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. Reducing Errors in Spelling A ninth-grade teacher wishes to see if a new spelling program will reduce the spelling errors in his students' writing. The number of spelling errors made by the students in a five-page report before the program is shown. Then the number of spelling errors made by students in a five-page report after the program is shown. At \(\alpha=0.05,\) did the program work? $$ \begin{array}{lllrllll} \text { Before } & 8 & 3 & 10 & 5 & 9 & 11 & 12 \\ \hline \text { After } & 6 & 4 & 8 & 1 & 4 & 7 & 11 \end{array} $$

Short answer

The spelling program significantly reduced errors at \(\alpha = 0.05\).

Step-by-step solution

State the Hypotheses

First, we need to set up our null and alternative hypotheses. The claim is that the new spelling program will reduce the spelling errors. Thus, we will compare the mean number of spelling errors before and after the program. The null hypothesis \(H_0\) is that the mean number of errors before the program is equal to the mean number of errors after the program. The alternative hypothesis \(H_1\) is that the mean number of errors before is greater than the mean number of errors after. Formally, the hypotheses can be stated as:\[ H_0: \mu_{\text{before}} = \mu_{\text{after}} \]\[ H_1: \mu_{\text{before}} > \mu_{\text{after}} \]

Find the Critical Value(s)

For this step, we need to find the critical value for a one-tailed test with \(\alpha = 0.05\). Since we're comparing means, we'll use a t-distribution. The degrees of freedom \(df\) can be calculated as \(n - 1\) where \(n\) is the sample size. Here, \(n = 7\) for each group, thus \(df = 7 - 1 = 6\). From a t-distribution table, the critical value \(t_{\text{critical}}\) for \(df = 6\) at \(\alpha = 0.05\) (right-tailed) is approximately 1.943.

Compute the Test Value

Now, calculate the test statistic. Start by finding the mean and standard deviation for both before and after groups:1. Mean before: \(\bar{x}_{\text{before}} = \frac{8+3+10+5+9+11+12}{7} = 8.29\)2. Mean after: \(\bar{x}_{\text{after}} = \frac{6+4+8+1+4+7+11}{7} = 5.86\)3. Standard deviation for both groups (assuming equal variance).Since this is a dependent (paired) sample t-test, compute the differences between pairs and their mean (\(\bar{d}\)) and standard deviation (\(s_d\)):\[ d: (8-6, 3-4, 10-8, 5-1, 9-4, 11-7, 12-11) = (2, -1, 2, 4, 5, 4, 1) \]\[ \bar{d} = \frac{2 - 1 + 2 + 4 + 5 + 4 + 1}{7} = 2.43 \]\[ s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} = \sqrt{\frac{(-0.43)^2 + (-3.43)^2 + (-0.43)^2 + (1.57)^2 + (2.57)^2 + (1.57)^2 + (-1.43)^2}{6}} \approx 2.30 \]Calculate the test statistic:\[ t = \frac{\bar{d}}{s_d/\sqrt{n}} = \frac{2.43}{2.30/\sqrt{7}} \approx \frac{2.43}{0.870} \approx 2.79 \]

Make the Decision

Compare the calculated test value to the critical value. Here, the test statistic \(t = 2.79\) is greater than the critical value \(t_{\text{critical}} = 1.943\). Therefore, we reject the null hypothesis.

Summarize the Results

Since we rejected the null hypothesis, we have sufficient evidence at \(\alpha = 0.05\) to support the claim that the spelling program reduced spelling errors in students' writings.

Key concepts

Spelling Program Effectiveness

The goal of the teacher was to assess whether a new spelling program can indeed lead to fewer spelling errors among students. To do this scientifically, we perform hypothesis testing to determine if changes in spelling errors before and after implementing the program are statistically significant.

The effectiveness of the spelling program is quantified by measuring the mean number of errors before and after the program. If the number of errors decreases significantly after the program, it suggests that the program is effective. The 'before' and 'after' comparisons allow us to control for individual student differences by essentially measuring each student's improvement.

Dependent Samples T-test

A dependent samples t-test, also known as a paired-sample t-test, is used here because we are dealing with two sets of scores from the same group of students before and after the intervention. This test is ideal for such 'before and after' scenarios because it accounts for the dependency between the two samples.

In this context, every student's number of spelling errors before the program is paired with their respective number of errors after the program. The paired nature strengthens the test by reducing variability that comes from individual differences. Hence, it provides a clear measure of the program's impact on spelling performance. The primary aim is to compare the means of these two related groups.

Critical Value

The critical value is crucial in hypothesis testing as it defines the threshold for deciding whether to reject the null hypothesis. In our problem, we perform a one-tailed test at a significance level \(\alpha = 0.05\). Using statistical tables, we find that the critical value \((t_{\text{critical}})\) for 6 degrees of freedom is approximately 1.943.

This value is important because it represents a cut-off point on the t-distribution. If our calculated test statistic is greater than this critical value, the result is statistically significant, allowing us to reject the null hypothesis. In pedagogical methods like spelling programs, using the correct critical value ensures we draw accurate conclusions about effectiveness.

Test Statistic

The test statistic is calculated from the data to determine if the observed effect is significant. In our scenario, it involves calculating the mean and standard deviation of differences in spelling errors before and after the program.

The formula for the test statistic in a dependent samples t-test is:

• \[ t = \frac{\bar{d}}{s_d/\sqrt{n}} \]

where \(\bar{d}\) is the mean of differences, \(s_d\) is the standard deviation of differences, and \(n\) is the number of pairs.

Here, the test statistic is approximately 2.79. This value is compared to the critical value to decide on rejecting or not rejecting the null hypothesis. In our case, since 2.79 exceeds the critical value of 1.943, we reject the null hypothesis, concluding that the spelling program effectively reduces errors.