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A company is drafting an advertising campaign that will involve endorsements by noted athletes. For the campaign to succeed, the endorser must be both highly respected and easily recognized. A random sample of 100 prospective customers is shown photos of various athletes. If the customer recognizes an athlete, then the customer is asked whether he or she respects the athlete. In the case of a top woman golfer, 16 of the 100 respondents recognized her picture and indicated that they also respected her. At the \(95 \%\) level of confidence, what is the true proportion with which this woman golfer is both recognized and respected?

Short Answer

Expert verified
The true proportion with which this woman golfer is both recognized and respected, at a 95% confidence level, is between 8.8% and 23.2%.

Step by step solution

01

Identify the Sample Size and Number of Success

From the problem, it is clear that the total sample size 'n' is 100 and the number of successful or positive outcomes 'X' (customers recognizing and respecting the athlete) is 16.
02

Calculate the sample proportion (PÌ‚)

The sample proportion (PÌ‚) is found by dividing the number of positive outcomes by the total sample size. In this case, PÌ‚ = X/n = 16/100 = 0.16.
03

Find the Standard Error of Proportion

The standard error of the proportion (SE) is calculated using the formula \(\sqrt{PÌ‚(1-PÌ‚)/n }\). So, SE = \(\sqrt{0.16 * 0.84 / 100} \approx 0.037 . \)
04

Find the Z value for 95% Confidence Level

The value of Z for a 95% confidence level is approximately \(1.96\). This value is found in the Z-table or standard normal distribution table.
05

Calculate the Confidence Interval for True Proportion

The 95% confidence interval for the true proportion (P) is given by the formula: P̂ ± Z*SE. Substituting the known values, we get \(0.16 ± 1.96 * 0.037\) which gives us the interval \([0.088 , 0.232]\). So, at the 95% confidence level, the true proportion with which this woman golfer is both recognized and respected is between 8.8% and 23.2%.

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Key Concepts

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

Sample Proportion
The sample proportion is a statistical measure used to estimate the proportion of a specific characteristic in a population based on a sample.To calculate the sample proportion, simply divide the number of successes (people who recognized and respected the athlete) by the total number of observations (total customers surveyed).In our case, the sample proportion (denoted as \( \hat{P} \)) is calculated as follows:
  • Number of successes: 16
  • Sample size: 100
  • Sample proportion: \( \hat{P} = \frac{16}{100} = 0.16 \)
This means that 16% of the sample recognizes and respects the woman golfer.The sample proportion is an essential part of many statistical analyses, as it gives a direct insight into the sample data relative to the entire population being studied.
Standard Error
Standard error is a key concept in statistics that gives us an idea of the variability in the sample proportion.It is calculated to understand how much the sample proportion (\( \hat{P} \)) can be expected to differ from the true population proportion.In simple terms, the standard error helps gauge the reliability of the sample proportion used to estimate the population parameter.To calculate the standard error of the sample proportion, use the formula: \[ SE = \sqrt{\frac{\hat{P} \times (1 - \hat{P})}{n}} \]For the golfer’s scenario:
  • \( \hat{P} = 0.16 \)
  • Sample size \( n = 100 \)
  • \( SE = \sqrt{\frac{0.16 \times 0.84}{100}} \approx 0.037 \)
This calculation shows that the variability in the sample is approximately 0.037, indicating how much the sample proportion could vary if different samples were taken from the population.
Z-score
The Z-score is a statistical measure that shows how many standard deviations an element is from the mean.In our context, it helps us determine the range of values that capture the true population proportion with a certain level of confidence.The Z-score is crucial when calculating confidence intervals, especially in determining how far the sample proportion could be from the true proportion.For a 95% confidence interval, a typical Z-score value used is 1.96.This information is typically obtained from a Z-table, which is derived from the standard normal distribution.Using the formula for the confidence interval: \( \hat{P} \pm Z \times SE \), where:
  • \( \hat{P} \) is the sample proportion (0.16),
  • \( Z = 1.96 \),
  • \( SE = 0.037 \).
The confidence interval provides a range in which the true population proportion is likely to lie, using the Z-score to account for variability.
Statistical Analysis
Statistical analysis involves using statistical methods to collect, examine, and interpret data to make informed decisions. In the context of the advertising campaign exercise, statistical analysis helps determine the unknown true population proportion of customers who recognize and respect the golfer based on sample data. Here, we utilize the following steps in the analysis:
  • Calculate the sample proportion
  • Determine the standard error to understand the variability of the sample proportion
  • Use the Z-score for a chosen confidence level (here, 95%)
  • Construct a confidence interval for the true proportion
Conducting this statistical analysis allows the company to make data-driven decisions. They can confidently predict the success of their advertising campaign based on the sample data and respective computations.

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Most popular questions from this chapter

The May \(30,2008,\) online article "Live with Your Parents After Graduation?" quoted a 2007 survey conducted by Monster-TRAK.com. The survey found that \(48 \%\) of college students planned to live at home after graduation. How large of a sample size would you need to estimate the true proportion of students that plan to live at home after graduation to within \(2 \%\) with \(98 \%\) confidence?

Find \(\alpha,\) the area of one tail, and the confidence coefficients of \(z\) that are used with each of the following levels of confidence. a. \(1-\alpha=0.90\) b. \(1-\alpha=0.95\) c. \(1-\alpha=0.99\)

Calculate the test statistic \(z \star\) used in testing the following: a. \(H_{o}: p=0.70\) vs. \(H_{a}: p>0.70,\) with the sample \(n=300\) and \(x=224\) b. \(H_{o}: p=0.50\) vs. \(H_{a}: p<0.50,\) with the sample \(n=450\) and \(x=207\) c. \(H_{o}: p=0.35\) vs. \(H_{a}: p \neq 0.35,\) with the sample \(n=280\) and \(x=94\) d. \(H_{o}: p=0.90\) vs. \(H_{a}: p>0.90,\) with the sample \(n=550\) and \(x=508\)

"You say tomato, burger lovers say ketchup!" According to a recent T.G.I. Friday's restaurants' random survey of 1027 Americans, approximately half \((47 \%)\) said that ketchup is their preferred burger condiment. The survey quoted a margin of error of plus or minus \(3.1 \% .\) a. Describe how this survey of 1027 Americans fits the properties of a binomial experiment. Specifically identify \(n,\) a trial, success, \(p,\) and \(x\). b. What is the point estimate for the proportion of all Americans who prefer ketchup on their burger? Is it a parameter or a statistic? c. Calculate the \(95 \%\) confidence maximum error of estimate for a binomial experiment of 1027 trials that results in an observed proportion of 0.47 d. How is the maximum error, found in part c, related to the \(3.1 \%\) margin of error quoted in the survey report? e. Find the \(95 \%\) confidence interval for the true proportion \(p\) based on a binomial experiment of 1027 trials that results in an observed proportion of 0.47.

Use a computer or calculator to find the \(p\) -value for the following hypothesis test: \(H_{o}: \sigma=12.4\) versus \(H_{a}: \sigma>12.4,\) if \(\chi^{2} \star=36.59\) for a sample of \(n=24\).

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