/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 "Smartest People Often Dumbest A... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

"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 degree- \(43 \%\) versus \(25 \%\)." 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
The short answer to the exercise will depend on the calculated Z-value. If the Z-value is greater than 1.645, the answer would be 'Yes, there is convincing evidence at a 0.05 significance level that the proportion experiencing a sunburn is higher for college graduates than it is for those without a high school degree'. If the Z-value is less than or equal to 1.645, the answer would be 'No, there is not convincing evidence at a 0.05 significance level that the proportion experiencing a sunburn is higher for college graduates than it is for those without a high school degree'.

Step by step solution

01

- Setup the Hypotheses

The null hypothesis (H0) indicates no difference between the two population proportions while the alternative hypothesis (H1) indicates a higher proportion among the college graduates. Therefore: \n\nH0: p1 - p2 = 0 (where p1 is the proportion of college graduates who experience sunburn and p2 is the proportion of non-high school graduates who experience sunburn) \n\nH1: p1 - p2 > 0 (indicating a higher proportion of sunburn among college graduates)
02

- Calculate the Sample Proportions

Calculate the sample proportions using the percentages provided: \n\np1 = \(43 \% = 0.43\) (college graduates) \np2 = \(25 \% = 0.25\) (non-high school graduates) \nAlso, it is told that each group has a size of 200.
03

- Calculate the Test Statistic

The test statistic (Z) can be calculated using the formula: \n\nZ = \(\frac {(p1 - p2) - 0} {\sqrt {\frac {(p1(1-p1))} {n1} + \frac {(p2(1-p2))} {n2}}} \), where n1 and n2 are the sample sizes. Plugging in the numbers from the given data we get the Z value.
04

- Determine the Critical Value and Decision Rule

For a level of significance (α) of 0.05 and a one-tailed test, the critical value from the Z table is approximately 1.645. If the resulting test statistic is greater than 1.645, the null hypothesis will be rejected in favor of the alternative hypothesis, indicating a significant difference in sunburn rates between the two groups.
05

- Make a Decision and Interpret the Result

After calculating Z, compare it to the critical value. If Z > 1.645, reject the null hypothesis and conclude that there is evidence to suggest that the proportion of college graduates experiencing sunburn is higher than the proportion of non-high school graduates experiencing sunburn. If Z ≤ 1.645, fail to reject the null hypothesis and conclude that there is no significant difference in sunburn rates among the two populations.

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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 is a foundational concept in hypothesis testing. It serves as the default or starting assumption, proposing that there is no effect or no difference between two compared groups. In the context of our sunburn study, the null hypothesis asserts that the incidence of sunburn is the same for both college graduates and non-high school graduates.

Formally, the null hypothesis is denoted as H0: p1 - p2 = 0, where p1 and p2 represent the population proportions of the two groups in question. If our test concludes that the data significantly deviates from this hypothesis, we may have grounds to reject it. Otherwise, we lack sufficient evidence and therefore retain it, maintaining that there's no significant difference.
Alternative Hypothesis
Conversely, the alternative hypothesis is the opposite of the null hypothesis, indicating that there is an effect or a difference between the groups being compared. In our sunburn example, the alternative hypothesis posits that college graduates have a higher incidence of sunburn compared to those without a high school degree.

The alternative hypothesis can be expressed as H1: p1 - p2 > 0, suggesting that the proportion p1 is greater than p2. It is what researchers are typically trying to gather evidence for, as finding support for the alternative hypothesis can indicate a need for a change in belief or policy.
Test Statistic
The test statistic is a calculated value that we compare to a critical value in order to make a decision about the hypotheses. It quantifies the difference between the observed sample result and what we would expect under the null hypothesis, standardized by the variability in the data.

For the sunburn study, the test statistic is a Z-score given by Z = \(\frac{(p1 - p2) - 0}{\sqrt{\frac{p1(1-p1)}{n1} + \frac{p2(1-p2)}{n2}}}\) where p1 and p2 are the sample proportions and n1 and n2 are the sizes of the respective samples. A larger absolute value of the test statistic indicates a larger difference between the groups and provides stronger evidence against the null hypothesis.
Level of Significance
The level of significance, often denoted by α (alpha), is the probability threshold at which we reject the null hypothesis. It reflects the risk we are willing to take of incorrectly rejecting the null hypothesis when it is actually true, an error known as a Type I error.

In our sunburn study, a .05 significance level means there’s a 5% chance we would reject the null hypothesis due to random sampling error when there is actually no difference in the population. The critical value associated with our chosen level of significance helps us to make the decision: if our test statistic exceeds this value, we reject the null hypothesis.
Sample Proportions
Sample proportions are estimates of the population proportions that we derive from our data. They provide insight into the characteristics of the population from which the sample was drawn. In our exercise, the calculated sample proportions are p1 = 43% and p2 = 25% for college graduates and non-high school graduates, respectively.

These sample proportions are essential in assessing the variability of the outcomes and are used to compute the test statistic. It's crucial to use a random sample, as it allows the sample proportions to be more representative of the population, More accurate sample proportions mean a more reliable test statistic and hence, a more trustworthy hypothesis test.

