Ch. 7 - Estimating Parameters and Determining Sample Sizes

Triola - Elementary Statistics 14th Edition

Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook

Triola 14th Edition

Ch. 7 - Estimating Parameters and Determining Sample Sizes

Problem 7.1.37b

Chapter 7, Problem 7.1.37b

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.

Verified step by step guidance

Step 1: Identify the formula for determining the required sample size for estimating a population proportion. The formula is:

n = (Z² * p * (1 - p)) / E²

, where

n

is the sample size,

Z

is the z-score corresponding to the confidence level,

p

is the estimated population proportion, and

E

is the margin of error.

Step 2: Determine the values for the variables in the formula. The confidence level is 90%, so the z-score corresponding to 90% confidence is approximately 1.645. The estimated population proportion

p

is 0.11 (11%), and the margin of error

E

is 0.03 (3%).

Step 3: Substitute the values into the formula. The equation becomes:

n = (1.645² * 0.11 * (1 - 0.11)) / 0.03²

Step 4: Simplify the numerator. Calculate

1.645²

, then multiply it by

0.11

and

(1 - 0.11)

. This will give you the value of the numerator.

Step 5: Simplify the denominator. Calculate

0.03²

. Then divide the numerator by the denominator to find the required sample size

n

. Round up to the nearest whole number, as sample size must be a whole number.

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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 estimated proportion of consumers who plan to purchase the new iPhone and the margin of error to determine how many individuals need to be surveyed to ensure that the results are reliable.

Confidence Level

The confidence level represents the degree of certainty that the true population parameter lies within the estimated range. A 90% confidence level means that if the survey were repeated multiple times, 90% of the time the calculated confidence interval would contain the true population percentage. This concept is crucial for understanding the reliability of the survey results.

Margin of Error

The margin of error indicates the range within which the true population percentage is expected to fall, based on the survey results. In this case, a margin of error of three percentage points means that the estimated percentage from the survey could be three points higher or lower than the actual percentage. This concept is essential for assessing the precision of the survey findings.

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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

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).

118

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Textbook Question

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.

420

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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.

167

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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?

112

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