/*! 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 34 A medical research group is recr... [FREE SOLUTION] | 91Ó°ÊÓ

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A medical research group is recruiting people to complete short surveys about their medical history. For example, one survey asks for information on a person's family history in regards to cancer. Another survey asks about what topics were discussed during the person's last visit to a hospital. So far, as people sign up, they complete an average of just 4 surveys, and the standard deviation of the number of surveys is about \(2.2 .\) The research group wants to try a new interface that they think will encourage new enrollees to complete more surveys, where they will randomize each enrollee to either get the new interface or the current interface. How many new enrollees do they need for each interface to detect an effect size of 0.5 surveys per enrollee, if the desired power level is \(80 \%\) ?

Short Answer

Expert verified
152 enrollees are needed for each interface to detect the effect size with 80% power.

Step by step solution

01

Understanding the Problem

We are tasked with determining the sample size required for each group (new interface vs. current interface) to detect an effect size in the number of completed surveys, given certain statistical parameters.
02

Determine Effect Size

The effect size is the difference in means divided by the standard deviation. We are given that the effect size is 0.5 surveys per enrollee. This means that the difference in the average number of completed surveys between the two groups is 0.5 surveys, and the standard deviation is 2.2 surveys.
03

Statistical Analysis for Power Calculation

The required statistical context involves using the concept of statistical power, which is the probability of correctly rejecting the null hypothesis when it is false. We desire a power of 80%, meaning there is a 20% chance of a Type II error (failing to detect a true effect).
04

Using the Formula for Sample Size Calculation

The formula for calculating sample size for comparing two means is given by: \[ n = \left( \frac{(Z_{\alpha/2} + Z_{\beta}) \times \sigma}{\text{Effect size}} \right)^2 \]where \( Z_{\alpha/2} \) is the critical value from the normal distribution for a two-tailed test at the specified significance level (usually 0.05), and \( Z_{\beta} \) is the critical value to achieve the desired power.
05

Substituting Values

First, determine the critical values from the standard normal distribution table:- For a 5% significance level (\( \alpha = 0.05 \)), \( Z_{\alpha/2} \approx 1.96 \) for a two-tailed test.- For 80% power (\( 1 - \beta = 0.80 \)), find \( Z_{\beta} \approx 0.84 \).Substitute these into the sample size formula along with \( \sigma = 2.2 \) and the effect size of 0.5:\[ n = \left( \frac{(1.96 + 0.84) \times 2.2}{0.5} \right)^2 \]
06

Calculate the Sample Size

Carrying out the calculations:\[ n = \left( \frac{2.8 \times 2.2}{0.5} \right)^2 \]\[ n = \left( \frac{6.16}{0.5} \right)^2 \]\[ n = (12.32)^2 \]\[ n \approx 151.85 \]Since the sample size must be a whole number, round up to ensure adequate power: \( n = 152 \). This is the sample size needed for each group.

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Key Concepts

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

Understanding Effect Size
Effect size is a key concept in sample size calculation that measures the magnitude of difference between two groups. It helps researchers understand how large a "practical" difference is between the new interface and the current one. In the exercise, the effect size is given as 0.5 surveys per enrollee. This implies that on average, the new interface aims to increase survey completion by 0.5 surveys compared to the existing interface. You calculate effect size by dividing the mean difference by the standard deviation. In scenarios like these, a moderate effect size indicates a noticeable change without being overly disparate, making it both measureable and meaningful in practical terms. By establishing a specific effect size, researchers can determine how many subjects need to be studied to detect this effect with a high degree of confidence.
Exploring Statistical Power
Statistical power is the probability that a study will detect an effect when there is one to be detected. In simpler terms, it's about knowing how likely you are to avoid a false negative, or Type II error. The researchers in the exercise desire a statistical power of 80%, meaning there's an 80% likelihood that they will correctly reject the null hypothesis (no effect) when the new interface actually does succeed in increasing survey completion. To maintain good statistical power, researchers need to balance between a small risk of false negatives and the ethical and practical constraints of gathering large data samples. An understanding of statistical power is crucial, as high power typically necessitates larger sample sizes, but pays off by increasing the reliability of the conclusions drawn.
Understanding Type II Error
Type II error, often symbolized by \( \beta \), occurs when we fail to reject a false null hypothesis, meaning we conclude there is no effect when, in fact, there is one. If the statistical power is 80%, the probability of making a Type II error is 20%. Though often less stressed than Type I errors (false positives), Type II errors can be equally misleading, resulting in a missed opportunity to recognize a beneficial treatment or intervention. Recognizing and accounting for this type of error is vital when planning experiments, especially when the stakes of undetected effects, like in medical research, can be significant. By minimizing Type II errors, researchers ensure that they don't overlook meaningful differences that deserve attention.
Reading the Normal Distribution Table
Normal distribution tables are fundamental tools for determining the critical values required in calculations like sample size determination. In the exercise, two critical Z values are needed: one for the significance level and another for the desired power. The value \( Z_{\alpha/2} \approx 1.96 \) corresponds to a significance level of 0.05, which is quite common in psychology and medical research, reflecting a conventional 5% chance of a Type I error. Meanwhile, \( Z_{\beta} \approx 0.84 \) is aligned with an 80% power level. The normal distribution table provides a way to obtain these thresholds by indicating the cumulative probability associated with each Z score. Understanding how to use these tables is pivotal for accurate data analysis and ensures the calculations made are based on validated statistical reasoning.