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Most popular questions from this chapter

The director of the Kaiser Family Foundation's Program for the Study of Entertainment Media and Health said, "It's not just teenagers who are wired up and tuned in, its babies in diapers as well." A study by Kaiser Foundation provided one of the first looks at media use among the very youngest children \(-\) those from 6 months to 6 years of age (Kaiser Family Foundation, \(2003,\) www .kff.org). Because previous research indicated that children who have a TV in their bedroom spend less time reading than other children, the authors of the Foundation study were interested in learning about the proportion of kids who have a TV in their bedroom. They collected data from two independent random samples of parents. One sample consisted of parents of children age 6 months to 3 years old. The second sample consisted of parents of children age 3 to 6 years old. They found that the proportion of children who had a TV in their bedroom was . 30 for the sample of children age 6 months to 3 years and .43 for the sample of children age 3 to 6 years old. Suppose that the two sample sizes were each 100 . a. Construct and interpret a \(95 \%\) confidence interval for the proportion of children age 6 months to 3 years who have a TV in their bedroom. Hint: This is a one-sample confidence interval. b. Construct and interpret a \(95 \%\) confidence interval for the proportion of children age 3 to 6 years who have a TV in their bedroom. c. Do the confidence intervals from Parts (a) and (b) overlap? What does this suggest about the two population proportions? d. Construct and interpret a \(95 \%\) confidence interval for the difference in the proportion that have TVs in the bedroom for children age 6 months to 3 years and for children age 3 to 6 years. e. Is the interval in Part (d) consistent with your answer in Part (c)? Explain.

A researcher at the Medical College of Virginia conducted a study of 60 randomly selected male soccer players and concluded that frequently "heading" the ball in soccer lowers players' IQs (USA Today, August 14 1995). The soccer players were divided into two groups, based on whether they averaged 10 or more headers per game. Mean IQs were reported in the article, but the sample sizes and standard deviations were not given. Suppose that these values were as given in the accompanying table. $$ \begin{array}{l|ccc} & & \text { Sample } & \text { Sample } \\ & n & \text { Mean } & \text { sd } \\ \hline \text { Fewer Than 1O Headers } & 35 & 112 & 10 \\ 10 \text { or More Headers } & 25 & 103 & 8 \\ \hline \end{array} $$ Do these data support the researcher's conclusion? Test the relevant hypotheses using \(\alpha=.05 .\) Can you conclude that heading the ball causes lower \(\mathrm{IQ}\) ?

A hotel chain is interested in evaluating reservation processes. Guests can reserve a room by using either a telephone system or an online system that is accessed through the hotel's web site. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. Based on these data, is it reasonable to conclude that the proportion who are satisfied is higher for those who reserve a room online? Test the appropriate hypotheses using \(\alpha=.05\)

Each person in a random sample of 228 male teenagers and a random sample of 306 female teenagers was asked how many hours he or she spent online in a typical week (Ipsos, January 25, 2006). The sample mean and standard deviation were 15.1 hours and 11.4 hours for males and 14.1 and 11.8 for females. a. The standard deviation for each of the samples is large, indicating a lot of variability in the responses to the question. Explain why it is not reasonable to think that the distribution of responses would be approximately normal for either the population of male teenagers or the population of female teenagers. Hint: The number of hours spent online in a typical week cannot be negative. b. Given your response to Part (a), would it be appropriate to use the two- sample \(t\) test to test the null hypothesis that there is no difference in the mean number of hours spent online in a typical week for male teenagers and female teenagers? Explain why or why not. c. If appropriate, carry out a test to determine if there is convincing evidence that the mean number of hours spent online in a typical week is greater for male teenagers than for female teenagers. Use a .05 significance level.

Do certain behaviors result in a severe drain on energy resources because a great deal of energy is expended in comparison to energy intake? The article "The Energetic Cost of Courtship and Aggression in a Plethodontid Salamander" (Ecology [1983]: 979-983) reported on one of the few studies concerned with behavior and energy expenditure. The accompanying table gives oxygen consumption \((\mathrm{mL} / \mathrm{g} / \mathrm{hr})\) for male-female salamander pairs. (The determination of consumption values is rather complicated. It is partly for this reason that so few studies of this type have been carried out.) $$ \begin{array}{lccc} \text { Behavior } & \begin{array}{c} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{c} \text { Sample } \\ \text { Mean } \end{array} & \begin{array}{c} \text { Sample } \\ \text { sd } \end{array} \\ \hline \text { Noncourting } & 11 & .072 & .0066 \\ \text { Courting } & 15 & .099 & .0071 \\ \hline \end{array} $$ a. The pooled \(t\) test is a test procedure for testing \(H_{0}: \mu_{1}-\mu_{2}=\) hypothesized value when it is reasonable to assume that the two population distributions are normal with equal standard deviations \(\left(\sigma_{1}=\right.\) \(\sigma_{2}\) ). The test statistic for the pooled \(t\) test is obtained by replacing both \(s_{1}\) and \(s_{2}\) in the two-sample \(t\) test statistic with \(s_{p}\) where \(s_{p}=\sqrt{\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}}\) When the population distributions are normal with equal standard deviations and \(H_{0}\) is true, the resulting pooled \(t\) statistic has a \(t\) distribution with \(\mathrm{df}=n_{1}+\) \(n_{2}-2 .\) For the reported data, the two sample standard deviations are similar. Use the pooled \(t\) test with \(\alpha=.05\) to determine whether the mean oxygen consumption for courting pairs is higher than the mean oxygen consumption for noncourting pairs. b. Would the conclusion in Part (a) have been different if the two-sample \(t\) test had been used rather than the pooled \(t\) test?

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