Question

As part of a larger pilot study, students at a Riverside, California middle school, will compare the learning of algebra by students using iPads versus students using the traditional algebra textbook with the same author and publisher. \(^{20}\) To remove teacher-to-teacher variation, the same teacher will teach both classes, and the iPad and textbook material are both provided by the same author and publisher. Suppose that after 1 month, 10 students were selected from each class and their scores on an algebra advancement test recorded. The summarized data follow. $$ \begin{array}{lcc} & \text { iPad } & \text { Textbook } \\ \hline \text { Mean } & 86.4 & 79.7 \\ \text { Standard Deviation } & 8.95 & 10.7 \\ \text { Sample Size } & 10 & 10 \end{array} $$ a. Use the summary data to test for a significant difference in advancement scores for the two groups using \(\alpha=.05 .\) b. Find a \(95 \%\) confidence interval for the difference in mean scores for the two groups. c. In light of parts a and b, what can we say about the efficacy of using an iPad versus a traditional textbook in learning algebra at the middle school level?

Short answer

Short Answer: Based on our analysis using a two-sample t-test, we fail to reject the null hypothesis, which means we cannot conclude that there is a significant difference in the advancement scores between the two groups. However, our 95% confidence interval ranges from -0.3 to 13.7, so it is possible that the mean difference could be positive. Further studies with larger samples and a longer timeframe may reveal significant differences in learning outcomes.

Step-by-step solution

State the null and alternative hypotheses

We want to test for a significant difference in scores for the two groups, so our null and alternative hypotheses are: \(H_0: \mu_{iPad} - \mu_{Textbook} = 0\) \(H_1: \mu_{iPad} - \mu_{Textbook} \neq 0\) Where \(\mu_{iPad}\) is the mean score of students using iPads and \(\mu_{Textbook}\) is the mean score of students using traditional textbooks.

Determine the appropriate test statistic and decision rule

Since we have two sample groups and we're given the means, standard deviations, and sample sizes, we will use the two-sample t-test. For this test, our test statistic (t-score) formula is: \(t=\frac{(\bar{x}_{iPad} - \bar{x}_{Textbook}) - 0}{\sqrt{\frac{s_{iPad}^{2}}{n_{iPad}} + \frac{s_{Textbook}^{2}}{n_{Textbook}}}}\) We're given significance level \(\alpha=0.05\). Since this is a two-tailed test, we divide \(\alpha/2 = 0.025\) for each tail. We will use a t-table or software to find the t-value corresponding to \(\alpha=0.025\) and degrees of freedom \(df = min(n_{iPad}, n_{Textbook}) - 1= min(10, 10) - 1 = 9\).

Calculate the test statistic

Using the given data and the t-score formula, we have: \(\bar{x}_{iPad} = 86.4\) \(\bar{x}_{Textbook} = 79.7\) \(s_{iPad} = 8.95\) \(s_{Textbook} = 10.7\) \(n_{iPad} = 10\) \(n_{Textbook} = 10\) \(t=\frac{(86.4 - 79.7) - 0}{\sqrt{\frac{8.95^{2}}{10} + \frac{10.7^{2}}{10}}}= 2.193\)

Compare the test statistic with the critical value

We look up the t-value for a 95% confidence level and 9 degrees of freedom in the t-table or using software, and get a critical value of \(2.262\). Since our calculated test statistic \(2.193\) is less than the critical value \(2.262\), we fail to reject the null hypothesis.

Calculate the 95% confidence interval for the difference in mean scores

To find the 95% confidence interval for the difference in mean scores, we use the following formula: \(CI = (\bar{x}_{iPad} - \bar{x}_{Textbook}) \pm t_{crit}*\sqrt{\frac{s_{iPad}^{2}}{n_{iPad}} + \frac{s_{Textbook}^{2}}{n_{Textbook}}}\) \(CI = (86.4 - 79.7) \pm 2.262*\sqrt{\frac{8.95^{2}}{10} + \frac{10.7^{2}}{10}}\) \(CI = (6.7 \pm 5.514)\) The 95% confidence interval is \((-0.3.0, 13.7)\).

Interpret the results

In light of parts a and b, we fail to reject the null hypothesis, which implies that there is no significant difference in the advancement scores between the iPad and textbook groups. However, this does not necessarily mean there's no benefit of using iPads for learning algebra. The confidence interval we calculated ranges from -0.3 to 13.7, so it is possible that the mean difference could be positive. Further studies with larger sample sizes and longer periods could potentially reveal significant differences over time.

Key concepts

Null and Alternative Hypotheses

When comparing two groups, like students using iPads versus textbooks, we set up hypotheses to help us analyze the data. Here's how it works:

The **null hypothesis** (\(H_0\)) claims that there's no difference between the groups. In this case, we assume:

• \(H_0: \mu_{iPad} - \mu_{Textbook} = 0\)

This means any difference seen is due to chance.

The **alternative hypothesis** (\(H_1\)) suggests there is a difference. It's what we suspect might be true:

• \(H_1: \mu_{iPad} - \mu_{Textbook} eq 0\)

This suggests that the learning tools do affect students' scores. By setting these hypotheses, we have a clear path for statistical testing. We'll use data to see if we can reject the null hypothesis in favor of the alternative, which will suggest that a real difference exists.

Confidence Interval

A confidence interval gives us a range in which we think the true difference in means lies. It helps in understanding the variability and certainty of our results.

In this study, we calculated a 95% confidence interval for the difference in scores using:

• The formula: \(CI = (\bar{x}_{iPad} - \bar{x}_{Textbook}) \pm t_{crit}*\sqrt{\frac{s_{iPad}^{2}}{n_{iPad}} + \frac{s_{Textbook}^{2}}{n_{Textbook}}}\)
• The values: Mean differences, standard deviations, and sample sizes.

The result was \((-0.3, 13.7)\). This range means we are 95% confident the true difference is between -0.3 and 13.7.

Importantly, because this interval includes zero, it suggests that the difference might not be statistically significant, though the potential for a positive difference is there. Thus, while the interval provides valuable insight, larger studies might be needed to clarify results.

Significance Level

The significance level, often denoted by \(\alpha\), is crucial in hypothesis testing. It represents the probability of rejecting a true null hypothesis, often set at 0.05 or 5%.

In this study, we used \(\alpha = 0.05\), implying we accept a 5% risk of claiming a difference exists when it actually doesn't. For a two-tailed test like ours, we check if our test statistic is in the extreme 5% (split into 2.5% in each tail of the distribution).

If our calculated test statistic falls beyond the critical value from a t-distribution table (here, 2.262 for 9 degrees of freedom), we would reject the null hypothesis.

• Our test statistic: 2.193
• Critical value: 2.262

Since our statistic is less than the critical value, we fail to reject the null hypothesis, suggesting the evidence isn't strong enough to assert a difference in student scores based on the learning method used.