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Data 2.1 on page 48 deals with a survey that asked whether people agree or disagree with the statement "There is only one true love for each person." The survey results in Table 2.1 show that 735 of the 2625 respondents agreed, 1812 disagreed, and 78 answered "don't know." (a) Find a \(90 \%\) confidence interval for the proportion of people who disagree with the statement. (b) Find a \(90 \%\) confidence interval for the proportion of people who "don't know." (c) Which estimate has the larger margin of error?

Short Answer

Expert verified
Please substitute the values of \(p1\), \(p2\) and \(n\) in their respective formulae to find confidence intervals for disagreement and 'don't know'. Then by comparing \(ME1\) and \(ME2\), one can determine which has a larger margin of error.

Step by step solution

01

Find the proportion of people who disagree

First, calculate the proportion of people who disagreed. This is obtained by dividing the number of people who disagreed (1812) by the total number of respondents (2625). Let's denote this proportion as \(p1\). \(p1 = \frac{1812}{2625}\)
02

Calculating 90% CI for disagreement

The formula for a confidence interval for a proportion is \(p \pm Z *\sqrt{p(1-p)/n}\) where Z is the z-score corresponding to the desired confidence level (for 90% confidence level, Z=1.645). Use this formula to calculate the upper and lower bounds for the confidence interval. The margin of error is given by: \(ME1 = Z*\sqrt{p1*(1-p1)/n}\) . Subtract and add this margin of error from/to \(p1\) to get the Confidence Interval for disagreement.
03

Calculate proportion of people who 'don't know'

Now calculate the proportion of people who answered 'don't know'. This is obtained by dividing the number of people who answered 'don't know' (78) by the total number of respondents (2625). Let's denote this proportion as \(p2\). \(p2 = \frac{78}{2625}\)
04

Calculating 90% CI for 'don't know'

Similar to before, use the formula for a confidence interval for a proportion to calculate the upper and lower bounds for the confidence interval. The margin of error is given by: \(ME2 = Z*\sqrt{p2*(1-p2)/n}\) . Subtract or add it from \(p2\) to get the Confidence Interval for 'don't know'.
05

Determine which estimate has the larger margin of error

Finally, to figure out which proportion has a larger margin of error, simply compare ME1 and ME2.

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Key Concepts

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

Proportion
Understanding the concept of 'proportion' is crucial when analyzing survey results. It represents the ratio of a part or share of the whole. For instance, if a survey question has multiple choice answers, the proportion of respondents selecting a specific option is calculated by dividing the number of individuals who chose that option by the total number of survey participants.

In the exercise provided, the proportion of people who disagreed with the statement 'There is only one true love for each person' is denoted as proportion 'p1'. This is found by dividing the number of disagreeing respondents (1812) by the total respondents (2625), resulting in the calculated proportion 'p1'. It is essential to identify this proportion accurately to assess the sentiment of the surveyed group towards the statement.
Margin of Error
The 'margin of error' (ME) plays a critical role in survey data analysis, providing an estimate of how much the survey results might differ from the true population value. In essence, it's the range within which we can expect the true proportion to lie, considering a certain confidence level.

In the exercise example, we calculate the margin of error for both groups of respondents - those who disagreed and those who answered 'don't know'. It incorporates the z-score for the given confidence level and the proportion statistics of the survey. It’s clear that the smaller the margin of error, the more accurate our confidence interval is likely to be. Understanding the margin of error helps give context to the variability of survey responses and the confidence we can have in the results obtained.
Z-Score
The z-score, in the context of confidence intervals, is a critical value from the standard normal distribution that corresponds to a desired confidence level. The z-score determines how many standard deviations an element is from the mean. In survey analysis, it is used to create a confidence interval around a proportion, allowing us to say that, based on our sample data, we are a certain percent confident the true proportion of the population falls within that interval.

The choice of z-score is related to the confidence level we desire; for instance, a 90% confidence level corresponds to a z-score of approximately 1.645. When constructing confidence intervals, this z-score is used to extend the range from the observed proportion, accounting for the variability expected in the population. Clear understanding of z-scores helps evaluate the reliability of survey results by defining the bounds within which we expect the true population parameter to fall.
Survey Data Analysis
Survey data analysis encompasses the entire process of collecting, evaluating, interpreting, and presenting data from survey responses. It entails calculating proportions, establishing confidence intervals, understanding margins of error, and much more. The goal is to accurately capture the sentiments or opinions of the survey population in a way that can be generalized to the entire group from which the sample was drawn.

Our example exercise requires analyzing the proportion of respondents who disagree or are unsure about the statement 'There is only one true love for each person'. We use statistical tools such as the z-score to estimate confidence intervals, thus giving us insight into the margin of error and the reliability of the survey results. Proper survey data analysis ensures that we draw meaningful and accurate conclusions from the sample data provided.

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Most popular questions from this chapter

Examine the results of a study \(^{45}\) investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics that has been linked to multiple health problems including hormone disruption. The study included 8,877 people who recorded all the food they ate over a 24 -hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in \(\mathrm{ng} / \mathrm{mL}\) ) in the urine: DEHP and DiNP. Find and interpret a \(95 \%\) confidence interval for the difference, \(\mu_{F}-\mu_{N},\) in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DiNP in the 3095 participants who had eaten fast food was \(\bar{x}_{F}=10.1\) with \(s_{F}=38.9\) while the mean for the 5782 participants who had not eaten fast food was \(\bar{x}_{N}=7.0\) with \(s_{N}=22.8\)

Close Confidants and Social Networking Sites Exercise 6.93 introduces a study \(^{48}\) in which 2006 randomly selected US adults (age 18 or older) were asked to give the number of people in the last six months "with whom you discussed matters that are important to you." The average number of close confidants for the full sample was \(2.2 .\) In addition, the study asked participants whether or not they had a profile on a social networking site. For the 947 participants using a social networking site, the average number of close confidants was 2.5 with a standard deviation of 1.4 , and for the other 1059 participants who do not use a social networking site, the average was 1.9 with a standard deviation of \(1.3 .\) Find and interpret a \(90 \%\) confidence interval for the difference in means between the two groups.

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. In another study to investigate the effect of women's tears on men, 16 men watch an erotic movie and then half sniff women's tears and half sniff a salt solution while brain activity is monitored.

Football Air Pressure During the NFL's 2014 AFC championship game, officials measured the air pressure on game balls following a tip that one team's balls were under-inflated. In exercise 6.124 we found that the 11 balls measured for the New England Patriots had a mean psi of 11.10 (well below the legal limit) and a standard deviation of 0.40. Patriot supporters could argue that the under-inflated balls were due to the elements and other outside effects. To test this the officials also measured 4 balls from the opposing team (Indianapolis Colts) to be used in comparison and found a mean psi of \(12.63,\) with a standard deviation of 0.12. There is no significant skewness or outliers in the data. Use the t-distribution to determine if the average air pressure in the New England Patriot's balls was significantly less than the average air pressure in the Indianapolis Colt's balls.

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.01

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