Question

Public Agenda conducted a survey of 1379 parents and 1342 students in grades \(6-12\) regarding the importance of science and mathematics in the school curriculum (Associated Press, February \(15,2 \mathrm{OO} 6\) ). It was reported that \(50 \%\) of students thought that understanding science and having strong math skills are essential for them to succeed in life after school, whereas \(62 \%\) of the parents thought it was crucial for today's students to learn science and higher-level math. The two samples - parents and students -were selected independently of one another. Is there sufficient evidence to conclude that the proportion of parents who regard science and mathematics as crucial is different than the corresponding proportion for students in grades \(6-12 ?\) Test the relevant hypotheses using a significance level of .05 .

Short answer

Depending on the value of the z-score calculated in Step 2 and compared to the critical value in Step 3, a conclusion is drawn whether to accept or reject the null hypothesis. If the z-score is greater than ± 1.96, the null hypothesis is rejected, implying there is a significant difference between the two proportions. If not, the null hypothesis is not rejected implying there is no significant difference between the two proportions. The final conclusion can only be given after doing the calculations.

Step-by-step solution

Formulate both hypotheses

The null hypothesis (\(H_0\)) is that there is no difference between the proportions of parents and students who think that understanding math and science is important, i.e., \(p_1 = p_2\). The alternative hypothesis (\(H_1\)) is that there is a significant difference in the proportions, i.e., \(p_1 \neq p_2\). Here, \(p_1\) represents the proportion of parents and \(p_2\) represents the proportion of students.

Compute the test statistic

The test statistic z is calculated using the formula: \(z = \frac{(p_1 - p_2) - 0}{sqrt(\bar{p}(1 - \bar{p}){[\frac{1}{n_1} + \frac{1}{n_2}]})}\), where \(\bar{p} = \frac{x_1 + x_2}{n_1 + n_2} \), \(x_1\) and \(x_2\) are the number of successes (yes responses) in each sample, and \(n_1\) and \(n_2\) are the sample sizes. For our case, \(n_1 = 1379\), \(n_2 = 1342\), \(x_1 = 0.62 * 1379\), \(x_2 = 0.50 * 1342\). After calculating these values, we can substitute them into the formula and find the z-score.

Determine the critical value and make decision

The critical value for a two-tailed test at a significance level of 0.05 is approximately ± 1.96. If the calculated z-score is greater than 1.96 or less than -1.96, we reject the null hypothesis. Otherwise, we fail to reject it.

Conclusion

Depend on the calculated z-score and the result of step 3, Draw a conclusion in context of the problem, whether there is sufficient evidence to conclude that the proportion of parents who think science and maths are crucial is different from the proportion of students.

Key concepts

Null Hypothesis

The null hypothesis, symbolized as \(H_0\), is the starting point of hypothesis testing. It reflects the assumption of no effect, no difference, or no change. In the context of our survey, the null hypothesis asserts that there is no difference between the proportion of parents and students who deem mathematics and science as critical for post-school success. Mathematically, this can be expressed as \(p_1 = p_2\), where \(p_1\) is the proportion of parents and \(p_2\) is the proportion of students.
For hypothesis testing, the null hypothesis acts as the default stance that researchers aim to challenge. It is accepted unless there is strong evidence to the contrary. This assumption of "no difference" ensures that any claim of significant variation is well-supported by data.
When conducting a hypothesis test, we strive to either "reject" or "fail to reject" the null hypothesis, based on the calculated evidence from the data. A common misconception is that "failing to reject" equates to proving the null hypothesis true. Instead, it simply means the evidence was insufficient to confirm a difference.

Alternative Hypothesis

The alternative hypothesis, or \(H_1\), presents a statement opposing the null hypothesis. In our scenario, it proposes that there is a significant difference between the attitudes of parents and students regarding the importance of math and science, expressed as \(p_1 eq p_2\).
The alternative hypothesis is instrumental in framing what you are seeking to prove. It represents the possibility that there indeed exists a notable difference or effect, initiating further exploration. If the data provides enough evidence, the alternative hypothesis becomes more plausible than the null.
In hypothesis testing, a critical part of the process is to determine whether we have enough proof to support this alternative perspective. It's important to note that supporting the alternative hypothesis involves rejecting the null hypothesis, indicating a significant finding.

Test Statistic

A test statistic is a standardized value derived from sample data. It allows comparison against a probability distribution to assess the evidence against a null hypothesis. In our survey task, we utilize the z-test statistic, suitable for comparing proportions. It calculates how far the observed difference in proportions (parent's and student's perceptions of math and science importance) is from the null hypothesis of no difference, in standard deviation units.
The formula for the z-statistic is:\[ z = \frac{(p_1 - p_2)}{\sqrt{\overline{p}(1 - \overline{p}) \left[ \frac{1}{n_1} + \frac{1}{n_2} \right]}} \]where:

• \(p_1\) and \(p_2\) are the sample proportions
• \(\overline{p}\) is the pooled sample proportion
• \(n_1\) and \(n_2\) are the sample sizes

By calculating the z-score, you can gauge the strength of evidence against the null hypothesis. A larger absolute value of this statistic suggests a stronger case against it.

Significance Level

The significance level, typically denoted as \(\alpha\), establishes a threshold for decision-making in hypothesis testing. It answers the question: "How much risk of making a Type I error am I willing to tolerate?" A Type I error occurs when the null hypothesis is incorrectly rejected. Therefore, the significance level embodies the probability of such a mistake.
For example, a significance level of 0.05 (commonly used) implies a 5% risk of rejecting the null hypothesis when it is, in fact, true. This means that the results must be so compelling that there's less than a 5% chance they happened due to random variability.
In our example, setting a significance level of 0.05 helped determine whether the computed z-score fell into the "rejection zone" of ±1.96 (approximate critical values for a two-tailed test). If the test statistic surpasses these critical values, the null hypothesis is rejected, confirming that there is a notable difference between the perspectives of parents and students on the criticality of maths and science.