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Of the 150 elements in a random sample, 45 are classified as "success." a. Explain why \(x\) and \(n\) are assigned the values 45 and \(150,\) respectively. b. Determine the value of \(p^{\prime} .\) Explain how \(p^{\prime}\) is found and the meaning of \(p^{\prime}\). For each of the following situations, find \(p^{\prime}\). c. \(x=24\) and \(n=250\) d. \(x=640\) and \(n=2050\) e. \(892\) of 1280 responded "Yes"

Short Answer

Expert verified
a. \(x = 45\) and \(n = 150\) are used to calculate the sample proportion, which estimates the probability of success. b. \(p' = 0.3\), and is calculated by dividing \(x\) by \(n\). It represents an estimated probability of success in the sample. c. \(p' = 0.096\). d. \(p' = 0.312\). e. \(p' = 0.697\).

Step by step solution

01

Explanation of \(x\) and \(n\)

\(x\) and \(n\) are assigned the values of 45 and 150 because they represent the number of positive outcomes (successes) and the total number of outcomes in the sample, respectively. They are used to find the sample proportion, \(p'\).
02

Calculation of \(p'\) for the First Sample

To calculate the value of \(p'\), divide the number of successes \(x\) by the total number of trials \(n\). This gives \(p' = \frac{45}{150} = 0.3\). This means that in our sample, the estimated probability of success is 0.3.
03

Calculation of \(p'\) for the Subsequent Samples

For the other scenarios, repeat the process from Step 2 to calculate \(p'\) each time. For instance, in situation 'c', \(x = 24\) and \(n = 250\), so \(p' = \frac{24}{250} = 0.096\). Similarly for 'd' and 'e', \(p' = \frac{640}{2050} = 0.312\) and \(p' = \frac{892}{1280} = 0.697\), respectively.

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Key Concepts

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

Random Sampling
When we talk about random sampling in statistics, we're discussing a process that ensures every individual or element within the population has an equal chance of being selected. This method is crucial because it helps eliminate bias, making the results more reliable and truly representative of the wider population. Imagine a jar filled with different colored balls; random sampling would be like picking balls without looking, ensuring each color has an equal chance of being chosen.

Random sampling is used in many scenarios:
  • Polls and surveys, where people's opinions are gathered.
  • Quality control in manufacturing, to test the efficiency of products.
  • Scientific experiments, where samples are tested for research purposes.
By applying random sampling, statisticians and researchers can ensure that their findings are credible and that conclusions can be generalized back to the entire population.
Successes in Statistics
In statistics, "success" doesn't necessarily mean something good is happening; it's about the outcome you're interested in studying. This outcome is what drives the analysis. For the exercise we discussed, a success is when one of the elements in the sample is classified as meeting a certain condition or criterion, like answering 'Yes' in a survey.

"Success" in statistics involves:
  • Defining clearly what a "success" is within your study context.
  • Counting the number of "successes" in your random sample, represented by the variable \(x\).
  • Using this count of successes to make further calculations, such as the probability of success.
By understanding what a success represents in your specific situation, you can accurately measure and interpret your statistical results.
Probability Estimation
Probability estimation is the art of using statistics to predict how likely an event is to occur. In our context, it's about estimating the probability of "success" through the sample proportion, \(p'\). The formula to calculate \(p'\) is straightforward: just divide the number of successes \(x\) by the total number of trials \(n\).

Here's how probability estimation works:
  • Determine the number of successes, \(x\), and total number of events, \(n\).
  • Calculate sample proportion as \(p' = \frac{x}{n}\).
  • Use this proportion to estimate the likelihood of "success" in the population.
This sample proportion \(p'\) gives us an estimate of how often we might expect a similar outcome in a broader population. It's an essential tool for decision-making and prediction in statistics.

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Most popular questions from this chapter

Bright-Lite claims that its 60 -watt light bulb burns with a length of life that is approximately normally distributed with a standard deviation of 81 hours. A sample of 101 bulbs had a variance of \(8075 .\) Is this sufficient evidence to reject Bright-Lite's claim in favor of the alternative,"the standard deviation is larger than 81 hours," at the 0.05 level of significance?

Just one serving a month of kale or collard greens or more than two servings of carrots a week can reduce the risk of glaucoma by more than \(60 \%,\) according to a UCLA study of 1000 women. Using a \(90 \%\) confidence interval for the true binomial proportion based on this random sample of 1000 binomial trials and an observed proportion of \(0.60,\) estimate the proportion of glaucoma risk reduction in women who eat the recommended servings of kale, collard greens, or carrots.

The Pizza Shack in Exercise 9.177 has completed its sampling and the results are in! On Tuesday afternoon, they sampled 15 customers and 9 preferred the new pizza crust. On Friday evening, they sampled 200 customers and 120 preferred the new pizza crust. Help the manager interpret the meaning of these results. Use a one-tailed test with \(H_{a}: p>0.50\) and \(\alpha=0.02 .\) Use \(z\) as the test statistic. a. Is there sufficient evidence to conclude a significant preference for the new crust based on Tuesday's customers? b. Is there sufficient evidence to conclude a significant preference for the new crust based on Friday's customers? c. since the percentage of customers preferring the new crust was the same, \(p^{\prime}=0.60\) in both samplings, explain why the answers in parts a and b are not the same.

A machine is considered to be operating in an acceptable manner if it produces \(0.5 \%\) or fewer defective parts. It is not performing in an acceptable manner if more than \(0.5 \%\) of its production is defective. The hypothesis \(H_{o}: p=0.005\) is tested against the hypothesis \(H_{a}: p>0.005\) by taking a random sample of 50 parts produced by the machine. The null hypothesis is rejected if two or more defective parts are found in the sample. Find the probability of the type I error.

A manufacturer of television sets claims that the maintenance expenditures for its product will average no more than \(110\)dollar during the first year following the expiration of the warranty. A consumer group has asked you to substantiate or discredit the claim. The results of a random sample of 50 owners of such television sets showed that the mean expenditure was \(131.60\)dollar and the standard deviation was \(42.46\)dollar At the 0.01 level of significance, should you conclude that the manufacturer's claim is true or not likely to be true?

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