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Ch. 7 - Estimating Parameters and Determining Sample Sizes
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
Chapter 7, Problem 7.2.34b

Finite Population Correction Factor If a simple random sample of size n is selected without replacement from a finite population of size (n>0.05N), and the sample size is more than 5% of the population size , better results can be obtained by using the finite population correction factor, which involves multiplying the margin of error E by [Image]. Refer to the weights of the M&M candies in Data Set 38 “Candies” in Appendix B.


b. Use only the red M&Ms and treat that sample as a simple random sample selected from the population of the 345 M&Ms listed in the data set. Find the 95% confidence interval estimate of the mean weight of all 345 M&Ms. Compare the result to the actual mean of the population of all 345 M&Ms.

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Step 1: Identify the given values and conditions. From the problem, the population size (N) is 345, and the sample size (n) is such that n > 0.05N, meaning the sample size is more than 5% of the population. This condition necessitates the use of the finite population correction factor.
Step 2: Recall the formula for the finite population correction factor (FPCF), which is: \( \sqrt{\frac{N - n}{N - 1}} \). Here, N is the population size, and n is the sample size. This factor will be used to adjust the margin of error.
Step 3: Calculate the margin of error (E) for the confidence interval. The formula for E is: \( E = z \cdot \frac{s}{\sqrt{n}} \), where z is the z-score corresponding to the 95% confidence level (typically 1.96), s is the sample standard deviation, and n is the sample size. Compute this value first without applying the correction factor.
Step 4: Adjust the margin of error using the finite population correction factor. Multiply the previously calculated margin of error (E) by the correction factor \( \sqrt{\frac{N - n}{N - 1}} \). This gives the corrected margin of error.
Step 5: Construct the 95% confidence interval for the mean weight of the M&Ms. The formula is: \( \text{Confidence Interval} = \bar{x} \pm E \), where \( \bar{x} \) is the sample mean and E is the corrected margin of error. Compare the resulting confidence interval to the actual mean of the population to evaluate the accuracy of the estimate.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Finite Population Correction Factor

The Finite Population Correction Factor (FPC) is used when sampling without replacement from a finite population. It adjusts the margin of error to account for the reduced variability in the sample when a significant portion of the population is sampled. Specifically, when the sample size exceeds 5% of the population, the FPC helps provide a more accurate estimate of the population parameters.
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Population Standard Deviation Known

Confidence Interval

A confidence interval is a range of values, derived from sample statistics, that is likely to contain the true population parameter with a specified level of confidence, typically 95%. It is calculated using the sample mean, the standard error, and a critical value from the normal distribution. This interval provides insight into the precision of the sample estimate and the uncertainty associated with it.
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Introduction to Confidence Intervals

Simple Random Sampling

Simple random sampling is a fundamental sampling technique where each member of the population has an equal chance of being selected. This method ensures that the sample is representative of the population, minimizing bias. In the context of the question, treating the selected red M&Ms as a simple random sample allows for valid statistical inferences about the mean weight of all M&Ms in the population.
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Sampling Distribution of Sample Proportion
Related Practice
Textbook Question

Mint Specs Listed below are weights (grams) from a simple random sample of pennies produced after 1983 (from Data Set 40 “Coin Weights” in Appendix B).


b. How does the result compare to the confidence interval found in Exercise 14 in Section 7-3?


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Textbook Question

Wiggle Your Ears Find the sample size needed to estimate the percentage of adults who can wiggle their ears. Use a margin of error of 3 percentage points and use a confidence level of 99%.


b. Assume that 22% of adults can wiggle their ears (based on data from Soul Publishing).

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Textbook Question

Smart Phone Apple is planning for the launch of a new and improved iPhone. The marketing team wants to know the worldwide percentage of consumers who intend to purchase the new model, so a survey is being planned. How many people must be surveyed in order to be 90% confident that the estimated percentage is within three percentage points of the true population percentage?


b. Assume that 11% of consumers have a smartphone and plan to upgrade to a new model.


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Textbook Question

E-Cigarettes A New York Times article reported that a survey conducted in 2014 included 36,000 adults, with 3.7% of them being regular users of e-cigarettes. Because e-cigarette use is relatively new, there is a need to obtain today’s usage rate. How many adults must be surveyed now if we want a confidence level of 95% and a margin of error of 1.5 percentage points?


b. Use the results from the 2014 survey.


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Textbook Question

Voting Survey In a survey of 1002 people, 70% said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records show that 61% of eligible voters actually did vote.


b. Find a 95% confidence interval estimate of the percentage of people who say that they voted.


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Textbook Question

Caffeine in Soft Drinks Listed below are measured amounts of caffeine (mg per 12 oz of drink) obtained in one can from each of 20 brands (7UP, A&W Root Beer, Cherry Coke, . . . , TaB).


b. Given that Exercise 20 in Section 7-2 used the same data for a 99% confidence interval based on use of the t distribution, and given that the data do not appear to be from a normally distributed population, which confidence interval is likely to be better: The confidence interval from part (a) or the confidence interval found in Exercise 20 in Section 7-2?


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