/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 The Monterey Bay Aquarium, found... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The Monterey Bay Aquarium, founded in 1984 , is situated on the beautiful coast of Monterey Bay in the historic Cannery Row district. In 1985 , the aquarium began a survey program that involved randomly sampling visitors as they exit for the day. The survey includes visitor demographic information, use of social media, and opinions on their aquarium visit. In 2015, the survey included 356 visitors over age 65 , of which 52 used a mobile device such as an Android phone or iPad during their visit. \({ }^{11}\) (a) What is the margin of error of the large-sample \(95 \%\) confidence interval for the proportion of visitors over 65 who used a mobile device during their visit? (b) How large a sample is needed to get the common \(\pm 3\) percentage point margin of error? Use the 2015 sample as a pilot study to get \(p^{*}\).

Short Answer

Expert verified
(a) The margin of error is about 3.51%. (b) A sample size of 506 is needed.

Step by step solution

01

Calculate the sample proportion

To find the sample proportion of visitors over 65 who used a mobile device, divide the number of users by the total number of visitors over 65. So, \( p = \frac{52}{356} \approx 0.1461 \).
02

Determine the margin of error formula

The margin of error for a large-sample \( 95\% \) confidence interval is calculated using the formula: \( E = Z_{95\%} \times \sqrt{\frac{p(1-p)}{n}} \), where \( Z_{95\%} \approx 1.96 \) for \( 95\% \) confidence, \( p \) is the sample proportion, and \( n \) is the sample size.
03

Calculate the margin of error

Substitute the values into the formula: \( E = 1.96 \times \sqrt{\frac{0.1461 \times (1-0.1461)}{356}} \). Calculating this gives \( E \approx 0.0351 \), or about \( 3.51\% \).
04

Use the formula for desired sample size

For a margin of error (E) of ±3 percentage points (\(0.03\)), we'll use the formula: \( n = \frac{Z_{95\%}^2 \times p^*(1-p^*)}{E^2} \). Substitute \( p^* = 0.1461 \) and \( E = 0.03 \), yielding \( n = \frac{1.96^2 \times 0.1461 \times 0.8539}{0.03^2} \).
05

Calculate the required sample size

Perform the calculation in the previous step: \( n \approx \frac{1.96^2 \times 0.1461 \times 0.8539}{0.0009} \approx 505.86 \). Rounding up, a sample size of 506 is needed.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Margin of Error
The margin of error is a key concept in statistics that indicates the range within which the true proportion of a population is likely to fall. When conducting surveys like the one at Monterey Bay Aquarium, it's crucial to know how accurate the results are.

The margin of error depends on the confidence level you choose. In our example, the confidence level is 95%. This means we are 95% confident that the true proportion of visitors over 65 using a mobile device falls within the calculated range.

To find the margin of error, we use the formula:
  • \[ E = Z_{95 ext{%}} \times \sqrt{\frac{p(1-p)}{n}} \]
  • Here, \(E\) is the margin of error, \(Z_{95\%} \approx 1.96\) is the z-score for a 95% confidence level, \(p\) is the sample proportion, and \(n\) is the sample size.
In this case, after doing the math, the margin of error turned out to be about 3.51%. This means you can expect a variation of 3.51% above or below the calculated proportion in the true population.
Confidence Interval
A confidence interval provides a range of values that likely contains the population parameter. When the Monterey Bay Aquarium obtained survey results, they sought a 95% confidence interval for the proportion of visitors over 65 using mobile devices.

The confidence interval is constructed around a central value known as the sample proportion, which in this instance is approximately 0.1461. This represents the proportion of surveyed individuals who used a mobile device.

To construct the confidence interval, we add and subtract the margin of error from the sample proportion:
  • Lower limit: \( p - E = 0.1461 - 0.0351 \approx 0.1110 \)
  • Upper limit: \( p + E = 0.1461 + 0.0351 \approx 0.1812 \)
Thus, the 95% confidence interval ranges from approximately 11.10% to 18.12%. This interval embodies our 95% assurance that the true proportion of mobile device users among visitors over 65 years old falls within these limits.
Sample Size Calculation
Determining the right sample size is vital for obtaining accurate survey results. With the Monterey Bay Aquarium survey, the challenge was ensuring the margin of error remained within ±3 percentage points.

