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A sample of 20 students who had recently taken elementary statistics yielded the following information on brand of calculator owned \((\mathrm{T}=\) Texas Instruments, \(\mathrm{H}=\) Hewlett Packard, \(\mathrm{C}=\) Casio, \(\mathrm{S}=\) Sharp): \(\begin{array}{llllllllll}\text { T } & \text { T } & H & \text { T } & \text { C } & \text { T } & \text { T } & \text { S } & \text { C } & \text { H }\end{array}\) \(\begin{array}{lllllllllll}S & S & T & H & C & T & T & T & H & T\end{array}\) a. Estimate the true proportion of all such students who own a Texas Instruments calculator. b. Of the 10 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. 0.5 b. 0.3

Step by step solution

01

Count the Total Number of Students

There is a provided list of 20 students. Each entry in the list represents one student who owns one brand of calculator. Confirm that there are indeed 20 total entries across both lists.
02

Count the Students with a TI Calculator

Go through the list and count the number of 'T' entries, which represent the students who own a Texas Instruments calculator. Count them one by one to ensure accuracy.
03

Calculate the Proportion of TI Owners

Divide the number of students who own a Texas Instruments calculator by the total number of students to estimate the proportion. If 10 out of 20 students own a TI calculator, the proportion is calculated as \( \frac{10}{20} = 0.5 \).
04

Calculate the Number of Students Without a TI Graphing Calculator

Subtract the number of TI owners with a graphing calculator from the total number of TI owners. If 10 students own a TI calculator and 4 of those have a graphing calculator, then \( 10 - 4 = 6 \) students do not have a TI graphing calculator.
05

Calculate the Proportion of Non-TI Graphing Calculator Owners

Calculate the proportion of students without a TI graphing calculator by dividing the number from Step 4 by the total number of students. The proportion is \( \frac{6}{20} = 0.3 \).

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Key Concepts

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

Elementary Statistics
Understanding elementary statistics is crucial for analyzing data. When dealing with statistical problems, we often aim to estimate unknown parameters, like the proportion of students owning a particular brand of calculator. This branch of mathematics helps us make sense of raw data and draw conclusions.
One core element of statistics is recognizing the difference between population and sample data. A population includes all members of a specified group, whereas a sample is a subset meant to represent that group. For instance, if we want to determine the proportion of students owning Texas Instruments calculators in a high school, we may only survey a sample rather than the entire student body. This makes our investigation more manageable and less resource-intensive.
In this problem, the sample comprises 20 students, which acts as a small-scale representation of a larger group. We compute the proportion of Texas Instruments owners within this sample to infer about the potential trend in the larger population. Remember, statistics is about generalizing results from a sample to a larger group in a valid manner.
Data Analysis
Data analysis involves processing and interpreting numerical data to extract meaningful insights. In statistics, analyzing data means breaking down the information, interpreting the figures, and determining patterns or relationships. This exercise involves counting occurrences and calculating proportions to predict trends.
Let’s consider steps in data analysis in this context:
  • Collect Data: The data is already provided in the form of calculator brands owned by 20 students.
  • Summarize Data: By counting how many students own each brand, you can summarize the dataset. Here, you count the number of 'T's for Texas Instruments' ownership.
  • Analyze Results: Compute the proportion of Texas Instruments owners by dividing the count by the total number of students (e.g., \( \frac{10}{20} = 0.5 \)), which helps in understanding the dataset's structure.
  • Make Interpretations: From the proportion calculated, one can infer that about 50% of the students in this sample prefer Texas Instruments, suggesting the brand's prominence among them.
Data analysis further involves checking for any peculiarities, understanding possible biases, and ensuring that our conclusions are logically derived from the data at hand.
Sampling
Sampling is the process of selecting a portion of the population to represent the whole. It enables us to gather data and make informed guesses about the entire group without having to conduct a full-scale survey.
In this exercise, a sample of 20 students is used instead of surveying all students in the school. This sample should be a good representative of the population. Here are some points to ensure effective sampling:
  • Representativeness: The sample should reflect the diversity and characteristics of the larger population. Random sampling techniques help achieve this, ensuring each student has an equal chance of being selected.
  • Sample Size: Though larger samples provide better estimates, they are also more resource-consuming. The sample of 20 students can give a rough estimate, but increasing the number of students could provide more accurate results.
  • Avoid Bias: Sampling bias can skew results. For instance, if only students known to prefer specific brands were chosen, it would affect the inference about the population's brand preference.
Ultimately, careful sampling helps statisticians make reliable predictions and uncover trends without the impracticality of surveying entire populations.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) represent a random sample from a Rayleigh distribution with pdf $$ f(x ; \theta)=\frac{x}{\theta} e^{-x^{2} /(2 \theta)} \quad x>0 $$ a. It can be shown that \(E\left(X^{2}\right)=2 \theta\). Use this fact to construct an unbiased estimator of \(\theta\) based on \(\sum X_{i}^{2}\) (and use rules of expected value to show that it is unbiased). b. Estimate \(\theta\) from the following \(n=10\) observations on vibratory stress of a turbine blade under specified conditions: \(\begin{array}{lllll}16.88 & 10.23 & 4.59 & 6.66 & 13.68 \\ 14.23 & 19.87 & 9.40 & 6.51 & 10.95\end{array}\)

