/*! 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 101 Use data from a study designed t... [FREE SOLUTION] | 91Ó°ÊÓ

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Use data from a study designed to examine the effect of doing synchronized movements (such as marching in step or doing synchronized dance steps) and the effect of exertion on many different variables, such as pain tolerance and attitudes toward others. In the study, 264 high school students in Brazil were randomly assigned to one of four groups reflecting whether or not movements were synchronized (Synch= yes or no) and level of activity (Exertion= high or low). \(^{49}\) Participants rated how close they felt to others in their group both before (CloseBefore) and after (CloseAfter) the activity, using a 7-point scale (1=least close to \(7=\) most close ). Participants also had their pain tolerance measured using pressure from a blood pressure cuff, by indicating when the pressure became too uncomfortable (up to a maximum pressure of \(300 \mathrm{mmHg}\) ). Higher numbers for this Pain Tolerance measure indicate higher pain tolerance. The full dataset is available in SynchronizedMovement. For each of the following problems: (a) Give notation for the quantity we are estimating, and define any relevant parameters. (b) Use StatKey or other technology to find the value of the sample statistic. Give the correct notation with your answer. (c) Use StatKey or other technology to find the standard error for the estimate. (d) Use the standard error to give a \(95 \%\) confidence interval for the quantity we are estimating. (e) Interpret the confidence interval in context. Are Males or Females More Likely to Go to Maximum Pressure? The study recorded whether participants were female or male \((\operatorname{Sex}=\mathrm{F}\) or \(\mathrm{M})\), and we see that 33 of the 165 females and 42 of the 99 males allowed the pressure to reach its maximum level of \(300 \mathrm{mmHg}\) after treatment, without ever saying that the pain was too much. Use this information to estimate the difference in proportion of people who would allow the pressure to reach its maximum level after treatment, between females and males.

Short Answer

Expert verified
The estimated difference in the proportions of males and females who would allow the pressure to reach its maximum level after treatment is approximately 0.22. The 95% confidence interval for this estimate is from 0.09 to 0.35.

Step by step solution

01

Define Parameters

For this problem, the parameters are the actual proportions of males and females who can withstand maximum pressure. Let's denote \( p_M \) for males and \( p_F \) for females.
02

Calculate Sample Statistic

According to the information, 33 out of 165 females and 42 out of 99 males can tolerate maximum pressure. We can calculate the sample proportions for each group using the formula: \( p = \frac{x}{n} \), where x is the number of successes and n is the total sample size. We find, \( \hat{p}_F = \frac{33}{165} = 0.20 \) and \( \hat{p}_M = \frac{42}{99} = 0.42 \). Thus, the point estimate of the difference between the two proportions is \( \hat{p}_M - \hat{p}_F = 0.42 - 0.20 = 0.22 \)
03

Calculate Standard Error

The standard error for the difference between two proportions is calculated with the formula: SE = \( \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \). Substituting the values we have: SE = \( \sqrt{\frac{0.42(1-0.42)}{99} + \frac{0.20(1-0.20)}{165}} = 0.07 \)
04

Construct Confidence Interval

A 95% confidence interval can be constructed using the formula: \( \hat{p}_1 - \hat{p}_2 \pm Z \times SE \) where Z value for 95% confidence is approximately 1.96. So, the confidence interval is \( 0.22 \pm 1.96 \times 0.07 = [0.09, 0.35] \)
05

Interpret Confidence Interval

The interval [0.09, 0.35] can be interpreted as: We are 95% confident that the true difference in proportion of people who would allow the pressure to reach its maximum level after treatment, between males and females, is between 9% and 35%.

