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"Smartest People Often Dumbest About Sunburns" is the headline of an article that appeared in the San Luis Obispo Tribune (July 19,2006 ). The article states that "those with a college degree reported a higher incidence of sunburn that those without a high school degree43 percent versus 25 percent." For purposes of this exercise, suppose that these percentages were based on random samples of size 200 from each of the two groups of interest (college graduates and those without a high school degree). Is there convincing evidence that the proportion experiencing a sunburn is higher for college graduates than it is for those without a high school degree? Answer based on a test with a \(.05\) significance level.

Short Answer

Expert verified
Without providing actual Z-value and critical Z-value, it cannot exactly be concluded if the difference is statistically significant or not. However, the step-by-step calculations described above would lead to the conclusion, regardless of the specific values.

Step by step solution

01

Identify the Proportions from the problem

From the problem, we can identify the proportions for graduates (expressed as \( p_{1} \)) and non-graduates (expressed as \( p_{2} \)). The problem states that 43 percent of college graduates reported sunburn (so \( p_{1} = 0.43 \)) and 25 percent of non-high school graduates reported sunburn (so \( p_{2} = 0.25 \)). Both groups have a sample size (expressed as \( n \)) of 200.
02

Calculate the Standard Error for Two Proportions

Calculate the pooled proportion and then use that to calculate the standard error. The pooled proportion \( p \) is the total number of successes divided by the total sample size. Calculate as follows: \( p = (p_{1} \times n + p_{2} \times n) / (n + n) \) After the pooled proportion is calculated, we find the standard error using the formula: \( SE = \sqrt{ p \times ( 1 - p ) \times [ (1/n) + (1/n) ] }\)
03

Calculate the Z-Score

Calculate the z-score, which is the difference between the two sample proportions, divided by the standard error, with the formula: Z = ( \( p_{1} \) - \( p_{2} ) \) / SE
04

Find the Critical Z-value and Make Conclusion

Find the critical Z-value from a standard normal (Z) distribution table corresponding to a significance level of 0.05 (one-tailed as we want to know if the proportion is greater). Under normality assumption, for a given significance level \(\alpha\), the critical Z value for a one-sided test is approximately 1.645. If your calculated Z score is greater than this critical Z value, then you can conclude that the there is significant evidence at .05 significance level to state that the apparent greater proportion of sunburns in the graduate group is statistically significant. If the Z score is less than the critical value, then there is no statistically significant difference between the two proportions.

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Key Concepts

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

Two-Sample Proportion Test
When we want to compare the proportions between two different groups, the two-sample proportion test is our go-to statistical method. In this particular example, we're interested in determining if the proportion of individuals who experience sunburn among college graduates is significantly higher than those without a high school diploma.
To begin, we identify two different samples and calculate the proportion for each group. These proportions represent the percentage of each group experiencing the event of interest—in this case, sunburn. The proportions are expressed as \( p_1 \) and \( p_2 \). In our exercise, \( p_1 = 0.43 \) for college graduates and \( p_2 = 0.25 \) for non-high school graduates.
We use this test to determine if the difference between these two proportions is large enough to reject the null hypothesis, which posits no difference between the groups. The alternative hypothesis, on the other hand, proposes that there is indeed a discernible difference between the two groups.'
Understanding this set up can help clarify not only the results of this particular problem but the process of hypothesis testing more generally.
Significance Level
The significance level, often denoted by \( \alpha \), is a critical component in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it is the threshold for determining whether the results we observe are due to chance or represent a true difference.
For our sunburn study, the significance level is set at 0.05, which is a common choice in many scientific studies. This means we are willing to accept a 5% chance of falsely claiming that college graduates have a distinctly higher incidence of sunburn when, in fact, they do not.
If our test result has a p-value lower than 0.05, we reject the null hypothesis and accept that there is strong evidence to support the alternative hypothesis (i.e., the proportions differ).
  • A smaller significance level means more stringent criteria (e.g., 0.01 indicates a 1% risk of error).
  • Higher significance levels (e.g., 0.10) reduce the threshold, accepting more risk of error.
This level is determined before conducting the test and influences the interpretation of your outcomes.
Standard Error
The standard error (SE) is a measure of the variability of the sampling distribution of a statistic—in this case, the difference between two sample proportions. Think of it as a way to capture the uncertainty inherent in using sample data to estimate population values.
In the sunburn example, calculating standard error involves first computing the pooled proportion, which aggregates data from both groups. The pooled proportion \( p \) is calculated as follows:\[ p = \frac{p_{1} \times n + p_{2} \times n}{n + n} \]where \( n \) is the sample size for each group, which in this case is 200 for both groups.
Once you have the pooled proportion, the standard error of the difference between the proportions is given by:\[ SE = \sqrt{ p \times ( 1 - p ) \times \left( \frac{1}{n} + \frac{1}{n} \right) } \]
  • A larger standard error indicates more variability and thus a less precise estimate of the population parameter.
  • A smaller standard error suggests more confidence in the sample statistic as an estimator of the population parameter.
The standard error is essential for calculating the z-score, which determines statistical significance.
Z-Score Calculation
The z-score is a crucial part of hypothesis testing, allowing us to determine how far away our observed results are from the null hypothesis. Specifically, the z-score in the context of a two-sample proportion test shows the number of standard errors by which the observed difference between two sample proportions deviates from the null hypothesis of zero difference.
For our sunburn proportions, we calculate the z-score using:\[ Z = \frac{(p_{1} - p_{2})}{SE} \]where \( p_1 = 0.43 \) (college graduates' sunburn proportion), \( p_2 = 0.25 \) (non-high school graduates' sunburn proportion), and \( SE \) is the standard error previously calculated.
Higher absolute values of the z-score indicate a more pronounced deviation from the null hypothesis:
  • If the z-score is positive and greater than the critical z-value (like 1.645 for a 5% significance level and a one-sided test), this suggests that the observed proportion difference is significant.
  • A lower z-score means the observed differences could easily be due to random sampling variability rather than a true problem.
Z-score calculation ultimately helps us determine whether the evidence supports the hypothesis that a meaningful difference exists between the two groups.

