/*! 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 17 This problem is based on informa... [FREE SOLUTION] | 91Ó°ÊÓ

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This problem is based on information taken from The Merck Manual (a reference manual used in most medical and nursing schools). Hypertension is defined as a blood pressure reading over \(140 \mathrm{~mm}\) Hg systolic and/or over \(90 \mathrm{~mm}\) Hg diastolic. Hypertension, if not corrected, can cause long- term health problems. In the college-age population \((18-24\) years), about \(9.2 \%\) have hypertension. Suppose that a blood donor program is taking place in a college dormitory this week (final exams week). Before each student gives blood, the nurse takes a blood pressure reading. Of 196 donors, it is found that 29 have hypertension. Do these data indicate that the population proportion of students with hypertension during final exams week is higher than \(9.2 \%\) ? Use a \(5 \%\) level of significance.

Short Answer

Expert verified
The data suggest that the proportion of students with hypertension during exams is higher than 9.2%, at a 5% significance level.

Step by step solution

01

Define the Hypotheses

In this problem, we start by defining our null hypothesis \(H_0\) and alternative hypothesis \(H_1\). \(H_0\): The proportion of students with hypertension is \(p = 0.092\). \(H_1\): The proportion of students with hypertension is greater than \(0.092\). This is a right-tailed test because we are testing if the proportion is higher than \(0.092\).
02

Calculate the Test Statistic

Use the formula for the z-test for proportions: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]where \(\hat{p}\) is the sample proportion, \(p_0\) is the population proportion under the null hypothesis, and \(n\) is the sample size. Here, \(\hat{p} = \frac{29}{196}\), \(p_0 = 0.092\), and \(n = 196\). First, compute \(\hat{p} = \frac{29}{196} \approx 0.148\). Then calculate the test statistic: \[ z = \frac{0.148 - 0.092}{\sqrt{\frac{0.092(1 - 0.092)}{196}}} \approx 2.447 \]
03

Find Critical Value and Make a Decision

At a \(5\%\) significance level, determine the critical value from the z-table for a right-tailed test. The critical value z for \(\alpha = 0.05\) is approximately 1.645. Compare the calculated \(z\) statistic with the critical value. Since the calculated \(z = 2.447\) is greater than 1.645, we reject the null hypothesis.
04

Conclusion

We conclude that there is sufficient evidence to suggest that the proportion of students with hypertension during final exams week is higher than \(9.2\%\) at the \(5\%\) level of significance.

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Key Concepts

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

Proportion Hypothesis Tests
Proportion hypothesis tests are an essential part of statistics when dealing with questions about population proportions. You use these tests to make an informed guess about the population proportion based on sample data. It starts with defining two hypotheses:
  • The null hypothesis (H_0): This is a statement of no effect or no difference. In our exercise, this hypothesis is that the college students with hypertension proportion is equal to a specified value, 9.2%.
  • The alternative hypothesis (H_1): This statement counters the null hypothesis. For the exercise, we ask if the proportion is greater than 9.2%, indicating a right-tailed test.
After stating these hypotheses, the test aims to determine how likely it is that the sample data happened by random chance if the null hypothesis is true. This involves calculating a test statistic, which in our case refers to how much the sample proportion differs from the hypothesized population proportion. Then, by comparing this value to a known threshold or critical value, you decide whether to reject the null hypothesis or not.
Z-Test for Proportions
The Z-test for proportions is a statistical test used to determine whether there is a significant difference between an observed sample proportion and a known population proportion. Here, this test involves the calculation of a "z-statistic."To calculate this z-statistic, we need certain elements: the sample proportion (\hat{p}), the hypothesized population proportion (p_0), and the sample size (n). The formula is straightforward:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]
  • The sample proportion \hat{p} is calculated by dividing the number of successful instances by the total sample size; in this case, 29 students out of 196 had hypertension.
  • The target proportion, p_0, represents the null hypothesis: 9.2% or 0.092.
After plugging these values into the equation, the z-statistic calculated gives insight into how far and in which direction our sample proportion is from the null hypothesis, measured in terms of standard deviation units. This value is crucial for comparing against the critical value to draw a conclusion.
Significance Level
The significance level, often denoted as \( \alpha \), represents the probability of rejecting the null hypothesis when it is actually true. It helps define the threshold for making this critical decision. In hypothesis testing, a common level of significance is 5% (or \( \alpha = 0.05 \)).When you perform a hypothesis test like our example, you compare the calculated test statistic with a critical value corresponding to the chosen significance level. The critical value depends on the type of test (right-tailed, left-tailed, or two-tailed).
  • In a right-tailed test as in the exercise, the setup implies interest in whether the sample statistic exceeds the hypothesized population value.
  • For \( \alpha = 0.05 \), the critical value from a standard normal distribution (z-distribution) table is approximately 1.645.
The decision rule is straightforward: if the calculated z-statistic is greater than the critical value, you reject the null hypothesis. Our example showed a z-statistic of 2.447, indicating that at a 5% significance level, there was enough evidence to reject the null hypothesis and conclude that students' hypertension rates during exam week were indeed higher."

