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Generation Next Born between 1980 and 1990, Generation Next was the topic of Exercise \(8.64 .^{17}\) In a survey of 500 female and 500 male students in Generation Next, 345 of the females and 365 of the males reported that they decided to attend college in order to make more money. a. Is there a significant difference in the population proportions of female and male students who decided to attend college in order to make more money? Use \(\alpha=.01\) b. Can you think of any reason why a statistically significant difference in these population proportions might be of practical importance? To whom might this difference be important?

Short Answer

Expert verified
Answer: No, there is not enough evidence to conclude that there is a significant difference in the population proportions of female and male students who decided to attend college in order to make more money. Explanation: Using a hypothesis test for the difference between two proportions, the p-value (0.221) was found to be greater than the alpha level (0.01). As a result, we fail to reject the null hypothesis, which states that there is no significant difference in the population proportions between female and male students.

Step by step solution

01

State the null and alternative hypotheses

Let \(p_1\) be the proportion of female students who decided to attend college to make more money, and \(p_2\) be the proportion of male students. The null hypothesis \(H_0\) and alternative hypothesis \(H_1\) are: \(H_0: p_1 = p_2\) \(H_1: p_1 \neq p_2\)
02

Calculate the pooled proportion

We will calculate the pooled proportion, \(p\) as follows: \(p = \frac{(number\ of\ successes\ in\ group\ 1) + (number\ of\ successes\ in\ group\ 2)}{(total\ sample\ size\ in\ group\ 1) + (total\ sample\ size\ in\ group\ 2)}\) \(p = \frac{345+365}{500+500}\) \(p = \frac{710}{1000}\) \(p = 0.71\)
03

Determine the test statistic

Now, we will calculate the difference in proportions and the test statistic, \(z\). The difference in proportions, \(\widehat{p}_1 - \widehat{p}_2\), is: \(\widehat{p}_1 - \widehat{p}_2 = \frac{345}{500} - \frac{365}{500} = -0.04\) Next, let's find the standard error, \(SE\): \(SE = \sqrt{\frac{p(1-p)}{n_1} + \frac{p(1-p)}{n_2}}\) \(SE = \sqrt{\frac{0.71(1-0.71)}{500} + \frac{0.71(1-0.71)}{500}}\) \(SE \approx 0.032669\) The test statistic, \(z\), is given by: \(z = \frac{(\widehat{p}_1 - \widehat{p}_2)}{SE}\) \(z = \frac{-0.04}{0.032669} \approx -1.224\)
04

Determine the critical value and p-value

The critical value for a two-tailed test at α = 0.01 is \(z_{0.005} = 2.576\). Since the test statistic falls within the -2.576 and 2.576 range, we fail to reject the null hypothesis. We can also calculate the p-value for the z-score \(-1.224\). The p-value is approximately 0.221.
05

Make a decision based on the p-value and α

Since the p-value (0.221) is greater than α (0.01), we fail to reject the null hypothesis. Therefore, based on the hypothesis test, there is not enough evidence to conclude that there is a significant difference in the population proportions of female and male students who decided to attend college in order to make more money. For part b of the question, if there were a statistically significant difference in these population proportions, it may suggest that one gender value education more for the financial benefits. This information could be important for educational institutions and policymakers when designing marketing materials, financial aid packages, or outreach programs targeting different genders.

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Key Concepts

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

Gender Proportion Comparison
When comparing proportions between two different groups, we're looking at whether there is a meaningful difference in the percentage or fraction of some characteristic.
In this context, the comparison involves determining if a higher or lower proportion of female students than male students decided to attend college for financial reasons.
Hypothesis tests often frame these comparisons by setting up null and alternative hypotheses.
  • The null hypothesis ( \(H_0\) ) typically states there is no difference ( \(p_1 = p_2\) ).
  • The alternative hypothesis ( \(H_1\) ) proposes there is a difference ( \(p_1 eq p_2\) ).
Breaking down these definitions provides a foundation to better understand subsequent steps, such as significance levels and test statistics.
Significance Level
The significance level, represented by the Greek letter \(\alpha\) , helps determine the threshold for 'significance' in hypothesis testing.
It indicates the probability of rejecting the null hypothesis, if it is indeed true, and thus, helps in managing false positives.
Typically chosen values are 0.01, 0.05, or 0.10.
In our exercise, \(\alpha = 0.01\) , which is quite strict.
This means there's only a 1% risk of wrongly stating there's a significant difference in gender-based motives for attending college due to wanting to make more money, assuming there's no actual difference.
  • A smaller \(\alpha\) means better control of Type I errors (false positives).
  • However, setting \(\alpha\) too low could increase the risk of Type II errors (false negatives), where one might miss detecting a true difference.
Choosing the right significance level requires balancing these probabilities based on the context and implications of the study.
Z-Test
The z-test is a statistical method used to determine whether there is a significant difference between sample data points.
It is especially suitable for comparing proportions, such as in our exercise where female and male motivations for attending college are evaluated.
The steps involve calculating the test statistic, which is derived from the difference in sample proportions. Test Statistic (z) = \frac{(\widehat{p}_1 - \widehat{p}_2)}{SE}
  • \(\widehat{p}_1\) and \(\widehat{p}_2\) are the sample proportions.
  • \(SE\) denotes the standard error, representing the variability of the sample statistic.
In this exercise, the calculated z-statistic of approximately -1.224 indicates how many standard errors the sample proportion difference is away from zero.
The closer the z-score is to zero, the less likely the samples differ significantly. By comparing this value to a critical value (-2.576 to 2.576 for \(\alpha = 0.01\)) or using a p-value, one can determine if the difference is statistically meaningful.
P-Value
The p-value is a numerical indication of evidence against the null hypothesis.
It tells us how likely it is that the observed data would occur, assuming the null hypothesis is true. A smaller p-value signals that the observed result is not just due to random chance.
For our exercise, the computed p-value was about 0.221, meaning there's a 22.1% probability the observed gender difference is due to random variation.
  • If \(p \le \alpha\) (here, less than 0.01), we reject the null hypothesis.
  • If \(p > \alpha\) , as in our case, we cannot reject it.
Thus, this p-value means we lack sufficient evidence to conclude a significant gender difference in motivations.
It's crucial for researchers to use p-values thoughtfully, neither overstating nor downplaying observed effects in real-world contexts.

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Most popular questions from this chapter

Independent random samples of \(n_{1}=140\) and \(n_{2}=140\) observations were randomly selected from binomial populations 1 and \(2,\) respectively. Sample 1 had 74 successes, and sample 2 had 81 successes. a. Suppose you have no preconceived idea as to whic parameter, \(p_{1}\) or \(p_{2}\), is the larger, but you want onl to detect a difference between the two parameters one exists. What should you choose as the alternative hypothesis for a statistical test? The null hypothesis? b. Calculate the standard error of the difference in the two sample proportions, \(\left(\hat{p}_{1}-\hat{p}_{2}\right) .\) Make sure to use the pooled estimate for the common value of \(p\) c. Calculate the test statistic that you would use for the test in part a. Based on your knowledge of the standard normal distribution, is this a likely or unlikely observation, assuming that \(H_{0}\) is true and the two population proportions are the same? d. \(p\) -value approach: Find the \(p\) -value for the test. Test for a significant difference in the population proportions at the \(1 \%\) significance level. e. Critical value approach: Find the rejection region when \(\alpha=.01 .\) Do the data provide sufficient evidence to indicate a difference in the population proportions?

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