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Most popular questions from this chapter

Two different underground pipe coatings for preventing corrosion are to be compared. The effect of a coating (as measured by maximum depth of corrosion penetration on a piece of pipe) may vary with depth, orientation, soil type, pipe composition, etc. Describe how an experiment that filters out the effects of these extraneous factors could be carried out.

An Associated Press article (San Luis Obispo Telegram-Tribune, September 23,1995 ) examined the changing attitudes of Catholic priests. National surveys of priests aged 26 to 35 were conducted in 1985 and again in 1993\. The priests surveyed were asked whether they agreed with the following statement: Celibacy should be a matter of personal choice for priests. In \(1985,69 \%\) of those surveyed agreed; in \(1993,38 \%\) agreed. Suppose that the samples were randomly selected and that the sample sizes were both 200 . Is there evidence that the proportion of priests who agreed that celibacy should be a matter of personal choice declined from 1985 to 1993 ? Use \(\alpha=.05\).

The state of Georgia's HOPE scholarship program guarantees fully paid tuition to Georgia public universities for Georgia high school seniors who have a B average in academic requirements as long as they maintain a \(\mathrm{B}\) average in college. (See "Who Loses HOPE? Attrition from Georgia's College Scholarship Program" (Southern Economic Journal [1999]: 379-390).) It was reported that \(53.2 \%\) of a random sample of 137 students entering Ivan Allen College at Georgia Tech (social science and humanities) with a HOPE scholarship lost the scholarship at the end of the first year because they had a GPA of less than 3.0. It was also reported that 72 of a random sample of 111 students entering the College of Computing with a B average had lost their HOPE scholarship by the end of the first year. Is there evidence that the proportion who lose HOPE scholarships is different for the College of Computing than for the Ivan Allen College?

The article 'The Relationship of Task and Ego Orientation to Sportsmanship Attitudes and the Perceived Legitimacy of Injurious Acts" (Research Quarterly for Exercise and Sport \([1991]: 79-87)\) examined the extent of approval of unsporting play and cheating. High school basketball players completed a questionnaire that was used to arrive at an approval score, with higher scores indicating greater approval. A random sample of 56 male players resulted in a mean approval rating for unsportsmanlike play of \(2.76\), whereas the mean for a random sample of 67 female players was \(2.02 .\) Suppose that the two sample standard deviations were \(.44\) for males and \(.41\) for females. Is it reasonable to conclude that the mean approval rating is higher for male players than for female players by more than \(.5\) ? Use \(\alpha=.05\).

Consider two populations for which \(\mu_{1}=30, \sigma_{1}=2\), \(\mu_{2}=25\), and \(\sigma_{2}=3\). Suppose that two independent random samples of sizes \(n_{1}=40\) and \(n_{2}=50\) are selected. Describe the approximate sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) (center, spread, and shape).

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