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Chapter 10: Problem 10

An insurance company is concerned that garage A charges more for repair work than garage B charges. It plans to send 25 cars to each garage and obtain separate estimates for the repairs needed for each car. a. How can the company do this and obtain independent samples? Explain in detail. b. How can the company do this and obtain dependent samples? Explain in detail.

Short Answer

Expert verified
a. The company can obtain independent samples by selecting two different sets of 25 cars and sending one set to garage A and the other to garage B. b. The company can obtain dependent samples by sending the same 25 cars to both garages for quotations.

Step by step solution

01

Independent Samples Explanation

In this case, the insurance company will need to randomly select 25 cars and send them to garage A, and then randomly select another new set of 25 cars and send them to garage B. Since the company does not use the same cars' estimates in both garages, these samples could be considered independent. The cost estimate of repair from garage A will not affect or have any relationship with the cost estimate of repair from garage B.
02

Dependent Samples Explanation

To obtain dependent samples, the insurance company needs to send the same 25 cars to both garages, A and B. In this case, you have a 'before and after' situation where the same car is evaluated by both garages. Therefore, measurements are taken on the same item two times, thus the cost estimation of one garage will be dependent on the cost estimation from the other garage. It is also called a paired sample because there is a one-to-one correspondence between the measurements. The car that was 1st in garage A is also 1st in garage B, the car that was 2nd in garage A is also 2nd in garage B, and so on.

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Key Concepts

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

Random Sampling
Imagine you are choosing items from a large pool, and each item has an equal chance of being chosen. That's random sampling! This method is essential in statistics to ensure that samples are unbiased and that any claims made from these samples can be generalized to the larger population.
To conduct random sampling, each car in this scenario should have the same chance of being sent to garage A or B. This can be done using:
  • A computer program to randomly select the cars
  • Picking cars from a shuffled list
  • Drawing lots if the list is not too large
Random sampling helps eliminate any bias and ensures that the results are not skewed by certain characteristics of the cars that could affect their repair costs.
Paired Samples
Consider paired samples as having a buddy system. Each data point has a corresponding match or pair. In this insurance company scenario, this could mean sending the same car to both garages to get repair estimates.
With paired samples, both estimates are linked to the same car. Hence, the results are dependent on each other. Here's why they're handy:
  • They help measure the agreement between two services or tools
  • They account for the natural variability in the items being measured
Paired samples are often used in situations where you want to compare the same thing under two conditions, like testing the same car in two different repair shops.
Experimental Design
Experimental design is the blueprint for a study. It dictates how to collect data, who to sample, and how to analyze the result. In our car repair example, experimental design will determine how to effectively assess the cost differences between the two garages.
The design process helps to ensure that the tests are scientific and unbiased. Key components include:
  • Clear objectives: What is the main question being answered?
  • Type of samples (independent vs. dependent)
  • Method of randomization to prevent selection bias
A good experimental design allows researchers to draw accurate conclusions while controlling for factors that could skew the data, such as differences between individual cars or repair processes.

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Most popular questions from this chapter

Two independent random samples of size 25 were taken from an English class and a chemistry class at a local community college. Students in both classes were asked to draw a 3 -inch line to the best of their ability without any measuring device (ruler, etc.). The following data resulted: $$\begin{array}{llll}\hline \text { English } & n=25 & \bar{x}=2.660 & s=0.617 \\\\\text { Chemistry } & \mathrm{n}=25 & \bar{x}=2.750 & \mathrm{s}=0.522 \\\\\hline\end{array}$$ At the 0.05 level of significance, is there a difference between the standard deviations for 3 -inch-line measurements from the English and chemistry classes?

Determine the \(p\) -value for each hypothesis test for the mean difference. a. \(\quad H_{o}: \mu_{d}=0\) and \(H_{a}: \mu_{d}>0,\) with \(n=20\) and \(t \star=1.86\) b. \(\quad H_{o}: \mu_{d}=0\) and \(H_{a}: \mu_{d} \neq 0,\) with \(n=20\) and \(t \star =-1.86\) c. \(\quad H_{o}: \mu_{d}=0\) and \(H_{a}: \mu_{d}<0,\) with \(n=29\) and \(t \star =-2.63\) d. \(\quad H_{o}: \mu_{d}=0.75\) and \(H_{a}: \mu_{d}>0.75,\) with \(n=10\) and \(t \star =3.57\)

At a large university, a mathematics placement exam is administered to all students. Samples of 36 male and 30 female students are randomly selected from this year's student body and the following scores recorded: $$\begin{array}{lllllllllll}\hline \text { Male } & 72 & 68 & 75 & 82 & 81 & 60 & 75 & 85 & 80 & 70 \\\& 71 & 84 & 68 & 85 & 82 & 80 & 54 & 81 & 86 & 79 \\\& 99 & 90 & 68 & 82 & 60 & 63 & 67 & 72 & 77 & 51 \\\& 61 & 71 & 81 & 74 & 79 & 76 & & & & \\\\\hline \text { Female } & 81 & 76 & 94 & 89 & 83 & 78 & 85 & 91 & 83 & 83 \\\& 84 & 80 & 84 & 88 & 77 & 74 & 63 & 69 & 80 & 82 \\\& 89 & 69 & 74 & 97 & 73 & 79 & 55 & 76 & 78 & 81 \\ \hline\end{array}$$ a. Describe each set of data with a histogram (use the same class intervals on both histograms), the mean, and the standard deviation. b. Construct \(95 \%\) confidence interval for the mean score for all male students. Do the same for all female students. c. Do the results found in part b show that the mean scores for males and females could be the same? Justify your answer. Be careful! d. Construct the \(95 \%\) confidence interval for the difference between the mean scores for male and female students. e. Do the results found in part d show that the mean scores for male and female students could be the same? Explain. f. Explain why the results in part b cannot be used to draw conclusions about the difference between the two means.

A study comparing attitudes toward death was conducted in which organ donors (individuals who had signed organ donor cards) were compared with nondonors. The study is reported in the journal Death Studies. Templer's Death Anxiety Scale (DAS) was administered to both groups. On this scale, high scores indicate high anxiety concerning death. The results were reported as follows.$$\begin{array}{lccc} & n & \text { Mean } & \text { Std. Dev. } \\\\\hline \text { Organ Donors } & 25 & 5.36 & 2.91 \\\\\text { Nonorgan Donors } & 69 & 7.62 & 3.45 \\\\\hline\end{array}$$Construct the \(95 \%\) confidence interval for the difference between the means, \(\mu_{\text {non }}-\mu_{\text {donor: }}\)

In a survey of 300 people from city \(A, 128\) prefer New Spring soap to all other brands of deodorant soap. In city \(\mathrm{B}, 149\) of 400 people prefer New Spring soap. Find the \(98 \%\) confidence interval for the difference in the proportions of people from the two cities who prefer New Spring soap.
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