Using a long rod that has length \(\mu\), you are going to lay out a square plot in which the length of each side is \(\mu\). Thus the area of the plot will be \(\mu^{2}\). However, you do not know the value of \(\mu\), so you decide to make \(n\) independent measurements \(X_{1}, X_{2}, \ldots X_{n}\) of the length. Assume that each \(X_{i}\) has mean \(\mu\) (unbiased measurements) and variance \(\sigma^{2}\). a. Show that \(\bar{X}^{2}\) is not an unbiased estimator for \(\mu^{2}\). \(\left[\right.\) Hint: For any rv \(Y, E\left(Y^{2}\right)=V(Y)+[E(Y)]^{2}\). Apply this with \(Y=\bar{X}\).] b. For what value of \(k\) is the estimator \(\bar{X}^{2}-k S^{2}\) unbiased for \(\mu^{2}\) ? [Hint: Compute \(E\left(\bar{X}^{2}-k S^{2}\right)\).]

Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from the shifted exponential pdf $$ f(x ; \lambda, \theta)=\left\\{\begin{array}{cl} \lambda e^{-\lambda(x-\theta)} & x \geq \theta \\ 0 & \text { otherwise } \end{array}\right. $$ Taking \(\theta=0\) gives the pdf of the exponential distribution considered previously (with positive density to the right of zero). An example of the shifted exponential distribution appeared in Example \(4.5\), in which the variable of interest was time headway in traffic flow and \(\theta=.5\) was the minimum possible time headway. a. Obtain the maximum likelihood estimators of \(\theta\) and \(\lambda\). b. If \(n=10\) time headway observations are made, resulting in the values \(3.11, .64,2.55,2.20,5.44,3.42,10.39,8.93\), \(17.82\), and \(1.30\), calculate the estimates of \(\theta\) and \(\lambda\).

The article from which the data of Exercise 1 was extracted also gave the accompanying strength observations for cylinders: \(\begin{array}{rrrrrrrrrr}6.1 & 5.8 & 7.8 & 7.1 & 7.2 & 9.2 & 6.6 & 8.3 & 7.0 & 8.3 \\ 7.8 & 8.1 & 7.4 & 8.5 & 8.9 & 9.8 & 9.7 & 14.1 & 12.6 & 11.2\end{array}\) Prior to obtaining data, denote the beam strengths by \(X_{1}, \ldots, X_{m}\) and the cylinder strengths by \(Y_{1}, \ldots, Y_{n^{*}}\). Suppose that the \(X_{i}\) s constitute a random sample from a distribution with mean \(\mu_{1}\) and standard deviation \(\sigma_{1}\) and that the \(Y_{i} \mathrm{~s}\) form a random sample (independent of the \(X_{i} \mathrm{~s}\) ) from another distribution with mean \(\mu_{2}\) and standard deviation \(\sigma_{2^{\circ}}\) a. Use rules of expected value to show that \(\bar{X}^{2}-\bar{Y}\) is an unbiased estimator of \(\mu_{1}-\mu_{2}\). Calculate the estimate for the given data. b. Use rules of variance from Chapter 5 to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a), and then compute the estimated standard error. c. Calculate a point estimate of the ratio \(\sigma_{1} / \sigma_{2}\) of the two standard deviations. d. Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference \(X-Y\) between beam strength and cylinder strength.

Let \(X\) denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of \(X\) is $$ f(x ; \theta)=\left\\{\begin{array}{cl} (\theta+1) x^{\theta} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ where \(-1<\theta\). A random sample of ten students yields data \(x_{1}=.92, x_{2}=.79, x_{3}=.90, x_{4}=.65, x_{5}=.86, x_{6}=.47\), \(x_{7}=.73, x_{8}=.97, x_{9}=.94, x_{10}=.77\). a. Use the method of moments to obtain an estimator of \(\theta\), and then compute the estimate for this data. b. Obtain the maximum likelihood estimator of \(\theta\), and then compute the estimate for the given data.

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