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A sample of 20 students who had recently taken elementary statistics yielded the following information on brand of calculator owned \((T=\) Texas Instruments, \(\mathrm{H}=\) Hewlett-Packard, \(\mathrm{C}=\) Casio, \(S=\) Sharp): \(\begin{array}{cccccccccc}\mathrm{T} & \mathrm{T} & \mathrm{H} & \mathrm{T} & \mathrm{C} & \mathrm{T} & \mathrm{T} & \mathrm{S} & \mathrm{C} & \mathrm{H} \\\ \mathrm{S} & \mathrm{S} & \mathrm{T} & \mathrm{H} & \mathrm{C} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{H} & \mathrm{T}\end{array}\) a. Estimate the true proportion of all such students who own a Texas Instruments calculator. b. Of the ten students who owned a TI calculator, 4 had graphing calculators. Estimate the proportion of students who do not own a TI graphing calculator.

Short Answer

Expert verified
a. 50% own a TI calculator. b. 30% do not own a TI graphing calculator.

Step by step solution

01

Count the Number of Texas Instruments Calculators

Review the given data to determine how many students own a Texas Instruments calculator. Look through the list: \( \mathrm{T}, \mathrm{T}, \mathrm{H}, \mathrm{T}, \mathrm{C}, \mathrm{T}, \mathrm{T}, \mathrm{S}, \mathrm{C}, \mathrm{H} \) \( \mathrm{S}, \mathrm{S}, \mathrm{T}, \mathrm{H}, \mathrm{C}, \mathrm{T}, \mathrm{T}, \mathrm{T}, \mathrm{H}, \mathrm{T} \). Count the occurrences of \( \mathrm{T} \). In total, there are 10 occurrences of \( \mathrm{T} \).
02

Calculate the Proportion of Texas Instruments Owners

The total sample size is 20. To find the proportion of students owning a Texas Instruments calculator, divide the number of Texas Instruments calculators by the total number of students: \( \frac{10}{20} = 0.5 \). This implies 50% own a Texas Instruments calculator.
03

Determine Non-Graphing TI Calculator Owners Proportion

Of the 10 Texas Instruments owners, 4 have graphing calculators, which means \( 10 - 4 = 6 \) do not have graphing calculators. Calculate the proportion of TI owners without graphing calculators: \( \frac{6}{20} = 0.3 \). Thus, 30% of the students do not own a TI graphing calculator.

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Key Concepts

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

Proportion Estimation
Proportion estimation is a statistical method used to approximate the percentage of a particular group within a larger population. It involves dividing the number of individuals in the group by the total number of individuals in the population. This simple yet effective tool allows statisticians and researchers to make inferences about a broader population based on the samples collected from it.

In the given exercise, students calculated the proportion of students who own a Texas Instruments calculator. By counting the number of occurrences of 'T,' which is 10 out of the 20 students sampled, they determined the proportion as \( \frac{10}{20} = 0.5 \). This tells us that 50% of the students in the sample use a Texas Instruments calculator.

Proportion estimation is crucial in various fields including market research, healthcare, and social sciences to predict behaviors or traits across large groups. It's a dynamic way of gauging information from a sample to infer about a whole. By using sample data correctly, researchers can reliably assume trends without studying the entire population, saving time and resources.
Data Analysis
Data analysis is the process of systematically applying statistical and logical techniques to illustrate, describe, condense, and evaluate data. It is a fundamental skill in any statistics-related field and is essential for deriving meaningful patterns and observations from raw data.

For the given exercise, the analysis started with a thorough review of the dataset. The students classified each calculator type—Texas Instruments, Hewlett-Packard, Casio, and Sharp—and counted occurrences to infer proportion statistics. This basic form of analysis uncovers how common each calculator brand is among a group of students, providing critical insights into their preferences.

