Question

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. Reducing Errors in Grammar A composition teacher wishes to see whether a new smartphone app will reduce the number of grammatical errors her students make when writing a two-page essay. She randomly selects six students, and the data are shown. At \(\alpha=0.025\) can it be concluded that the number of errors has been reduced? $$ \begin{array}{l|cccccc}{\text { Student }} & {1} & {2} & {3} & {4} & {5} & {6} \\\ \hline \text { Errors before } & {12} & {9} & {0} & {5} & {4} & {3} \\\ \hline \text { Errors after } & {9} & {6} & {1} & {3} & {2} & {3}\end{array} $$

Short answer

The evidence is insufficient to support the claim that the app reduces grammatical errors.

Step-by-step solution

State the Hypotheses and Identify the Claim

For this problem, we want to determine if the new smartphone app reduces the number of grammatical errors. Let \( \mu_d \) represent the mean difference in errors. We state the null hypothesis as \( H_0: \mu_d = 0 \) (no difference, the app does not reduce errors) and the alternative hypothesis as \( H_1: \mu_d > 0 \) (the app reduces errors). The claim here is in the alternative hypothesis that the number of errors has been reduced by the app.

Find the Critical Value

Since \( H_1: \mu_d > 0 \) is a one-tail test. Given \( \alpha = 0.025 \) and \( n = 6 \), we use the t-distribution with \( df = n-1 = 5 \). Using the t-table or calculator, we find the critical value for a right-tailed test at \( \alpha = 0.025 \) and \( df = 5 \) is approximately \( t_{critical} = 2.571 \).

Compute the Test Value

First, calculate the differences: \([-3, -3, 1, -2, -2, 0]\). Then calculate \( \bar{d} \), the mean of differences: \( \bar{d} = \frac{-3 + (-3) + 1 + (-2) + (-2) + 0}{6} = -1.5 \). Calculate the standard deviation of differences, \( s_d \), and then the test statistic \( t = \frac{\bar{d} - 0}{s_d/\sqrt{n}} \). This requires further calculations: find \( s_d = \sqrt{\frac{{\sum (d_i - \bar{d})^2}}{{n-1}}} = 1.8708 \). Thus, \( t = \frac{-1.5}{1.8708/\sqrt{6}} \approx -1.85 \).

Make the Decision

Since the computed test statistic \( t = -1.85 \) is not greater than the critical value \( t_{critical} = 2.571 \), we fail to reject the null hypothesis. Thus, there is insufficient evidence to support the claim that the smartphone app reduces grammatical errors.

Summarize the Results

At a significance level of \( \alpha = 0.025 \), the evidence does not suggest that the new smartphone app significantly reduces grammatical errors among students. Therefore, we conclude there is no significant reduction in grammatical errors attributed to the app.

Key concepts

t-distribution

In hypothesis testing, the t-distribution plays a crucial role, especially when dealing with small sample sizes. The t-distribution is similar to the normal distribution but comes with thicker tails. This feature helps it to handle the variability that comes with small samples. In the context of our exercise, we have a sample size of 6, which is considered small. Therefore, we use the t-distribution to find the critical value.

When using the t-distribution, we need to determine the degrees of freedom (df). For paired data like in our example, the degrees of freedom is calculated as the number of paired differences minus one:
\[ df = n - 1 \]
where represents the number of paired observations. This means for our study, df equals 5 (6 - 1). The correct critical value can then be found using a t-table or calculator with these degrees of freedom and the specified alpha level. The flexibility of the t-distribution makes it indispensable when we have less data to work with.

null hypothesis

The null hypothesis is a foundational concept in statistical hypothesis testing. It presents a statement that there is no effect or no difference. In our example involving the smartphone app, the null hypothesis is set as \( H_0: \mu_d = 0 \). This implies that the mean difference in grammatical errors before and after the app usage is zero. The null hypothesis assumes that the app has no influence in reducing errors.

The objective of hypothesis testing is to collect evidence from the sample data to determine whether to reject the null hypothesis. When the evidence is insufficient, we fail to reject it, suggesting there's no support for the alternative hypothesis that the app reduces errors. The null hypothesis essentially serves as the "status quo" or default assumption that must be disproved to claim any effect of the app.

critical value

The critical value acts as a threshold that separates the region where we accept the null hypothesis from the region where we reject it. This value is selected based on the level of significance (\( \alpha \)), as well as the desired tail of the test.

In our exercise, since the claim is that the smartphone app reduces errors, we are looking for evidence of improvement and thus, a one-tailed test is used. With a significance level \( \alpha = 0.025 \) and degrees of freedom \( df = 5 \), the critical value from the t-distribution, is approximately 2.571.

The insertion of the critical value into our process then allows us to directly compare it with the calculated test statistic to make our final decision. If the test statistic exceeds the critical value in a positive direction, we reject the null hypothesis, supporting the alternative.

statistical significance

Statistical significance is key to determining the validity of the results. It reflects whether the results observed are due to chance or indicate a real effect. This is evaluated against a pre-decided significance level, \( \alpha \), which in our case is set at 0.025.

When conducting the test, we calculate a test statistic from our sample data, representing the observed effect of the app on grammatical errors. If this statistic surpasses the critical value, it indicates the result is statistically significant, providing grounds to reject the null hypothesis in favor of the alternative hypothesis.

In the example given, the test statistic was found to be \( t = -1.85 \), which is not higher than the critical value, thus showing that the change is not statistically significant at the 0.025 level. This means there's not enough evidence to prove that the app drastically reduces grammatical errors.

• Statistical significance helps separate true effects from random noise in the data.
• It is not always synonymous with practical significance, which needs to be considered alongside real-world implications.