/*! 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 9 The article "Evaluation of' a Ve... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 9

The article "Evaluation of' a Ventilation Strategy to Prevent Barotrauma in Patients at High Risk for Acute Respiratory Distress Syndrome" (New Engl. \(J\). of Med., 1998: 355-358) reported on an experiment in which 120 patients with similar clinical features were randomly divided into a control group and a treatment group, each consisting of 60 patients. The sample mean ICU stay (days) and sample standard deviation for the treatment group were \(19.9\) and \(39.1\), respectively, whereas these values for the control group were \(13.7\) and \(15.8\). a. Calculate a point estimate for the difference between true average ICU stay for the treatment and control groups. Does this estimate suggest that there is a significant difference between true average stays under the two conditions? b. Answer the question posed in part (a) by carrying out a formal test of hypotheses. Is the result different from what you conjectured in part (a)? c. Does it appear that ICU stay for patients given the ventilation treatment is normally distributed? Explain your reasoning. d. Estimate true average length of stay for patients given the ventilation treatment in a way that conveys information about precision and reliability.

Short Answer

Expert verified
a) Point estimate for difference: 6.2 days. No significant difference. b) Fail to reject null hypothesis. c) Normal distribution unlikely due to large standard deviation compared to mean. d) 95% CI for treatment group: [9.81, 29.99] days.

Step by step solution

01

Calculate the Point Estimate for Difference

To calculate the point estimate for the difference between the true average ICU stay for the treatment and control groups, we subtract the control group's mean from the treatment group's mean. Therefore, the point estimate is \(19.9 - 13.7 = 6.2\) days.
02

State the Hypotheses for Hypothesis Testing

To formally test the hypothesis whether there is a significant difference between the two groups, we state our null and alternative hypotheses. The null hypothesis \(H_0\) states that there is no difference in means: \( \mu_1 = \mu_2 \). The alternative hypothesis \(H_a\) states that there is a difference: \( \mu_1 eq \mu_2 \).
03

Calculate the Test Statistic

The formula for the test statistic is given by \[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where \(\bar{x}_1 = 19.9, \bar{x}_2 = 13.7, s_1 = 39.1, s_2 = 15.8, n_1 = 60, n_2 = 60 \). Calculating, we find \( t = \frac{6.2}{\sqrt{\frac{39.1^2}{60} + \frac{15.8^2}{60}}} \approx 1.226 \).
04

Determine the Critical Value and Decision

Assuming a significance level of 0.05 and using a t-distribution with \(n_1 + n_2 - 2 = 118\) degrees of freedom, we determine the critical value for a two-tailed test. Looking at a t-table, the critical value is approximately \(t_{0.025, 118} \approx 1.98\). Since our calculated t-value \(1.226\) is less than the critical value \(1.98\), we fail to reject the null hypothesis.
05

Assess Normality for ICU Stay in Treatment Group

Check normality by comparing the sample mean (19.9) with the large sample standard deviation (39.1). Since the standard deviation is almost double the mean, it suggests the data may not be normally distributed, hinting at potential skewness or outliers.
06