The calculated proportions assist in comparing data and making predictions without having to review an entire dataset each time. Regularly used in business, public policy, and research, data analysis allows decision-makers to see the bigger picture and make informed conclusions about the dataset they're examining. This method helps to leverage collected data to boost efficiency, reveal trends, and create solutions tailored to specific needs.
Calculator Brands
Calculator brands such as Texas Instruments, Hewlett-Packard, Casio, and Sharp play a vital role in both academic and professional fields. Each brand offers a variety of models catering to different functionalities and user preferences, from basic calculators to advanced graphing tools.

In the provided data, Texas Instruments appears to be the dominant brand among the students. Known for their reliable series, especially in graphing calculators, Texas Instruments provide models widely used in high schools and universities. The exercise reveals an interesting behavior where not all TI-calculator users opt for graphing models; out of 10 users, 6 use non-graphing types.

Hewlett-Packard calculators also show a significant presence. They're praised for their robust build and computational power, often preferred in engineering and technical fields. Casio and Sharp offer a blend of functionality and affordability, making them popular for basic calculation needs.

Understanding the different calculator brands and their features helps students and professionals select a calculator based on specific activities and needs. Picking the right calculator can significantly impact user efficiency and ease in solving complex problems. Whether it's a recommendation for a course or personal preference, recognizing the strengths of each brand reinforces better learning and working practices.

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Most popular questions from this chapter

At time \(t=0\), there is one individual alive in a certain population. A pure birth process then unfolds as follows. The time until the first birth is exponentially distributed with parameter \(\lambda\). After the first birth, there are two individuals alive. The time until the first gives birth again is exponential with parameter \(\lambda\), and similarly for the second individual. Therefore, the time until the next birth is the minimum of two exponential \((\lambda)\) variables, which is exponential with parameter 2\lambda. Similarly, once the second birth has occurred, there are three individuals alive, so the time until the next birth is an exponential rv with parameter \(3 \lambda\), and so on (the memoryless property of the exponential distribution is being used here). Suppose the process is observed until the sixth birth has occurred and the successive birth times are \(25.2,41.7,51.2,55.5,59.5,61.8\) (from which you should calculate the times between successive births). Derive the mle of \(\lambda\).

Consider a random sample \(X_{1}, \ldots, X_{n}\) from the pdf $$ f(x ; \theta)=.5(1+\theta x) \quad-1 \leq x \leq 1 $$ where \(-1 \leq \theta \leq 1\) (this distribution arises in particle physics). Show that \(\hat{\theta}=3 \bar{X}\) is an unbiased estimator of \(\theta\). [Hint: First determine \(\mu=E(X)=E(\bar{X}) .\).]

Suppose the true average growth \(\mu\) of one type of plant during a 1-year period is identical to that of a second type, but the variance of growth for the first type is \(\sigma^{2}\), whereas for the second type, the variance is \(4 \sigma^{2}\). Let \(X_{1}, \ldots, X_{m}\) be \(m\) independent growth observations on the first type [so \(\left.E\left(X_{i}\right)=\mu, V\left(X_{i}\right)=\sigma^{2}\right]\), and let \(Y_{1}, \ldots, Y_{n}\) be \(n\) independent growth observations on the second type \(\left[E\left(Y_{i}\right)=\mu, V\left(Y_{i}\right)=4 \sigma^{2}\right]\). Let \(c\) be a numerical constant and consider the estimator \(\hat{\mu}=c \bar{X}+(1-c) \bar{Y}\). For any \(c\) between 0 and 1 this is a weighted average of the two sample means, e.g., \(.7 \bar{X}+.3 \bar{Y}\) a. Show that for any \(c\) the estimator is unbiased. b. For fixed \(m\) and \(n\), what value \(c\) minimizes \(V(\hat{\mu})\) ? [Hint: The estimator is a linear combination of the two sample means and these means are independent. Once you have an expression for the variance, differentiate with respect to \(c\).]

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For \(\theta>0\) consider a random sample from a uniform distribution on the interval from \(\theta\) to \(2 \theta\) (pdf \(1 / \theta\) for \(\theta

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