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Chapter 9: Problem 80

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?

Short Answer

Expert verified
The evidence does not support rejecting the instructor's claim of 60% enrichment.

Step by step solution

01

Define Null and Alternative Hypotheses

First, we need to set up our hypotheses. The null hypothesis \(H_0\) is that the true proportion of students who feel enriched is 0.60, or 60%. The alternative hypothesis \(H_a\) is that the true proportion is not 0.60, indicating a two-tailed test.
02

Collect Sample Data

We have surveyed 64 students and found that 34 feel enriched. This gives us a sample proportion \(\hat{p} = \frac{34}{64} = 0.53125\).
03

Calculate Standard Error

The standard error (SE) for the proportion is calculated using the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\), where \(p = 0.60\) and \(n = 64\). So, \(SE = \sqrt{\frac{0.60 \times 0.40}{64}} = 0.0612\).
04

Calculate the Test Statistic

The test statistic is calculated using \(z = \frac{\hat{p} - p}{SE}\), where \(\hat{p} = 0.53125\), \(p = 0.60\), and \(SE = 0.0612\). So, \(z = \frac{0.53125 - 0.60}{0.0612} = -1.123\).
05

Determine Critical Value or P-value

Since this is a two-tailed test, we typically use a significance level of \(\alpha = 0.05\). For \(z\)-tests, the critical value for a two-tailed test at this significance level is \(z_{\alpha/2} = 1.96\). Alternatively, we can calculate the p-value associated with our test statistic \(-1.123\).
06

Make a Decision

Since \(-1.96 < -1.123 < 1.96\), the test statistic does not fall in the critical region. The p-value for \(-1.123\) is greater than 0.05, indicating a lack of evidence to reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often symbolized as \( H_0 \), represents a statement that there is no effect or no difference, and it is the hypothesis that researchers often try to disprove. In this scenario, the null hypothesis is that the proportion of students feeling enriched by the class is exactly 60%, \( p = 0.60 \).

The null hypothesis is foundational in hypothesis testing as it provides a starting point. We presume \( H_0 \) to be true unless the data presents substantial evidence otherwise. The objectivity of \( H_0 \) helps keep analysis unbiased. To summarize:
  • The null hypothesis suggests no change or no effect.
  • It is assumed true until evidence suggests otherwise.
  • It sets the stage for testing using statistical methods.
Alternative Hypothesis
The alternative hypothesis, represented by \( H_a \) or \( H_1 \), contradicts the null hypothesis and suggests that there is an effect or a difference. In our example, the alternative hypothesis posits that the proportion of students who feel enriched is not 60%. This is a two-tailed alternative since it asserts any deviation (higher or lower) from 60%.

The role of the alternative hypothesis is to challenge the belief stated by the null hypothesis, giving researchers a structured avenue to test their theories:
  • It proposes a new claim (different from the null).
  • Guides the direction of the statistical test (one-tailed or two-tailed).
  • Helps to decide if variations in the data provide enough evidence to reject \( H_0 \).
Standard Error
The standard error (SE) is a measure that quantifies the variability or dispersion of sample proportions around the true population proportion. It's important in determining how much the sample mean is expected to vary from the actual population mean. In this case, we're interested in the standard error of the proportion.

The formula used here is \( SE = \sqrt{\frac{p(1-p)}{n}} \), where \( p \) is the proportion stated in the null hypothesis, and \( n \) is the sample size. For our problem, with \( p = 0.60 \) and \( n = 64 \), the SE calculates to 0.0612:
  • Determines how accurate the sample data is likely to be.
  • A smaller SE suggests more reliable estimates of the population parameter.
  • Helps to compute the test statistic, reflecting how extreme the sample result is.
Test Statistic
The test statistic is a standardized value used in hypothesis testing to decide whether to reject the null hypothesis. It puts your sample result into context of the standard distribution expected under the null hypothesis.

Calculated as \( z = \frac{\hat{p} - p}{SE} \), where \( \hat{p} \) is the sample proportion, \( p \) is the null hypothesis proportion, and \( SE \) is the standard error of the sample proportion. In our scenario, \( z = \frac{0.53125 - 0.60}{0.0612} = -1.123 \).
  • Allows us to determine how far the sample result is from the null hypothesis.
  • Aids in finding the p-value or comparing to critical values.
  • Central to making an inference about whether to reject or fail to reject the null hypothesis.

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