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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 \(140 \mathrm{~mm} \mathrm{Hg}\) systolic and/or over \(90 \mathrm{~mm}\) Hg diastolic. Hypertension, if not corrected, can cause longterm 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.

Step by step solution

01

Define the Hypotheses

We are conducting a hypothesis test to determine if the proportion of students with hypertension during final exams is greater than 9.2%. Thus, we define the null hypothesis (H_0) and the alternative hypothesis (H_a):\[ H_0: p = 0.092 \]\[ H_a: p > 0.092 \]where \( p \) is the true proportion of students with hypertension during exam week.

02

Collect the Sample Data

From the problem, we have a sample size of 196 students, with 29 of them having hypertension. Therefore, the sample proportion (\hat{p}) is \( \hat{p} = \frac{29}{196} \approx 0.147 \).

03

Calculate the Test Statistic

We will use a one-sample z-test for proportions to calculate the test statistic. The formula for the z-test statistic is:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]Substitute the values:\[ z = \frac{0.147 - 0.092}{\sqrt{\frac{0.092 \times (1-0.092)}{196}}} \]Calculate the result to find the value of \( z \).

04

Determine the Critical Value and Decision Rule

For a significance level of 5%, the critical value for a one-tailed z-test can be found using a z-table or standard normal distribution calculator. The critical value is approximately 1.645. If calculated \( z \)-value is greater than 1.645, we reject the null hypothesis.

05

Compare the Test Statistic to the Critical Value

Calculate \( z \) from Step 3 and compare it to the critical value of 1.645. If \( z \) exceeds the critical value, this indicates that the sample provides sufficient evidence to support the alternative hypothesis.

06

Conclusion

Based on the comparison in the previous step, if \( z > 1.645 \), we reject \( H_0 \) and conclude that there is significant evidence at the 5% significance level to suggest that more than 9.2% of college students have hypertension during final exams week. Otherwise, we do not reject the null hypothesis.

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Key Concepts

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

Proportion Test

A proportion test is a type of hypothesis test that evaluates whether the proportion of a certain characteristic in a population is equal to a specified value. In the context of hypothesis testing, it compares the sample proportion to a hypothesized population proportion to see if there is a significant difference.
For our exercise, we focused on whether the proportion of students with hypertension during final exams is greater than the usual 9.2%. This requires the use of a one-sample proportion test. The test compares the sample proportion, calculated from the data, to the known proportion stated in the hypothesis.

• Sample proportion ({ }hat{p}) represents the proportion of individuals in the sample with the characteristic being measured.
• Known population proportion (p_0) is the proportion we are testing against; in this case, it is 9.2%.

This test is essential for making data-driven decisions on whether significant evidence exists to claim a greater proportion of hypertension among students during exams.

Significance Level

The significance level in hypothesis testing is a threshold set by the researcher to determine the probability of rejecting the null hypothesis when it is true. It is denoted by { }α and commonly set at 5% (0.05).
This concept essentially helps in deciding how strict our test should be. A 5% significance level, as used in our problem, suggests that there is a 5% risk of concluding that the sample data reflect a true effect when there is none.

• If the P-value calculated from the test statistic is less than the α, we reject the null hypothesis.
• A 5% level means we are willing to accept a 5% chance of a Type I error, suggesting either type of hypothesis rejection when it's not warranted.

Using this level of significance ensures a balanced approach, providing ample evidence before declaring significant deviation from norms.

Z-Test

A Z-test for proportions is a statistical test used to determine whether there is a significant difference between a sample proportion and a known population proportion.
The test involves calculating a Z-test statistic, which indicates how many standard deviations the sample proportion is from the hypothesized population proportion. For our hypothesis test:

• The formula is given by: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]
• Where \( \hat{p} \) is the sample proportion and \( p_0 \) is the hypothesized population proportion.

The result is compared to a critical Z-value from the Z-table, informed by the chosen significance level. If the computed Z-statistic exceeds this critical value, it provides evidence to reject the null hypothesis.
In this exercise, the calculated Z-statistic was compared with the critical Z-value of 1.645 (for a 5% significance level). The comparison informs whether to maintain or reject our assumption about the population proportion.