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Key Concepts

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

Sample Proportion
When we talk about sample proportion, we are referring to a statistic that represents a portion of a population observed in a sample. In this study, you saw how sample proportions were calculated for determining how many males and females tolerated the maximum pressure applied. For both groups, the sample proportion is computed using the formula:
  • \( p = \frac{x}{n} \)
Here, \( x \) represents the number of successes, and \( n \) is the total sample size. For example, for females, 33 of 165 could withstand the pressure, giving a sample proportion \( \hat{p}_F = \frac{33}{165} = 0.20 \).
Similarly, for males, \( \hat{p}_M = \frac{42}{99} = 0.42 \).
These proportions help in understanding the characteristics of the sample allowing us to make inferences about the entire population.
Standard Error
The standard error is a measure that describes how much we expect the sample statistic to fluctuate if we were to take multiple samples. Think of it as a gauge for reliability; the smaller the standard error, the more reliable your sample statistic is.
The formula to calculate the standard error of the difference between two sample proportions is given by:
  • \[ SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \]
Here, \( \hat{p}_1 \) and \( \hat{p}_2 \) are the sample proportions, and \( n_1 \) and \( n_2 \) are the corresponding sample sizes.
In our case, substituting the values gives us a standard error of 0.07. This number tells us about the expected range of variability in the difference in pressure tolerance between males and females.
Difference of Proportions
When comparing two groups, like males and females in our study, the difference in their sample proportions shows how those groups are distinct based on the attribute in question. In this exercise, we focused on how likely each gender was to endure a specified pressure.
The difference in sample proportions between males and females is calculated as:
  • \( \hat{p}_M - \hat{p}_F = 0.42 - 0.20 = 0.22 \)
This difference of 0.22 or 22% suggests that a significantly greater proportion of males reach the maximum pressure limit than females.
We use this difference to create a confidence interval that provides a range where the true difference in population proportions is likely to be. By following the given calculations, we find a 95% confidence interval between 9% and 35%.

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Most popular questions from this chapter

Topical Painkiller Ointment The use of topical painkiller ointment or gel rather than pills for pain relief was approved just within the last few years in the US for prescription use only. \({ }^{13}\) Insurance records show that the average copayment for a month's supply of topical painkiller ointment for regular users is \$30. A sample of \(n=75\) regular users found a sample mean copayment of \(\$ 27.90\). (a) Identify each of 30 and 27.90 as a parameter or a statistic and give the appropriate notation for each. (b) If we take 1000 samples of size \(n=75\) from the population of all copayments for a month's supply of topical painkiller ointment for regular users and plot the sample means on a dotplot, describe the shape you would expect to see in the plot and where it would be centered. (c) How many dots will be on the dotplot you described in part (b)? What will each dot represent?

Predicting Election Results Throughout the US presidential election of \(2016,\) polls gave regular updates on the sample proportion supporting each candidate and the margin of error for the estimates. This attempt to predict the outcome of an election is a common use of polls. In each case below, the proportion of voters who intend to vote for each of two candidates is given as well as a margin of error for the estimates. Indicate whether we can be relatively confident that candidate A would win if the election were held at the time of the poll. (Assume the candidate who gets more than \(50 \%\) of the vote wins.) \(\begin{array}{lll}\text { (a) Candidate A: } 54 \% & \text { Candidate }\end{array}\) B: \(46 \%\) Margin of error: \(\pm 5 \%\) (b) Candidate A: \(52 \%\) Candidate B: \(48 \%\) Margin of error: \(\pm 1 \%\) \(\begin{array}{ll}\text { (c) Candidate A: } 53 \% & \text { Candidate }\end{array}\) B: \(47 \%\) Margin of error: \(\pm 2 \%\) \(\begin{array}{lll}\text { (d) Candidate A: } 58 \% & \text { Candidate }\end{array}\) B: \(42 \%\) Margin of error: \(\pm 10 \%\)

Downloading Apps for Your Smartphone A random sample of \(n=461\) smartphone users in the US in January 2015 found that 355 of them have downloaded an app. \(^{10}\) (a) Give notation for the parameter of interest, and define the parameter in this context. (b) Give notation for the quantity that gives the best estimate and give its value. (c) What would we have to do to calculate the parameter exactly?

Moose Drool Makes Grass More Appetizing Different species can interact in interesting ways. One type of grass produces the toxin ergovaline at levels about 1.0 part per million in order to keep grazing animals away. However, a recent study \(^{27}\) has found that the saliva from a moose counteracts these toxins and makes the grass more appetizing (for the moose). Scientists estimate that, after treatment with moose drool, mean level of the toxin ergovaline (in ppm) on the grass is \(0.183 .\) The standard error for this estimate is 0.016 . (a) Give notation for the quantity being estimated, and define any parameters used. (b) Give notation for the quantity that gives the best estimate, and give its value. (c) Give a \(95 \%\) confidence interval for the quantity being estimated. Interpret the interval in context.

In Exercises 3.51 to 3.56 , information about a sample is given. Assuming that the sampling distribution is symmetric and bell-shaped, use the information to give a \(95 \%\) confidence interval, and indicate the parameter being estimated. $$ \bar{x}=55 \text { and the standard error is } 1.5 . $$

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