Ch. 7 - Estimating Parameters and Determining Sample Sizes

Triola - Elementary Statistics 14th Edition

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Triola 14th Edition

Ch. 7 - Estimating Parameters and Determining Sample Sizes

Problem 7.1.36b

Chapter 7, Problem 7.1.36b

Online Gambling Some states now allow online gambling. As a marketing manager for a casino, you need to determine the percentage of adults in those states who gamble online. How many adults must you survey in order to be 99% confident that your estimate is in error by no more than two percentage points?

b. Assume that 18% of all adults gamble online (based on 2017 data from a Gambling Commission study in Great Britain).

Verified step by step guidance

Step 1: Identify the key components of the problem. The confidence level is 99%, the margin of error (E) is 2 percentage points (0.02), and the estimated proportion (p̂) is 18% (0.18). The formula for determining the required sample size is: n = (Z² * p̂ * (1 - p̂)) / E².

Step 2: Determine the Z-value corresponding to a 99% confidence level. For a 99% confidence level, the Z-value is approximately 2.576. This value is derived from the standard normal distribution table.

Step 3: Substitute the values into the formula. Using the formula n = (Z² * p̂ * (1 - p̂)) / E², substitute Z = 2.576, p̂ = 0.18, and E = 0.02. The formula becomes: n = ((2.576)² * 0.18 * (1 - 0.18)) / (0.02)².

Step 4: Simplify the expression inside the formula. First, calculate (1 - p̂), which is (1 - 0.18) = 0.82. Then calculate the numerator: (2.576)² * 0.18 * 0.82. Finally, calculate the denominator: (0.02)².

Step 5: Divide the numerator by the denominator to find the required sample size. Round up the result to the nearest whole number, as the sample size must be an integer.

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Key Concepts

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

Sample Size Determination

Sample size determination is a statistical method used to calculate the number of observations or replicates needed in a survey to achieve a desired level of confidence and precision. In this context, it involves using the margin of error, confidence level, and population proportion to find the minimum number of adults to survey to ensure that the estimate of online gambling participation is accurate within two percentage points.

Confidence Level

The confidence level represents the degree of certainty that the true population parameter lies within the confidence interval calculated from the sample data. A 99% confidence level indicates that if the same survey were conducted multiple times, 99% of the calculated intervals would contain the true proportion of adults who gamble online. This high level of confidence is crucial for making reliable marketing decisions.

Margin of Error

The margin of error quantifies the range within which the true population parameter is expected to fall, based on the sample data. In this scenario, a margin of error of two percentage points means that the estimate of the percentage of adults who gamble online could be two points higher or lower than the calculated sample proportion. This concept is essential for understanding the precision of the survey results.

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Finding the Minimum Sample Size Needed for a Confidence Interval

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

Mean Pulse Rate of Females Data Set 1 “Body Data” in Appendix B includes pulse rates of 147 randomly selected adult females, and those pulse rates vary from a low of 36 bpm to a high of 104 bpm. Find the minimum sample size required to estimate the mean pulse rate of adult females. Assume that we want 99% confidence that the sample mean is within 2 bpm of the population mean.

b. Assume that sigma=12.5 bpm, based on the value of s=12.5 bpm for the sample of 147 female pulse rates.

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Textbook Question

Minting Quarters Listed below are weights (grams) of quarters minted after 1964 (based on Data Set 40 “Coin Weights” in Appendix B).

b. Specifications require that the quarters have a weight of 5.670 g. What does the confidence interval suggest about that specification?

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Textbook Question

7. FRESHMAN 15 Here is a sample of amounts of weight change (kg) of college students in their freshman year (from Data Set 13 “Freshman 15” in Appendix B): 11, 3, 0, –2, where –2 represents a loss of 2 kg and positive values represent weight gained. Here are ten bootstrap samples:

{11, 11, 11, 0}, {11, –2, 0, 11}, {11, –2, 3, 0}, {3, –2, 0, 11}, {0, 0, 0, 3}, {3, –2, 3, –2}, {11, 3, –2, 0}, {–2, 3, –2, 3}, {–2, 0, –2, 3}, {3, 11, 11, 11}.

b. Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the standard deviation of the weight changes for the population.

132

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Textbook Question

Comparing Waiting Lines

The values listed below are waiting times (in minutes) of customers at the Bank of Providence, where customers may enter any one of three different lines that have formed at three teller windows. Construct a 95% confidence interval for the population standard deviation sigma.

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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.

101

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