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A binomial distribution is a way to represent the number of successes in a fixed number of trials with two possible outcomes. These trials are often referred to as "success" and "failure." The distribution provides a probability framework for questions like, "What is the probability that we will have a certain number of successes in 60 trials?"

Some important characteristics of a binomial distribution include:

• The total number of trials is fixed. Here, we have 60 trials.
• Each trial is independent, meaning the result of one trial does not affect the others.
• The probability of success, denoted as \( p \), is constant across all trials.

In this exercise, we're considering the situation where \( p = 0.18 \) for population proportion. Hence, if we had a large number of samples, the outcomes would follow a pattern described by the binomial distribution. But why do we need this? It's because determining if a normal distribution can approximate our sample proportion requires understanding these trials and probabilities.

The population proportion, commonly denoted as \( p \), represents the fraction of a population that has a particular attribute or characteristic. In simpler words, it tells us how many individuals in the total population "succeed" in the context of the problem.

For this exercise, the focus is on whether the actual proportion of successes in our population exceeds 0.18. This proportion serves as the foundation for setting up our hypotheses:

• The Null Hypothesis \( H_0: p = 0.18 \), assumes this proportion is at most 0.18.
• The Alternative Hypothesis \( H_a: p > 0.18 \), suggests that the proportion is greater than 0.18.

Understanding the population proportion helps us define what we're seeking to prove and serves as the basis for further calculations, like the z-score and the computation of the sample proportion \( \hat{p} \).

A z-score measures the number of standard deviations a data point is from the mean. In hypothesis testing, it's essential because it translates the sample statistics into a standard normal distribution.

In our exercise, the z-score is utilized to see how far our sample proportion \( \hat{p} \) (which is 0.3) diverges from the hypothesized population proportion \( p_0 \) (which is 0.18). The formula used is:

• \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]

This calculation gives us a standardized score, allowing us to determine the likelihood of observing a sample proportion like ours by random chance if the null hypothesis was true. In our example, the calculated z-score was approximately 2.30, indicating the sample proportion was 2.30 standard deviations above the mean of this normal distribution.

The P-value is a critical concept in hypothesis testing as it measures the probability that the observed results would occur under the null hypothesis. It essentially tells us how "weird" our observation is if everything else was normal.

In this scenario, we calculated the P-value using the {z-score we found earlier.

• Since our z-score was 2.30, we used a standard normal table to find the corresponding probability \( P(Z < 2.30) \).
• This yielded a result of approximately 0.9893. To understand how likely or unlikely our observation is, we took \( 1 - 0.9893 \), resulting in a P-value of about 0.0107.

For hypothesis testing, if the P-value is less than or equal to our chosen level of significance \( \alpha \), we'd reject \( H_0 \). In this instance, since 0.0107 is slightly higher than 0.01, we fail to reject the null hypothesis, implying insufficient evidence to support the claim that our population proportion is greater than 18%.