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Most popular questions from this chapter

Salmon Homser Lake, Oregon, has an Atlantic salmon catch and release program that has been very successful. The average fisherman's catch has been \(\mu=8.8\) Atlantic salmon per day. (Source: National Symposium on Catch and Release Fishing, Humboldt State University.) Suppose that a new quota system restricting the number of fishermen has been put into effect this season. A random sample of fishermen gave the following catches per day: \(\begin{array}{rrrrrrr} 12 & 6 & 11 & 12 & 5 & 0 & 2 \\ 7 & 8 & 7 & 6 & 3 & 12 & 12 \end{array}\) i. Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 7.36\) and \(s \approx 4.03\). ii. Assuming the catch per day has an approximately normal distribution, use a \(5 \%\) level of significance to test the claim that the population average catch per day is now different from \(8.8\).

If we fail to reject (i.e., "accept") the null hypothesis, does this mean that we have proved it to be true beyond all doubt? Explain your answer.

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. What is the value of the sample test statistic? (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha\) ? (e) State your conclusion in the context of the application. Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in \(\mathrm{mg} / 100 \mathrm{ml}\) ). \(\begin{array}{llllllll}93 & 88 & 82 & 105 & 99 & 110 & 84 & 89\end{array}\) The sample mean is \(\bar{x} \approx 93.8\). Let \(x\) be a random variable representing glucose readings taken from Gentle Ben. We may assume that \(x\) has a normal distribution, and we know from past experience that \(\sigma=12.5 .\) The mean glucose level for horses should be \(\mu=85 \mathrm{mg} / 100 \mathrm{ml}\) (Reference: Merck Veterinary Manual). Do these data indicate that Gentle Ben has an overall average glucose level higher than 85? Use \(\alpha=0.05\).

A random sample of \(n_{1}=288\) voters registered in the state of California showed that 141 voted in the last general election. A random sample of \(n_{2}=216\) registered voters in the state of Colorado showed that 125 voted in the most recent general election. (See reference in Problem 25.) Do these data indicate that the population proportion of voter turnout in Colorado is higher than that in California? Use a \(5 \%\) level of significance.

Is fishing better from a boat or from the shore? Pyramid Lake is located on the Paiute Indian Reservation in Nevada. Presidents, movie stars, and people who just want to catch fish go to Pyramid Lake for really large cutthroat trout. Let row \(B\) represent hours per fish caught fishing from the shore, and let row \(A\) represent hours per fish caught using a boat. The following data are paired by month from October through April. (Source: Pyramid Lake Fisheries, Paiute Reservation, Nevada.) \(\begin{array}{l|ccccccc} \hline & \text { Oct. } & \text { Nov. } & \text { Dec. } & \text { Jan. } & \text { Feb. } & \text { March } & \text { April } \\ \hline \text { B: Shore } & 1.6 & 1.8 & 2.0 & 3.2 & 3.9 & 3.6 & 3.3 \\ \hline \text { A: Boat } & 1.5 & 1.4 & 1.6 & 2.2 & 3.3 & 3.0 & 3.8 \\ \hline \end{array}\) Use a \(1 \%\) level of significance to test if there is a difference in the population mean hours per fish caught using a boat compared with fishing from the shore.

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