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Suppose that 10 patients with meningitis received treatment with large doses of penicillin. Three days later, temperatures were recorded, and the treatment was considered successful if there had been a reduction in a patient's temperature. Denoting success by \(S\) and failure by \(\mathrm{F}\), the 10 observations are \(\begin{array}{llllllllll}\text { S } & \text { S } & \text { F } & \text { S } & \text { S } & \text { S } & \text { F } & \text { F } & \text { S } & \text { S }\end{array}\) a. What is the value of the sample proportion of successes? b. Replace each \(S\) with a 1 and each \(F\) with a 0 . Then calculate \(\bar{x}\) for this numerically coded sample. How does \(\bar{x}\) compare to \(\hat{p}\) ? c. Suppose that it is decided to include 15 more patients in the study. How many of these would have to be S's to give \(\hat{p}=.80\) for the entire sample of 25 patients?

Short Answer

Expert verified
The sample proportion of successes is 0.70. Replacing S with 1 and F with 0 provides a mean value of 0.70, which is equivalent to the sample proportion. To achieve a sample proportion of 0.80 for the entire sample of 25, 13 out of the next 15 patients must have a successful outcome.

Step by step solution

01

Computing Sample Proportion

Count the number of successes (S's) out of the total sample. It equals to 7 out of 10. The sample proportion of successes \(\hat{p}\) is then \(\hat{p}=\frac{7}{10}=0.70\).
02

Computing Sample Mean and Comparison

Replace each S by 1 and F by 0. Compute the sample mean \(\bar{x}\). For example, the mean of {\[1, 1, 0, 1, 1, 1, 0, 0, 1, 1 \]will also equal to 0.70. It can be observed \(\bar{x} = \hat{p}\)
03

Compute Additional Successes Needed

We want to find out how many more successes (S's) are necessary out of 15 additional patients to make the proportion of successful outcomes 0.80, based on all 25 patients. Let's assume the necessary number of successful cases is x. Thus, we have:\[\frac{7 + x}{25} = 0.80\]Solving for x gives: \(x = 20 - 7 = 13\). So 13 out of 15 new patients need to have successful outcomes

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Key Concepts

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

Binary Outcomes
Binary outcomes refer to two possibilities in an observation or experiment. In our example, the treatment's outcome for a patient is either successful (denoted by S) or unsuccessful (denoted by F).
This simplifies assessments by allowing a clear distinction between two states.
Binary data is often recorded as 1s and 0s, which facilitates easier processing and analysis in statistical computations.
  • 1 represents success or the occurrence of an event.
  • 0 represents failure or the non-occurrence.
Binary outcomes are foundational in statistics and probability because they allow researchers to quantify and analyze the chance of different outcomes easily.
Sample Mean
In statistics, the sample mean is a way to summarize data by finding the average value of a set of numbers.
When you convert binary outcomes into numerical values (1s and 0s), the sample mean can be calculated by summing all these values and dividing by the total number of observations.
For instance, in our exercise, the sample mean \(\bar{x}\) is based on the binary conversion of the outcomes: \[\bar{x} = \frac{1 + 1 + 0 + 1 + 1 + 1 + 0 + 0 + 1 + 1}{10} = 0.70\].
The sample mean in this context directly correlates to the sample proportion, giving insight into the average level of success across the sample group.
Statistical Computation
Statistical computation is the process of analyzing and interpreting numerical data to gain insights or make predictions.
In our exercise, computing the sample proportion and sample mean are examples of basic statistical computations.
These calculations help summarize the data so we can make informed assumptions or predictions about a larger population.
In statistical computation, finding out how many additional successes are needed to reach a new sample proportion involves equations: \[\frac{7 + x}{25} = 0.80\].
Solving such equations helps determine required conditions or outcomes in future experiments.
Probability
Probability helps us understand the likelihood of different outcomes. In terms of binary outcomes, it gauges the chance of one outcome over the other.
Probability is expressed as a number between 0 and 1, with 0 meaning an event is impossible and 1 meaning it is certain.
The probability of success in our sample (which is also the sample proportion) is \(0.70\), meaning there's a 70% chance of treatment success for a given patient.
In planning experiments or treatments, understanding probability allows for the prediction of outcomes and assessment of risks and benefits.
It is a cornerstone of data-driven decision-making in scientific research and numerous other fields.

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Most popular questions from this chapter

The U.S. Census Bureau ( 2000 census) reported the following relative frequency distribution for travel time to work for a large sample of adults who did not work at home: $$ \begin{array}{cc} \begin{array}{c} \text { Travel Time } \\ \text { (minutes) } \end{array} & \text { Relative Frequency } \\ \hline 0 \text { to }<5 & .04 \\ 5 \text { to }<10 & .13 \\ 10 \text { to }<15 & .16 \\ 15 \text { to }<20 & .17 \\ 20 \text { to }<25 & .14 \\ 25 \text { to }<30 & .05 \\ 30 \text { to }<35 & .12 \\ 35 \text { to }<40 & .03 \\ 40 \text { to }<45 & .03 \\ 45 \text { to }<60 & .06 \\ 60 \text { to }<90 & .05 \\ 90 \text { or more } & .02 \\ \hline \end{array} $$ a. Draw the histogram for the travel time distribution. In constructing the histogram, assume that the last interval in the relative frequency distribution ( 90 or more) ends at 200 ; so the last interval is 90 to \(<200\). Be sure to use the density scale to determine the heights of the bars in the histogram because not all the intervals have the same width. b. Describe the interesting features of the histogram from Part (a), including center, shape, and spread. c. Based on the histogram from Part (a), would it be appropriate to use the Empirical Rule to make statements about the travel time distribution? Explain why or why not. d. The approximate mean and standard deviation for the travel time distribution are 27 minutes and 24 minutes, respectively. Based on this mean and standard deviation and the fact that travel time cannot be negative, explain why the travel time distribution could not be well approximated by a normal curve. e. Use the mean and standard deviation given in Part (d) and Chebyshev's Rule to make a statement about i. the percentage of travel times that were between 0 and 75 minutes ii. the percentage of travel times that were between 0 and 47 minutes f. How well do the statements in Part (e) based on Chebyshev's Rule agree with the actual percentages for the travel time distribution? (Hint: You can estimate the actual percentages from the given relative frequency distribution.)

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