91Ó°ÊÓ

In hypothesis testing, we start with an assumption or claim that we wish to test. This initial assumption is called the null hypothesis, represented by \(H_0\). The null hypothesis sets a baseline that there is no effect or no difference. In the context of our exercise, the null hypothesis states that the proportion of parents who feel that not enough math and science are being taught in high schools today is still \(52 \%\). We can write it mathematically as: \[H_0: p = 0.52\]

Here, \(p\) denotes the population proportion. The null hypothesis is critical because it forms the basis for the statistical test. The goal is to find if there's enough evidence to reject it.

While the null hypothesis represents no effect or no difference, the alternative hypothesis, represented by \(H_a\), proposes what we are looking to provide evidence for. It’s an alternative claim to the null hypothesis. In our exercise, we want to see if the proportion of parents who feel that not enough math and science are taught now is different from \(1994\). Mathematically, it is represented as: \[H_a: p \eq 0.52\]

This indicates a two-tailed test since we are checking for any difference, whether an increase or decrease, in the proportion. The alternative hypothesis is essential as it defines what is being tested against the null hypothesis.

To test our hypotheses, we need to gather some data. This data is often represented using a sample proportion, which is the proportion of the sample with a particular characteristic. In our exercise, we found that \(256\) out of \(800\) parents feel that not enough math and science are being taught. We calculate the sample proportion as follows: \[\bar{p} = \frac{256}{800} = 0.32\]

Here, \(\bar{p}\) is the sample proportion. This calculation provides an estimate of the population proportion based on the sample. The sample proportion is then used to calculate the test statistic.

The test statistic helps us determine how far our sample statistic is from the null hypothesis value, in terms of standard errors. For proportions, the test statistic follows a standard normal distribution. It is given by the formula: \[z = \frac{\bar{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\]

In this formula, \(\bar{p}\) is the sample proportion, \(p_0\) is the population proportion under the null hypothesis, and \(n\) is the sample size. Substituting our values, we get: \[z = \frac{0.32 - 0.52}{\sqrt{\frac{0.52 \times (1 - 0.52)}{800}}} \approx -11.098\]

The test statistic tells us how many standard deviations the sample proportion is from the null hypothesis proportion. A larger absolute value of the test statistic indicates stronger evidence against the null hypothesis.

The critical value is a threshold that determines the cut-off for rejecting the null hypothesis. For a significance level \(\alpha = 0.05\), it defines the boundary in the probability distribution beyond which the null hypothesis is rejected. In a two-tailed test, the critical values are \(\pm 1.96\).

We compare our test statistic to these critical values. If the test statistic is less than \(-1.96\) or greater than \(1.96\), we reject the null hypothesis. In our exercise:

• Test Statistic: \(-11.098\)
• Critical Value: \(-1.96 \text{ and } 1.96\)

Since \(-11.098\) is less than \(-1.96\), we reject the null hypothesis. This conclusion signifies that there is significant evidence that the proportion of parents who feel a lack of math and science taught in high school has changed from \(1994\).