/*! 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 11 A hospital administrator hoping ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A \(95 \%\) confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statements are true or false, and explain your reasoning. (a) We are \(95 \%\) confident that the average waiting time of these 64 emergency room patients is between 128 and 147 minutes. (b) We are \(95 \%\) confident that the average waiting time of all patients at this hospital's emergency room is between 128 and 147 minutes. (c) \(95 \%\) of random samples have a sample mean between 128 and 147 minutes. (d) A \(99 \%\) confidence interval would be narrower than the \(95 \%\) confidence interval since we need to be more sure of our estimate. (e) The margin of error is 9.5 and the sample mean is 137.5 . (f) In order to decrease the margin of error of a \(95 \%\) confidence interval to half of what it is now, we would need to double the sample size. (Hint: the margin of error for a mean scales in the same way with sample size as the margin of error for a proportion.)

Short Answer

Expert verified
(a) False. (b) True. (c) False. (d) False. (e) True. (f) False.

Step by step solution

01

Evaluating Statement (a)

The confidence interval provided allows us to estimate the average waiting time of the entire population of the hospital’s emergency room patients, not just the specific sample of 64 patients. Thus, we are confident that the range 128 to 147 minutes contains the true mean of all ER patients, not just these 64.
02

Evaluating Statement (b)

This statement correctly uses the confidence interval to make an inference about the average waiting time for the entire population of ER patients at the hospital. We use the sample mean to estimate the population mean. Therefore, this statement is true.
03

Evaluating Statement (c)

This statement is false. The confidence interval refers to the range in which we believe the true mean of the population falls, not where sample means fall. If multiple random samples were taken, some sample means might lie outside this interval.
04

Evaluating Statement (d)

This statement is false. A 99% confidence interval would actually be wider, not narrower, than a 95% confidence interval to account for increased certainty, requiring a broader range to ensure that the true mean is captured with higher confidence.
05

Evaluating Statement (e)

To calculate the margin of error: the confidence interval's endpoints are 128 and 147. The sample mean is the midpoint: \( (128 + 147)/2 = 137.5 \). The margin of error is half the width of the interval: \( (147 - 128)/2 = 9.5 \). Therefore, this statement is true.
06

Evaluating Statement (f)

The margin of error for a mean scales with the square root of the sample size. Thus, to halve the margin of error, we need a sample size four times higher, not double (since the square root of 4 is 2). Therefore, this statement is false.

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Key Concepts

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

Margin of Error
The margin of error is a key component in statistics, especially when dealing with confidence intervals. It provides us with a range around the sample mean where the true population mean is likely to be found. Basically, it's the "wiggle room" we allow ourselves in our estimates. This helps account for potential sampling errors.

In our problem, the margin of error is calculated based on the difference between the upper and lower limits of the confidence interval. For a confidence interval that spans from 128 to 147 minutes, the margin of error is half the difference between these two numbers: \[ \text{Margin of Error} = \frac{(147 - 128)}{2} = 9.5 \text{ minutes} \]

This means that the estimated true population mean could be 9.5 minutes lower or higher than the calculated sample mean of 137.5 minutes.
Sample Mean
The sample mean is an average calculated from a set of data points in a sample. It is a significant estimator for deriving the population mean when it's difficult to collect data from the entire population.

In our example, the sample mean is derived from a group of 64 patients in the emergency room. Once calculated, we find the result to be 137.5 minutes. The calculation used is: \[ \text{Sample Mean} = \frac{128 + 147}{2} = 137.5 \]

This value acts as a center point for our confidence interval, indicating our best estimate for the average waiting time based on the data we collected from the sample.
Population Mean
The population mean is the average of a set of values for the whole population. It represents the true central value that we want to estimate using sample data. Calculating the exact population mean often involves too much data or feasibility challenges. Therefore, we estimate it using the sample mean and confidence intervals.

In our emergency room example, we don't have the data for every single patient, but by using the sample mean and the calculated margin of error, we can infer that the true average time patients have to wait is likely between 128 and 147 minutes. The confidence interval gives us a statistical assurance (in this case, at a 95% confidence level) about where that population mean falls.
Confidence Level
The confidence level indicates the degree of certainty (or probability) that the true population parameter lies within the calculated confidence interval. It tells us how 'confident' we are about our interval estimation.

In the hospital's study, a 95% confidence level implies that if we were to take 100 different random samples of the same size, approximately 95 of those confidence intervals would be expected to contain the true population mean. This concept doesn't mean the true mean is between two numbers 95% of the time. Instead, it's about the reliability of our estimation process.

A key aspect to remember is that increasing the confidence level from 95% to, say, 99% would result in a wider interval, not a narrower one. This is because a higher confidence level requires a broader range to ensure it captures the true population mean, thus shifting the trade-off between precision and certainty.

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Most popular questions from this chapter

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail. Construct a \(95 \%\) confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail.

Write the null and alternative hypotheses in words and then symbols for each of the following situations. (a) A tutoring company would like to understand if most students tend to improve their grades (or not) after they use their services. They sample 200 of the students who used their service in the past year and ask them if their grades have improved or declined from the previous year. (b) Employers at a firm are worried about the effect of March Madness, a basketball championship held each spring in the US, on employee productivity. They estimate that on a regular business day employees spend on average 15 minutes of company time checking personal email, making personal phone calls, etc. They also collect data on how much company time employees spend on such non-business activities during March Madness. They want to determine if these data provide convincing evidence that employee productivity changed during March Madness.

A study suggests that \(60 \%\) of college student spend 10 or more hours per week communicating with others online. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. You randomly sample 160 students from your dorm and find that \(70 \%\) spent 10 or more hours a week communicating with others online. A friend of yours, who offers to help you with the hypothesis test, comes up with the following set of hypotheses. Indicate any errors you see. $$ \begin{array}{l} H_{0}: \hat{p}<0.6 \\ H_{A}: \hat{p}>0.7 \end{array} $$

As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. (a) What population is under consideration in the data set? (b) What parameter is being estimated? (c) What is the point estimate for the parameter? (d) What is the name of the statistic can we use to measure the uncertainty of the point estimate? (e) Compute the value from part (d) for this context. (f) The historical rate of defects is \(10 \%\). Should the engineer be surprised by the observed rate of defects during the current week? (g) Suppose the true population value was found to be \(10 \%\). If we use this proportion to recompute the value in part (e) using \(p=0.1\) instead of \(\hat{p},\) does the resulting value change much?

Write the null and alternative hypotheses in words and using symbols for each of the following situations. (a) Since 2008 , chain restaurants in California have been required to display calorie counts of each menu item. Prior to menus displaying calorie counts, the average calorie intake of diners at a restaurant was 1100 calories. After calorie counts started to be displayed on menus, a nutritionist collected data on the number of calories consumed at this restaurant from a random sample of diners. Do these data provide convincing evidence of a difference in the average calorie intake of a diners at this restaurant? (b) The state of Wisconsin would like to understand the fraction of its adult residents that consumed alcohol

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