/*! 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 44 A good book An English professor... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A good book An English professor is attempting to estimate the mean number of novels that the student body reads during their time in college. He is conducting an exit survey with seniors. He hopes to have a margin of error of 3 books with \(95 \%\) confidence. From reading previous studies, he expects a large standard deviation and is going assume it is \(10 .\) How many students should he survey?

Short Answer

Expert verified
The professor should survey 43 students.

Step by step solution

01

Identify the Required Values

To determine the required sample size, we identify the given values. The professor seeks a margin of error (E) of 3 books, a confidence level of 95%, and a standard deviation (σ) of 10.
02

Find the Critical Z-Value

Given the confidence level of 95%, we look up the Z-value in a Z-table (standard normal distribution table). The critical Z-value for a 95% confidence level is approximately 1.96.
03

Apply the Formula for Sample Size

The formula for determining sample size when estimating a population mean is \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] where Z is the Z-value, σ is the standard deviation, and E is the margin of error. Substituting the known values, we have \[ n = \left( \frac{1.96 \cdot 10}{3} \right)^2 \].
04

Simplify the Expression

Calculate inside the parentheses first: \[ \frac{1.96 \cdot 10}{3} = 6.53 \].
05

Square the Result

Square the value obtained in Step 4: \[ n = 6.53^2 = 42.70 \].
06

Round up to the Nearest Whole Number

Since the sample size must be a whole number and rounding down would compromise the margin of error, we round up to the nearest whole number: \( n = 43 \).

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 91Ó°ÊÓ!

Key Concepts

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

Confidence Interval
A confidence interval is a range of values that provides an estimate for a population parameter. It is constructed in such a way that the true parameter value is expected to lie within this range a certain percentage of the time. In our example, the professor aims for a 95% confidence interval. This means that if he conducted the survey multiple times, 95% of those intervals would contain the true mean number of books read by students.
To construct a confidence interval, you need a couple of key elements: a point estimate (like the sample mean), a margin of error, and the standard deviation. The interval is centered around the point estimate and extends in both directions by the size of the margin of error. This ensures that the interval is symmetric around the point estimate.
  • 95% confidence level is common, meaning you can be 95% confident in your estimate.
  • The width of the confidence interval is twice the margin of error.
Using confidence intervals helps to understand the range in which the true mean could fall, and it gives a sense of certainty about the estimate.
Margin of Error
The margin of error (E) indicates the range of values above and below the sample mean that the true population parameter is likely to be found. In simpler terms, it describes how much you expect the survey results to reflect the true population value.
In our particular example, the professor desires a margin of error of 3 books. This simply means that the predicted average number of books read by students might be within 3 books of the actual average. This indicator of accuracy is vital when interpreting survey results.
  • The margin of error is determined by both the confidence level and the standard deviation.
  • Reducing the margin of error requires a larger sample size.
  • A smaller margin of error narrows the confidence interval, leading to a more precise estimate.
In practice, a precise margin of error tells you about the precision of your confidence interval and overall estimate.
Standard Deviation
Standard deviation (\( \ sigma \)) measures the amount of variability or dispersion in a set of data points. A larger standard deviation indicates that data points are spread out over a wider range of values. Conversely, a smaller standard deviation indicates that data points tend to cluster closer to the mean.
In the professor's example, he assumes a standard deviation of 10 based on previous studies, indicating significant variability in the number of books students read. This is crucial because higher variability (or larger standard deviation) often requires a larger sample size to achieve the desired confidence level and margin of error.
  • Standard deviation is critical in determining the spread of your data.
  • Higher standard deviation values usually entail larger sample sizes for accurate results.
  • The calculation of sample size directly relies on the estimated standard deviation.
Understanding your data's standard deviation is a cornerstone for planning effective studies and surveys.
Z-Value
The Z-value, also known as the Z-score, indicates how many standard deviations an element is from the mean. In statistics, it represents the critical value that corresponds to a chosen confidence level. In the professor's survey, a 95% confidence level corresponds to a Z-value of approximately 1.96.
The Z-value is pivotal because it helps to scale the margin of error and adjust the confidence interval width accordingly. Different confidence levels will have different Z-values, affecting the confidence interval margin.
  • Z-values are found using the standard normal distribution.
  • A higher Z-value results in a wider confidence interval, whereas a lower one results in a narrower interval.
  • Z-values are essential for accurate confidence interval calculations.
Accurate determination of Z-values is crucial for calculating confidence intervals and ensuring a reliable representation of your data.

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

Salmon A specialty food company sells whole King Salmon to various customers. The mean weight of these salmon is 35 pounds with a standard deviation of 2 pounds. The company ships them to restaurants in boxes of 4 salmon, to grocery stores in cartons of 16 salmon, and to discount outlet stores in pallets of 100 salmon. To forecast costs, the shipping department nceds to cstimate the standard deviation of the mean weight of the salmon in each type of shipment a) Find the standard deviations of the mean weight of the salmon in each type of shipment. b) The distribution of the salmon weights turns out to be skewed to the high end. Would the distribution of shipping weights be better characterized by a Normal model for the boxes or pallets? Explain.

Accuracy The researchers in the previous problem also measured the accuracy of the two types of tips. The bow was aimed at a target and the distance was measured from the center. a) Here are the data from the six wooden-tipped shots. Find and interpret a \(95 \%\) confidence interval for the measure of accuracy (measured in cm). \(9.3 \quad 16.7\) \(7.1 \quad 14 \quad 1\) 1.2 b) Here are the data from the six stone-tipped shots. Find and interpret a \(95 \%\) interval for the measure of accuracy (measured in cm). \(\begin{array}{cccccc}4.9 & 21.1 & 7 & 1.8 & 5.4 & 8.6\end{array}\)

\- Home sales again In the previous exercise, you found a \(95 \%\) confidence interval to estimate the average loss in home value. a) Suppose the standard deviation of the losses had been S3000 instead of \$1500. What would the larger standard deviation do to the width of the confidence interval (assuming the same level of confidence)? b) Your classmate suggests that the margin of error in the interval could be reduced if the confidence level were changed to \(90 \%\) instead of \(95 \% .\) Do you agree with this statement? Why or why not? c) Instead of changing the level of confidence, would it be more statistically appropriate to draw a bigger sample?

Chips Ahoy In \(1998,\) as an advertising campaign, the Nabisco Company announced a " 1000 Chips Challenge," claiming that every 18 -ounce bag of their Chips Ahoy cookies contained at least 1000 chocolate chips. Dedicated Statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies, and counted the chocolate chips. Some of their data are given below. (Chance, \(12,\) no. 1[1999] ) $$\begin{array}{llllllll} 1219 & 1214 & 1087 & 1200 & 1419 & 1121 & 1325 & 1345 \\ 1244 & 1258 & 1356 & 1132 & 1191 & 1270 & 1295 & 1135 \end{array}$$ a) Check the assumptions and conditions for inference. Comment on any concerns you have. b) Create a \(95 \%\) confidence interval for the average number of chips in bags of Chips Ahoy cookies. c) What does this evidence say about Nabisco's claim? Use your confidence interval to test an appropriate hypothesis and state your conclusion.

\(t\) -models, part II Using the \(t\) tables, software, or a calculator, estimate a) the critical value of \(t\) for a \(95 \%\) confidence interval with df \(=7\) b) the critical value of \(t\) for a \(99 \%\) confidence interval with \(\mathrm{df}=102\) c) the P-value for \(t \leq 2.19\) with 41 degrees of freedom. d) the P-value for \(|t|>2.33\) with 12 degrees of freedom.

See all solutions

Recommended explanations on Math Textbooks

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.