91Ó°ÊÓ

In hypothesis testing, the null hypothesis (\textbf{H\textsubscript{0}}) is the statement that there is no effect or no difference, and it reflects the assumption that any kind of discrepancy or variation from the observed data is due to random chance. In the given exercise, the null hypothesis is formulated to reflect that the proportions of the two samples are equal or that the first sample proportion is greater than or equal to the second sample proportion. Mathematically, it is expressed as:

\textbf{H\textsubscript{0}}: \(\textstyle p_{1} \ge p_{2}\)

This hypothesis serves as a starting point for statistical testing and will only be rejected if there is strong evidence against it.

The alternative hypothesis (\textbf{H\textsubscript{a}}) contradicts the null hypothesis and represents what we seek to prove. It states that there is a significant effect or a difference between the groups being compared. In the context of the given exercise, the alternative hypothesis suggests that the proportion for the first sample is less than the proportion for the second sample. This is stated as:

\textbf{H\textsubscript{a}}: \(\textstyle p_{1} < p_{2}\)

The alternative hypothesis is usually what the researcher aims to support. If evidence significantly favors this hypothesis, it may lead to the rejection of the null hypothesis.

The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to determine whether to reject the null hypothesis. For the given exercise, the test statistic for comparing two sample proportions is computed using the formula:

\[\textstyle z = \frac{ \hat{p}_{1} - \hat{p}_{2}}{\sqrt{ \hat{p}(1 - \hat{p})( \frac{1}{n_{1}} + \frac{1}{n_{2}})} } \]

Where \(\textstyle \hat{p}_{1} \)\textstyle and \(\textstyle \hat{p}_{2} \)\textstyle are the sample proportions, and \(\textstyle \hat{p} \)\textstyle is the combined proportion. For this specific problem:

We first find \(\textstyle \hat{p}_{1} = \frac{109}{475} = 0.229 \) and \(\textstyle \hat{p}_{2} = \frac{78}{325} = 0.240 \).

Then, calculate the combined proportion \(\textstyle \hat{p} = \frac{187}{800} = 0.23375 \).

Plugging these values into the test statistic formula, we get:

\[\textstyle z = \frac{ 0.229 - 0.240 }{\sqrt{ 0.23375 (1 - 0.23375) ( \frac{1}{475} + \frac{1}{325} ) } } \approx -0.285. \]

The critical value is a threshold that determines whether the test statistic sufficiently rejects the null hypothesis. For a given level of significance \(\textstyle \alpha \)\textstyle, the critical value helps to decide the cut-off point. In a one-tailed test (like this exercise, where \(\textstyle \alpha = 0.05\)\textstyle) the critical value needs to be fetched from the standard normal distribution (Z-Table) and it is found to be:

For \(\textstyle \alpha = 0.05\)\textstyle, the critical value \(\textstyle z_c = -1.645. \)

If our computed test statistic (\(\textstyle z \)) is less than or equal to this critical value, we would reject the null hypothesis. Otherwise, we fail to reject it. This critical value method helps us determine the region of rejection for the null hypothesis.

The P-value is the probability that the observed data (or something more extreme) would occur if the null hypothesis were true. It provides a measure of the evidence against the null hypothesis. The smaller the P-value, the stronger the evidence against \(\textstyle \mathbf{H\textsubscript{0}}. \)

For the given exercise, we computed the test statistic \(\textstyle z \)\textstyle as approximately \(\textstyle -0.285 \)\textstyle. We look this value up in the standard normal distribution table to find the corresponding P-value:

The P-value corresponding to \(\textstyle z \textapprox -0.285\)\textstyle is about 0.388.

In hypothesis testing, if the P-value is less than the significance level \(\textstyle \alpha \)\textstyle (0.05 in this case), we reject the null hypothesis. Since 0.388 is greater than 0.05, we fail to reject \(\textstyle \mathbf{H\textsubscript{0}} \)\textstyle. This means there is not enough evidence to support that the proportion of the first sample is less than that of the second sample.