural variations between the pairs. This gives us a clearer view of the effectiveness of the teaching methods of the two schools.

Once the differences are calculated, the test examines if these observed differences are statistically significant or if they could have occurred by chance. The test assumes that the differences are normally distributed and calculates a t-statistic to help us make a decision. Overall, the paired t-test is a robust method when one wants to compare groups that are naturally paired.
Significance Level
In hypothesis testing, the significance level, often denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It's essentially the threshold we set for deciding whether an observed effect is statistically significant.

In this exercise, we used a significance level of 0.05. This means that in 5 out of 100 tests, we might conclude that there is a difference between schools when there is none. Setting this level helps balance the trade-off between Type I errors (wrongly rejecting the null hypothesis) and power (the ability to detect a true effect).

The significance level plays a crucial role in determining the critical value from the t-distribution table, which we compare against our test statistic. If the test statistic is more extreme than the critical value, we reject the null hypothesis. Here, a two-tailed test was used because the alternative hypothesis suggested schools could differ in effectiveness in either direction.
Null and Alternative Hypotheses
Formulating hypotheses is a fundamental step in hypothesis testing, helping us to make a well-defined statement that can be tested statistically. For this problem, our null hypothesis \( H_0 \) is that schools A and B are equally effective at teaching reading skills, which is expressed as \( \mu_A = \mu_B \). The null hypothesis represents a baseline assumption of no effect or no difference.

The alternative hypothesis \( H_1 \) contradicts the null hypothesis, suggesting that there is a difference in effectiveness: \( \mu_A eq \mu_B \). It's a two-tailed hypothesis, meaning that we aren't specifying which school is better, just that they are not the same.

The null and alternative hypotheses guide our data analysis. We use the test results to determine whether the observed data provides enough evidence to reject the null hypothesis in favor of the alternative. In this case, because the test statistic did not exceed the critical value, we failed to reject the null hypothesis, suggesting no significant differences between the schools at the set significance level.
Statistical Differences
Statistical differences refer to whether observed differences in a sample capture a real effect or are due to random variation. In the context of this exercise, it's about determining if the differences in reading scores between twins in Schools A and B reflect a genuine difference in teaching effectiveness, rather than being a product of chance alone.

To detect statistical differences, the paired t-test helps by turning the observed differences into a t-statistic. This synthesis of observed data, sample size, and variability, allows comparison to a theoretical distribution (t-distribution) to assess if the result is significant.

A statistical difference implies that the difference is unlikely to have occurred by chance. However, in our exercise, the t-statistic did not reach the critical value at the 0.05 significance level. Thus, we could not infer a significant difference in school effectiveness. In practical terms, the differences in test scores, while varying, didn't convincingly show one school's teaching methods as superior to the other's in this setup.

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Most popular questions from this chapter

A cognitive aptitude test consists of putting together a puzzle. Eleven people in group A took the test in a competitive setting (first and second to finish received a prize). Twelve people in group B took the test in a noncompetitive setting. The results follow (in minutes required to complete the puzzle). $$ \begin{array}{l|cccccccccccc} \hline \text { Group A } & 7 & 12 & 10 & 15 & 22 & 17 & 18 & 13 & 8 & 16 & 11 & \\ \hline \text { Group B } & 9 & 16 & 30 & 11 & 33 & 28 & 19 & 14 & 24 & 27 & 31 & 29 \\ \hline \end{array} $$ Use a \(5 \%\) level of significance to test the claim that there is no difference in the distributions of time to complete the test.

Does the average length of time to earn a doctorate differ from one field to another? Independent random samples from large graduate schools gave the following averages for length of registered time (in years) from bachelor's degree to doctorate. Sample A was taken from the humanities field, and sample \(\mathrm{B}\) from the social sciences field. $$ \begin{array}{l|cccccccccccc} \hline \text { Field A } & 8.9 & 8.3 & 7.2 & 6.4 & 8.0 & 7.5 & 7.1 & 6.0 & 9.2 & 8.7 & 7.5 & \\ \hline \text { Field B } & 7.6 & 7.9 & 6.2 & 5.8 & 7.8 & 8.3 & 8.5 & 7.0 & 6.3 & 5.4 & 5.9 & 7.7 \\ \hline \end{array} $$ Use a \(1 \%\) level of significance to test the claim that there is no difference in the distributions of time to complete a doctorate for the two fields.