Estimate the True Average ICU Stay with Confidence Interval

To estimate the true average ICU stay for patients in the treatment group, calculate a confidence interval (CI). Using the formula \(\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}\) where \(\bar{x} = 19.9\), \(s = 39.1\), \(n = 60\), and \(t_{0.025, 59}\) is approximately 2.00, the 95% CI is \(19.9 \pm 2.00 \times \frac{39.1}{\sqrt{60}}\). Thus, the CI is approximately \(19.9 \pm 10.09\), or \([9.81, 29.99]\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Point Estimate
To understand hypothesis testing, it's crucial to grasp what a point estimate is. A point estimate uses sample data to calculate a single value, which serves as our best guess for an unknown population parameter. In this exercise, we calculate the point estimate for the difference between the true average ICU stay of the treatment and control groups. We do this by subtracting the control group's mean from the treatment group's mean:
  • Treatment Group Mean = 19.9 days
  • Control Group Mean = 13.7 days
  • Point Estimate = 19.9 - 13.7 = 6.2 days
This suggests that on average, ICU stays under the treatment conditions are 6.2 days longer than under the control conditions. It's worth mentioning that while this point estimate gives us a useful snapshot, it doesn't speak to statistical significance or reliability, which are addressed later with testing and intervals.
Confidence Interval
Confidence intervals provide a range of values which likely include the true population parameter. They give more information than a simple point estimate as they convey both the estimate's precision and reliability. For example, a 95% confidence interval tells us that if we were to conduct the experiment 100 times, the true average would fall within this interval in 95 of those experiments. In this exercise, we calculated the 95% confidence interval for the true average ICU stay of patients given the treatment. Using the formula for a confidence interval: \[ \bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}} \]where:
  • \(\bar{x} = 19.9\) (sample mean)
  • \(s = 39.1\) (sample standard deviation)
  • \(n = 60\) (sample size)
  • \(t_{0.025, 59} \approx 2.00\)
The calculated interval is \([9.81, 29.99]\) days. This means we are 95% confident the true mean ICU stay under treatment lies in this range.
Normal Distribution
Data that follow a normal distribution will appear as a symmetrical bell-shaped curve when plotted. This is a fundamental assumption in many statistical analyses because certain mathematical properties of the normal distribution play into how we conduct hypothesis tests and construct confidence intervals. For the ICU stay data, we assess normality to determine whether the statistical testing methods are appropriately applied. The sample mean for the treatment group is 19.9 days, with a standard deviation of 39.1 days. The fact that the standard deviation is more than twice the mean suggests the data may not be normally distributed, possibly due to skewness or the presence of outliers. Breaking this normality assumption can affect the validity of further inferential statistics.
Test Statistic
A test statistic is a standardized value derived from sample data during a hypothesis test. It allows us to decide the likelihood of observing the sample data under the null hypothesis. In the exercise, we were interested in checking if there was a significant difference between the two group means.We calculated the test statistic using the formula: \[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where:
  • \(\bar{x}_1 = 19.9\) (treatment group mean)
  • \(\bar{x}_2 = 13.7\) (control group mean)
  • \(s_1 = 39.1\) (treatment group standard deviation)
  • \(s_2 = 15.8\) (control group standard deviation)
  • \(n_1 = n_2 = 60\) (sample sizes)
The computed test statistic \(t\approx 1.226\) was less than the critical value \(1.98\) for a 5% significance level, leading us to conclude that there is no statistically significant difference, as we fail to reject the null hypothesis. This step is critical in hypothesis testing, as it influences decision-making based on statistical evidence.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Olestra is a fat substitute approved by the FDA for use in snack foods. Because there have been anecdotal reports of gastrointestinal problems associated with olestra consumption, a randomized, double-blind, placebo- controlled experiment was carried out to compare olestra potato chips to regular potato chips with respect to GI symptoms ("Gastrointestinal Symptoms Following Consumption of Olestra or Regular Triglyceride Potato Chips," \(J\). of the Amer: Med. Assoc., 1998: 150-152). Among 529 individuals in the TG control group, \(17.6 \%\) experienced an adverse GI event, whereas among the 563 individuals in the olestra treatment group, \(15.8 \%\) experienced such an event. a. Carry out a test of hypotheses at the \(5 \%\) significance level to decide whether the incidence rate of GI problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the \(\mathrm{TG}\) control treatment. b. If the true percentages for the two treatments were \(15 \%\) and \(20 \%\), respectively, what sample sizes \((m=n)\) would be necessary to detect such a difference with probability 90 ?

Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor air pollution. The article "Domestic Fuels, Indoor Air Pollution, and Children's Health" (Annals of the N.Y. Academy of Sciences, 2008: 209-217) presented information on various pulmonary characteristics in samples of children whose households in India used either biomass fuel or liquefied petroleum gas (LPG). For the 755 children in biomass households, the sample mean peak expiratory flow (a person's maximum speed of expiration) was \(3.30 \mathrm{~L} / \mathrm{s}\), and the sample standard deviation was \(1.20\). For the 750 children whose households used liquefied petroleum gas, the sample mean PEF was \(4.25\) and the sample standard deviation was 1.75. a. Calculate a confidence interval at the \(95 \%\) confidence level for the population mean PEF for children in biomass households and then do likewise for children in LPG households. What is the simultaneous confidence level for the two intervals? b. Carry out a test of hypotheses at significance level \(.01\) to decide whether true average PEF is lower for children in biomass households than it is for children in LPG households (the cited article included a \(P\)-value for this test). c. \(\mathrm{FEV}_{1}\), the forced expiratory volume in 1 second, is another measure of pulmonary function. The cited article reported that for the biomass households the sample mean \(\mathrm{FEV}_{1}\) was \(2.3 \mathrm{~L} / \mathrm{s}\) and the sample standard deviation was \(.5 \mathrm{~L} / \mathrm{s}\). If this information is used to compute a \(95 \%\) CI for population mean \(\mathrm{FEV}_{1}\), would the simultaneous confidence level for this interval and the first interval calculated in (a) be the same as the simultaneous confidence level determined there? Explain.

