91Ó°ÊÓ

In determining the "goodness" of a test question, a teacher will often compare the percentage of better students who answer it correctly with the percentage of poorer students who answer it correctly. One expects that the proportion of better students who will answer the question correctly is greater than the proportion of poorer students who will answer it correctly. On the last test, 35 of the students with the top 60 grades and 27 of the students with the bottom 60 grades answered a certain question correctly. Did the students with the top grades do significantly better on this question? Use \(\alpha=0.05.\) a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Step by step solution

01

Identify the parameters

Identify the sample sizes and number of successes for each group. For the 'better' students, \(n_1=60\) and \(x_1=35.\) For the 'poorer' students, \(n_2=60\) and \(x_2=27.\)

02

Compute sample proportions

Compute the sample proportions for each group: \(\hat{p}_1 = x_1 / n_1\) and \(\hat{p}_2 = x_2 / n_2\).

03

Define the null and alternative hypotheses

For this two-tailed test, the null hypothesis \(H_0\) assumes that the population proportions for better students ( \(p_1\) ) and poorer students ( \(p_2\) ) are equal, i.e., \(p_1 = p_2\). The alternative hypothesis \(H_a\) assumes that the population proportions are not equal, i.e., \(p_1 \neq p_2\).

04

Compute test statistic using pooled proportion

Compute the pooled sample proportion ( \(\hat{p}\) ): \(\hat{p} = (x_1 + x_2) / (n_1 + n_2)\)Then, calculate the z-score:\(z = (\hat{p}_1 - \hat{p}_2) / \sqrt{\hat{p}(1 - \hat{p})(1/n_1 + 1/n_2)}\).

05

Step 5a: Compute the \(\p\) -value

First find the z-score's corresponding probability from the standard normal distribution table (let's denote it by \(P(z)\)). This gives the probability that the difference in proportions is less than what was observed. However, as this is a two-tailed test, the \(p\) -value is \(2 * (1 - P(z))\). Compare the \(p\) -value with \(\alpha = 0.05\). If \(p\) -value \( < \alpha\), reject \(H_0\).

06

Step 5b: Classical approach

Alternatively, determine the critical values for the 0.05 level of significance for a two-tailed test, using the standard normal distribution. If the calculated z-score falls in the rejection region (i.e., it is more extreme than the critical values), reject the null hypothesis \(H_0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Significant Difference

When we talk about a significant difference in hypothesis testing, we're looking to find out if one group's results are distinctively different from another group's. In this exercise, the teacher wants to know if the top students performed significantly better than the poorer students on a test question.

We use statistical tools to measure if a difference is notable or if it might have happened just by chance. If we say the difference is significant, it means it’s unlikely to have occurred randomly.

Determining significance in this scenario involves comparing the performance of two groups of students to see if the observed differences could likely arise by random variations alone.

• A significant difference suggests a true disparity in group performance.
• This is assessed by comparing calculated statistical values to predetermined thresholds.

Two-Tailed Test

A two-tailed test is used when we want to find out if there is any difference at all between two groups, without specifying the direction.

In our example, the null hypothesis is that both groups of students performed equally well, meaning any difference could go in any direction—it could be either better or worse for either group.

Why use a two-tailed test? It doesn’t assume that one group's performance could only be better; it checks for any sort of difference.

• It considers the possibility of differences in both directions.
• This is more conservative and informative as it ensures we check all possible outcomes.

P-Value

The p-value helps us determine the strength of our results in hypothesis testing. It essentially tells us how likely it is to observe our results, or more extreme, purely by random chance.

In our test, after computing the z-score, we check its corresponding probability, denoted as the p-value.

If this value is less than 0.05, it suggests that the difference we observed is unlikely due to chance alone, and thus, it is statistically significant. This helps us decide whether to reject our null hypothesis.

• P-value is a crucial part of deciding the outcome of the test.
• A smaller p-value indicates stronger evidence against the null hypothesis.

Classical Approach

The classical approach, sometimes called the critical value approach, involves comparing our test statistic (e.g., z-score) to a critical value from a statistical distribution.

In our context, we find critical values for a two-tailed test at a 0.05 significance level.

If our calculated z-score is greater than these critical values in either direction (positive or negative), we reject the null hypothesis. This method provides a visual and intuitive decision criterion.

• It focuses on critical boundaries of acceptance or rejection for hypotheses.
• Helps visualize the extremity of test statistics in relation to assumed true values.