The following data represent annual percentage returns on Vanguard Total Bond Index for a sequence of recent years. This fund represents nearly all publicly traded U.S. bonds (Reference: Morningstar Mutual Fund Analysis). $$ \begin{array}{llllllllllll} 7.1 & 9.7 & -2.7 & 18.2 & 3.6 & 9.4 & 8.6 & -0.8 & 11.4 & 8.4 & 8.3 & 0.8 \end{array} $$ (i) Convert this sequence of numbers to a sequence of symbols \(\mathrm{A}\) and \(\mathrm{B}\), where A indicates a value above the median and B a value below the median. (ii) Test the sequence for randomness about the median. Use \(\alpha=0.05\).

A psychology professor is studying the relation between overcrowding and violent behavior in a rat colony. Eight colonies with different degrees of overcrowding are being studied. By using a television monitor, lab assistants record incidents of violence. Each colony has been ranked for crowding and violence. A rank of 1 means most crowded or most violent. The results for the eight colonies are given in the following table, with \(x\) being the population density rank and \(y\) the violence rank. $$ \begin{array}{l|rrrrrrrr} \hline \text { Colony } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline x \text { rank } & 3 & 5 & 6 & 1 & 8 & 7 & 4 & 2 \\ y \text { rank } & 1 & 3 & 5 & 2 & 8 & 6 & 4 & 7 \\ \hline \end{array} $$ Using a \(0.05\) level of significance, test the claim that lower crowding ranks mean lower violence ranks (i.e., the variables have a monotone-increasing relationship).

For each successive presidential term from Franklin Pierce (the 14th president, elected in 1853 ) to George W. Bush (43rd president), the party affiliation controlling the White House is shown below, where R designates Republican and D designates Democrat (Reference: The New York Times Almanac). Historical Note: We start this sequence with the 14 th president because earlier presidents belonged to political parties such as the Federalist or Wigg (not Democratic or Republican) party. In cases in which a president died in office or resigned, the period during which the vice president finished the term is not counted as a new term. The one exception is the case in which Lincoln (a Republican) was assassinated and the vice president Johnson (a Democrat) finished the term. $$ \begin{array}{llllllllllllllllllll} \text { D } & \text { D } & \text { R } & \text { R } & \text { D } & \text { R } & \text { R } & \text { R } & \text { R } & \text { D } & \text { R } & \text { D } & \text { R } & \text { R } & \text { R } & \text { R } & \text { D } & \text { D } & \text { R } & \text { R } \\ \text { D } & \text { D } & \text { D } & \text { D } & \text { D } & \text { R } & \text { R } & \text { D } & \text { D } & \text { R } & \text { R } & \text { D } & \text { R } & \text { R } & \text { R } & \text { D } & \text { D } & \text { R } & & \end{array} $$ Test the sequence for randomness at the \(5 \%\) level of significance. Use the following outline. (a) State the null and alternate hypotheses. (b) Find the number of runs \(R, n_{1}\), and \(n_{2} .\) Let \(n_{1}=\) number of Republicans and \(n_{2}=\) number of Democrats. (c) In this case, \(n_{1}=21\), so we cannot use Table 10 of Appendix II to find the critical values. Whenever either \(n_{1}\) or \(n_{2}\) exceeds 20, the number of runs \(R\) has a distribution that is approximately normal, with $$ \mu_{R}=\frac{2 n_{1} n_{2}}{n_{1}+n_{2}}+1 \text { and } \quad \sigma_{R}=\sqrt{\frac{\left(2 n_{1} n_{2}\right)\left(2 n_{1} n_{2}-n_{1}-n_{2}\right)}{\left(n_{1}+n_{2}\right)^{2}\left(n_{1}+n_{2}-1\right)}} $$ We convert the number of runs \(R\) to a \(z\) value, and then use the normal distribution to find the critical values. Convert the sample test statistic \(R\) to \(z\) using the formula $$ z=\frac{R-\mu_{R}}{\sigma_{R}} $$ (d) The critical values of a normal distribution for a two-tailed test with level of significance \(\alpha=0.05\) are \(-1.96\) and \(1.96\) (see Table \(5(\mathrm{c})\) of Appendix II. Reject \(H_{0}\) if the sample test statistic \(z \leq-1.96\) or if the sample test statistic \(z \geq 1.96\). Otherwise, do not reject \(H_{0}\). $$ \begin{array}{c|c|c} \text { Sample } z \leq-1.96 & -1.96<\text { sample } z<1.96 & \text { Sample } z \geq 1.96 \\ \hline \text { Reject } H_{0} & \text { Fail to reject } H_{0} . & \text { Reject } H_{0} \end{array} $$ Using this decision process, do you reject or fail to reject \(H_{0}\) at the \(5 \%\) level of significance? What is the \(P\) -value for this two-tailed test? At the \(5 \%\) level of significance, do you reach the same conclusion using the \(P\) -value that you reach using critical values? Explain. (e) Interpret your results in the context of the application.

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