To find this sample size, we can use a specific formula:
  • \[ n = \frac{Z_{95\%}^2 \times p^*(1-p^*)}{E^2} \]
  • Here, \(n\) is the required sample size, \(p^* = 0.1461\) is the estimated proportion from the pilot study, \(Z_{95\%} = 1.96\), and \(E = 0.03\) is the desired margin of error.
After substituting the values, the calculated sample size is roughly 506 visitors.

By aiming for this sample size, the aquarium can be reasonably sure that their findings will accurately represent the broader visitor population, maintaining a margin of error of just ±3 percentage points.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Exercise \(22.25\) describes a Harris Poll survey of smokers in which 848 of a sample of 1010 smokers agreed that smoking would probably shorten their lives. Harris announces a margin of error of \(\pm 3\) percentage points for all samples of about this size. Opinion polls announce the margin of error for \(95 \%\) confidence. (a) What is the actual margin of error (in percent) for the large-sample confidence interval from this sample? (b) The margin of error is largest when \(\mathrm{p}^{\wedge \hat{p}}=0.5\). What would the margin of error (in percent) be if the sample had resulted in \(\mathrm{p}^{\wedge \hat{p}}=0.5\) ? (c) Why do you think that Harris announces a \(\pm 3\) margin of error for all samples of about this size?

A random digit dialing telephone survey of 880 drivers asked, "Recalling the last 10 traffic lights you drove through, how many of them were red when you entered the intersections?" Of the 880 respondents, 171 admitted that at least one light had been red. \({ }^{24}\) (a) Give a \(95 \%\) confidence interval for the proportion of all drivers who ran one or more of the last 10 red lights they met. (b) Nonresponse is a practical problem for this survey-only \(21.6 \%\) of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: do you think more or fewer than 171 of the 880 respondents really ran a red light? Why?

An opinion poll asks an SRS of 100 college seniors how they view their job prospects. In all, 53 say Good. The large-sample \(95 \%\) confidence interval for estimating the proportion of all college seniors who think their job prospects are good is (a) \(0.530 \pm 0.082 .\) (b) \(0.530 \pm 0.098 .\) (c) \(0.530 \pm 0.049\).

Does the order in which wine is presented make a difference? Several choices of wine are presented one at a time and in sequence, and the subject is then asked to choose the preferred wine at the end of the sequence. In this study, subjects were asked to taste two wine samples in sequence. Both samples given to a subject were the same wine, although subjects were expecting to taste two different samples of a particular variety. Of the 32 subjects in the study, 22 selected the wine presented first, when presented with two identical wine samples. 29 (a) Do the data give good reason to conclude that the subjects are not equally likely to choose either of the two positions when presented with two identical wine samples in sequence? (b) The subjects were recruited in Ontario, Canada, via advertisements to participate in a study of "attitudes and values toward wine." Can we generalize our conclusions to all wine tasters? Explain

A high-school teacher in a lowincome urban school in Worcester, Massachusetts, used cash incentives to encourage student learning in his AP statistics class. \({ }^{28}\) In 2010,15 of the 61 students enrolled in his class scored a 5 on the AP statistics exam. Worldwide, the proportion of students who scored a 5 in 2010 was \(0.15\). Is this evidence that the proportion of students who would score a 5 on the AP statistics exam when taught by the teacher in Worcester using cash incentives is higher than the worldwide proportion of \(0.15\) ? (a) State hypotheses, find the \(P\)-value, and give your conclusions in the context of the problem. Do you have any reservations about using the \(z\) test for proportions for this data? (b) Does this study provide evidence that cash incentives cause an increase in the proportion of \(5 \mathrm{~s}\) on the AP statistics exam? Explain your answer.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.