/*! 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 64 When two lights close together b... [FREE SOLUTION] | 91Ó°ÊÓ

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When two lights close together blink alternately, we "see" one light moving back and forth if the time between blinks is short. What is the longest interval of time between blinks that preserves the illusion of motion? Ask subjects to turn a knob that slows the blinking until they "see" two lights rather than one light moving. A report gives the results in the form "mean plus or minus the standard error of the mean." \({ }^{22}\) Data for 12 subjects are summarized as \(251 \pm 45\) (in milliseconds). (a) Find the sample standard deviation \(s_{x}\) for these measurements. Show your work. (b) A hasty reader believes that the interval given in the report is a \(95 \%\) confidence interval for the population mean. Find the actual confidence level for the given interval.

Short Answer

Expert verified
(a) The sample standard deviation is approximately 155.88 ms. (b) The actual confidence level is less than 95%.

Step by step solution

01

Understand the given data

We are provided with a set of measurements from 12 subjects. These measurements are represented as a mean of 251 milliseconds with a standard error of 45 milliseconds.
02

Calculate the Sample Standard Deviation

The standard error of the mean (SEM) is calculated as \(\frac{s}{\sqrt{n}}\), where \(s\) is the sample standard deviation and \(n\) is the number of observations. We rearrange this formula to solve for the sample standard deviation: \(s = \text{SEM} \times \sqrt{n}\). Here, \(\text{SEM} = 45\) and \(n = 12\). Therefore, \(s = 45 \times \sqrt{12}\). Calculate \(\sqrt{12} \approx 3.464\), which results in \(s \approx 45 \times 3.464 \approx 155.88\).
03

Calculate the Confidence Interval

The given report provides an interval of \(251 \pm 45\), which suggests a margin based on the standard error. This is an approximation typically used to convey some confidence level, like a confidence interval.
04

Determine the actual confidence level

Since the confidence interval is expressed using the standard error and assuming it is intended to reflect a normal distribution, we use the formula for a confidence interval: \( \text{CI} = \bar{x} \pm z \times \frac{s}{\sqrt{n}} \). Here, \(z\) relates to the actual confidence level achieved with the given interval width. The interval width is \(90\) (as \(2 \times 45\)). Given \(s = 155.88\) and \(n = 12\), calculate \(\frac{s}{\sqrt{n}}\) again to ensure accuracy for needed computations, then find \(z\) that satisfies the desired interval width from which the corresponding confidence level via standard Z-tables is checked.

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Key Concepts

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

Sample Standard Deviation
In statistics, understanding the concept of sample standard deviation is crucial when analyzing data. The sample standard deviation (\(s_x\)) provides an estimate of the spread or variability of a dataset. It is particularly useful when dealing with small sample sizes, as it helps to quantify how much individual data points tend to deviate from the sample mean.
The formula to calculate the sample standard deviation is:\[ s = \text{SEM} \times \sqrt{n} \]where \(\text{SEM}\) is the standard error of the mean and \(n\) is the number of observations in the sample.
In the exercise provided, the standard error is 45 milliseconds and there are 12 observations. To find the sample standard deviation, we use \(s = 45 \times \sqrt{12}\), where \(\sqrt{12} \approx 3.464\).This calculation gives us \(s \approx 155.88\) milliseconds.
This value of 155.88 milliseconds indicates the degree of variation in the blink interval perception among the subjects.
Confidence Interval
A confidence interval is a range of values that is used to estimate a population parameter with a certain level of confidence. It's represented as "mean plus or minus a margin of error." In this exercise, the margin of error is provided as the standard error (SEM), calculated to estimate the variability in sample means if we took multiple samples from the same population.
The formula to determine a confidence interval is:\[ \text{CI} = \bar{x} \pm z \times \frac{s}{\sqrt{n}} \]Here, \(\bar{x}\) is the sample mean, \(z\) is the z-score that corresponds to the desired confidence level, and \(\frac{s}{\sqrt{n}}\) is the standard error.
The exercise presents a confidence interval as \(251 \pm 45\), which implies that the observed mean might differ from the true population mean by 45 milliseconds.
However, if we want to confirm the actual level of confidence of this interval, we need to compute the z-score from this width (90 milliseconds, as it's twice the SEM). This requires understanding the underlying distribution and utilizing statistical tables to interpret the confidence corresponding to the given interval.
Standard Error
The standard error (SE) is a key concept in statistics, reflecting the expected variability of a sample mean across different samples. It provides insight into how precise our estimate of the population mean is likely to be.
In the context of the exercise, the standard error is given as 45 milliseconds. This number helps us understand the typical difference we might expect between the sample mean and the true population mean if we gathered several samples of the same size.
Mathematically, standard error is calculated by:\[ \text{SE} = \frac{s}{\sqrt{n}} \]where \(s\) is the sample standard deviation, and \(n\) is the sample size. With an SE of 45 and a calculated sample standard deviation of 155.88 milliseconds, we gain insight into the spread of blink interval measurements from the subjects.
Standard error is integral to constructing confidence intervals and performing hypothesis tests, as it provides a tool for estimating how close the sample statistics are to the actual population parameters. By understanding SE, we can better assess the reliability and accuracy of our statistical estimates.

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Most popular questions from this chapter

You have measured the systolic blood pressure of an SRS of 25 company employees. A \(95 \%\) confidence interval for the mean systolic blood pressure for the employees of this company is \((122,138) .\) Which of the following statements is true? (a) \(95 \%\) of the sample of employees have a systolic blood pressure between 122 and 138 . (b) \(95 \%\) of the population of employees have a systolic blood pressure between 122 and 138 . (c) If the procedure were repeated many times, \(95 \%\) of the resulting confidence intervals would contain the population mean systolic blood pressure. (d) If the procedure were repeated many times, \(95 \%\) of the time the population mean systolic blood pressure would be between 122 and 138 . (e) If the procedure were repeated many times, \(95 \%\) of the time the sample mean systolic blood pressure would be between 122 and 138 .

Tonya wants to estimate what proportion of her school's seniors plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom. (a) Identify the population and parameter of interest. (b) Check conditions for constructing a confidence interval for the parameter. (c) Construct a \(90 \%\) confidence interval for \(p\). Show your method. (d) Interpret the interval in context.

High school students who take the SAT Math exam a second time generally score higher than on their first try. Past data suggest that the score increase has a standard deviation of about 50 points. How large a sample of high school students would be needed to estimate the mean change in SAT score to within 2 points with \(95 \%\) confidence? Show your work.

One reason for using a \(t\) distribution instead of the standard Normal curve to find critical values when calculating a level \(C\) confidence interval for a population mean is that (a) \(z\) can be used only for large samples. (b) \(z\) requires that you know the population standard deviation \(\sigma\). (c) \(z\) requires that you can regard your data as an \(\mathrm{SRS}\) from the population. (d) \(z\) requires that the sample size is at most \(10 \%\) of the population size. (e) a \(z\) critical value will lead to a wider interval than a \(t\) critical value.

Check whether each of the conditions is met for calculating a confidence interval for the population proportion \(\bar{p}\). Latoya wants to estimate what proportion of the seniors at her boarding high school like the cafeteria food. She interviews an SRS of 50 of the 175 seniors living in the dormitory. She finds that 14 think the cafeteria food is good.

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