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Most popular questions from this chapter

A professor who teaches a large introductory statistics class (197 students) with eight discussion sections would like to test if student performance differs by discussion section, where each discussion section has a different teaching assistant. The summary table below shows the average final exam score for each discussion section as well as the standard deviation of scores and the number of students in each section. $$ \begin{array}{rrrrrrrrrr} \hline & \text { Sec 1 } & \text { Sec 2 } & \text { Sec 3 } & \text { Sec 4 } & \text { Sec 5 } & \text { Sec 6 } & \text { Sec 7 } & \text { Sec 8 } \\ \hline n_{i} & 33 & 19 & 10 & 29 & 33 & 10 & 32 & 31 \\ \bar{x}_{i} & 92.94 & 91.11 & 91.80 & 92.45 & 89.30 & 88.30 & 90.12 & 93.35 \\\ s_{i} & 4.21 & 5.58 & 3.43 & 5.92 & 9.32 & 7.27 & 6.93 & 4.57 \\ \hline \end{array} $$ The ANOVA output below can be used to test for differences between the average scores from the different discussion sections. $$ \begin{array}{lrrrrr} \hline & \text { Df } & \text { Sum Sq } & \text { Mean Sq } & \text { F value } & \operatorname{Pr}(>\mathrm{F}) \\ \hline \text { section } & 7 & 525.01 & 75.00 & 1.87 & 0.0767 \\ \text { Residuals } & 189 & 7584.11 & 40.13 & & \\ \hline \end{array} $$ Conduct a hypothesis test to determine if these data provide convincing evidence that the average score varies across some (or all) groups. Check conditions and describe any assumptions you must make to proceed with the test.

Georgianna claims that in a small city renowned for its music school, the average child takes less than 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. (a) Evaluate Georgianna's claim (or that the opposite might be true) using a hypothesis test. (b) Construct a \(95 \%\) confidence interval for the number of years students in this city take piano lessons, and interpret it in context of the data. (c) Do your results from the hypothesis test and the confidence interval agree? Explain your reasoning.

Researchers interested in lead exposure due to car exhaust sampled the blood of 52 police officers subjected to constant inhalation of automobile exhaust fumes while working traffic enforcement in a primarily urban environment. The blood samples of these officers had an average lead concentration of \(124.32 \mu \mathrm{g} / \mathrm{l}\) and a SD of \(37.74 \mu \mathrm{g} / \mathrm{l} ;\) a previous study of individuals from a nearby suburb, with no history of exposure, found an average blood level concentration of \(35 \mu \mathrm{g} / \mathrm{l} .\) (a) Write down the hypotheses that would be appropriate for testing if the police officers appear to have been exposed to a different concentration of lead. (b) Explicitly state and check all conditions necessary for inference on these data. (c) Regardless of your answers in part (b), test the hypothesis that the downtown police officers have a higher lead exposure than the group in the previous study. Interpret your results in context.

We considered the change in the number of days exceeding \(90^{\circ} \mathrm{F}\) from 1948 and 2018 at 197 randomly sampled locations from the NOAA database in Exercise \(7.19 .\) The mean and standard deviation of the reported differences are 2.9 days and 17.2 days. (a) Calculate a \(90 \%\) confidence interval for the average difference between number of days exceeding \(90^{\circ} \mathrm{F}\) between 1948 and 2018 . We've already checked the conditions for you. (b) Interpret the interval in context. (c) Does the confidence interval provide convincing evidence that there were more days exceeding \(90^{\circ} \mathrm{F}\) in 2018 than in 1948 at NOAA stations? Explain.

Determine if the following statements are true or false, and explain your reasoning for statements you identify as false. (a) When comparing means of two samples where \(n_{1}=20\) and \(n_{2}=40,\) we can use the normal model for the difference in means since \(n_{2} \geq 30\). (b) As the degrees of freedom increases, the \(t\) -distribution approaches normality. (c) We use a pooled standard error for calculating the standard error of the difference between means when sample sizes of groups are equal to each other.

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