Wait staff at restaurants have employed various strategies to increase tips. An article in the Sept. 5, 2005, New Yorker reported that "In one study a waitress received \(50 \%\) more in tips when she introduced herself by name than when she didn't." Consider the following (fictitious) data on tip amount as a percentage of the bill: Introduction: \(\quad m=50 \quad \bar{x}=22.63 \quad s_{1}=7.82\) No introduction: \(\quad n=50 \quad \bar{y}=14.15 \quad s_{2}=6.10\) Does this data suggest that an introduction increases tips on average by more than \(50 \%\) ? State and test the relevant hypotheses. [Hint: Consider the parameter \(\left.\theta=\mu_{1}-1.5 \mu_{2} .\right]\)

Damage to grapes from bird predation is a serious problem for grape growers. The article "Experimental Method to Investigate and Monitor Bird Behavior and Damage to Vineyards" (Amer. \(J\). of Enology and Viticulture, 2004: 288-291) reported on an experiment involving a bird-feeder table, time-lapse video, and artificial foods. Information was collected for two different bird species at both the experimental location and at a natural vineyard setting. Consider the following data on time \((\mathrm{sec})\) spent on a single visit to the location. $$ \begin{array}{llrrc} \text { Species } & \text { Location } & n & \bar{x} & \text { SE mean } \\ \hline \text { Blackbirds } & \text { Exptl } & 65 & 13.4 & 2.05 \\ \text { Blackbirds } & \text { Natural } & 50 & 9.7 & 1.76 \\ \text { Silvereyes } & \text { Exptl } & 34 & 49.4 & 4.78 \\ \text { Silvereyes } & \text { Natural } & 46 & 38.4 & 5.06 \\ \hline \end{array} $$ a. Calculate an upper confidence bound for the true average time that blackbirds spend on a single visit at the experimental location. b. Does it appear that true average time spent by blackbirds at the experimental location exceeds the true average time birds of this type spend at the natural location? Carry out a test of appropriate hypotheses. c. Estimate the difference between the true average time blackbirds spend at the natural location and true average time that silvereyes spend at the natural location, and do so in a way that conveys information about reliability and precision. [Note: The sample medians reported in the article all seemed significantly smaller than the means, suggesting substantial population distribution skewness. The authors actually used the distribution-free test procedure presented in Section 2 of Chapter 15.]

Fusible interlinings are being used with increasing frequency to support outer fabrics and improve the shape and drape of various pieces of clothing. The article "Compatibility of Outer and Fusible Interlining Fabrics in Tailored Garments" (Textile Res. J., 1997: \(137-142\) ) gave the accompanying data on extensibility (\%) at \(100 \mathrm{gm} / \mathrm{cm}\) for both high-quality (H) fabric and poor-quality (P) fabric specimens. \(\begin{array}{rrrrrrrrrr}\mathrm{H} & 1.2 & .9 & .7 & 1.0 & 1.7 & 1.7 & 1.1 & .9 & 1.7 \\ & 1.9 & 1.3 & 2.1 & 1.6 & 1.8 & 1.4 & 1.3 & 1.9 & 1.6 \\ & .8 & 2.0 & 1.7 & 1.6 & 2.3 & 2.0 & & & \\ \mathrm{P} & 1.6 & 1.5 & 1.1 & 2.1 & 1.5 & 1.3 & 1.0 & 2.6 & \end{array}\) a. Construct normal probability plots to verify the plausibility of both samples having been selected from normal population distributions. b. Construct a comparative boxplot. Does it suggest that there is a difference between true average extensibility for high-quality fabric specimens and that for poor-quality specimens? c. The sample mean and standard deviation for the highquality sample are \(1.508\) and \(.444\), respectively, and those for the poor-quality sample are \(1.588\) and \(.530\). Use the two-sample \(t\) test to decide whether true average extensibility differs for the two types of fabric.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.