/*! 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 35 We examine the effect of differe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

We examine the effect of different inputs on determining the sample size needed to obtain a specific margin of error when finding a confidence interval for a proportion. Find the sample size needed to give a margin of error to estimate a proportion within \(\pm 3 \%\) with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. (Assume no prior knowledge about the population proportion \(p\).) Comment on the relationship between the sample size and the confidence level desired.

Short Answer

Expert verified
The required sample sizes for 99%, 95% and 90% confidence intervals are 1860, 1068, and 751 respectively. The larger the desired confidence level, the larger the required sample size.

Step by step solution

01

Calculate the Z-Score

The first step is to calculate the Z-score for each confidence level. The Z-Scores for 99% confidence, 95% confidence and 90% confidence are approximately 2.58, 1.96 and 1.645 respectively.
02

Apply the Margin of error formula

We rearrange the error bound formula: \(E = z * \sqrt{\frac{p(1-p)}{n}}\) to be in the form \(n = (\frac{z}{E})^2 * p(1-p)\) and use it to calculate the sample size \(n\), applying the worst case scenario \(p=0.5\) for each confidence level. For 99% confidence interval, \(n =(\frac{2.58}{0.03})^2 * 0.5*0.5 = 1860\). Similarly for 95% confidence interval, \(n = (\frac{1.96}{0.03})^2 * 0.5*0.5 = 1068\), and for 90% confidence interval, \(n =(\frac{1.645}{0.03})^2 * 0.5*0.5 = 751\).
03

Round up to the nearest whole number

Sample size should always be a whole number. So, if you got a decimal in step 2, round up to the nearest whole number because you can't really have a fraction of a sample. Therefore, the sample sizes are 1860 for 99%, 1068 for 95% and 751 for 90%.
04

Analysis of the relationship between sample size and confidence level

With all else being equal, the higher the desired confidence level, the larger the required sample size. This is because increasing the confidence level requires a wider confidence interval, which requires a larger sample size.

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.

Confidence Interval
When statisticians estimate the characteristics of a population, like the average height or the proportion of people favoring a certain policy, they use a statistic known as a confidence interval (CI). This interval gives a range within which we can say, with a specified probability, that the true population parameter lies. The most common confidence levels are 90%, 95%, and 99%.

For example, a 95% CI for the population proportion might be expressed as [0.45, 0.55]. This means we are 95% confident that the true proportion is between 45% and 55%. However, exactly 95% confidence doesn't mean the parameter will definitely fall within the range; there's still a 5% chance it won't. This possible error is where the concept of margin of error comes into play.

Imagine you're trying to estimate the usage of a new educational app among college students. If your 95% CI is [0.20, 0.40], then you can be quite confident that the true proportion of college students using the app is between 20% and 40%. The confidence level and the CI are crucial for researchers and analysts since they provide both a measure and an indicator of the reliability of an estimate.
Margin of Error
The margin of error (ME) is a statistic that expresses the amount of random sampling error in a survey's results. It represents how much the responses in the sample can deviate from the true population value and is directly connected to the confidence interval.

Using the formula for margin of error, which is usually expressed as ME = z * \(\sqrt{\frac{p(1-p)}{n}}\), you can see it depends on three main factors: the Z-score, the population proportion (p), and the sample size (n). In the context of our textbook problem, with an ME of \(\pm 3\%\), we determine how big of a sample we need to be certain about our estimate within that \(\pm 3\%\) range.

Returning to our educational app estimation, if we claim that 30% of college students use the app with a margin of error of \(\pm 5\%\), it means that the true proportion could be as low as 25% or as high as 35%. A smaller margin of error requires a larger sample size, but it provides a closer estimate of the population parameter, which can be very significant in studies that require precise measurements.
Z-score
A Z-score, also known as a standard score, is a statistical measurement that describes a value's relationship to the mean of a group of values. It's measured in terms of standard deviations from the mean. In the context of confidence intervals, the Z-score determines how far out from the mean you have to go to capture the desired proportion of the data.

The higher the Z-score, the further out from the mean, and the wider the confidence interval. This is why, as seen in the textbook solution, different confidence levels have different Z-scores. For a 99% confidence level, we use a Z-score of 2.58 because it reflects that only 1% of the data lies outside the 99% confidence interval, thus requiring a larger sample size to ensure that the estimated parameter falls within this strict criterion.

Understanding Z-scores is key if you're working with normal distributions. Since many statistical methods assume data is normally distributed or can be approximated as such, Z-scores and the standard normal distribution are foundational concepts in statistics. For students grappling with sample size determinations, knowing how the confidence level alters the Z-score, and in turn the required sample size, is crucial for designing reliable studies and experiments.

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

Left-Handed Lawyers Approximately \(10 \%\) of Americans are left-handed (we will treat this as a known population parameter). A study on the relationship between handedness and profession found that in a random sample of 105 lawyers, 16 of them were left-handed. \({ }^{13}\) Test the hypothesis that the proportion of left-handed lawyers differs from the proportion of left-handed Americans. (a) Clearly state the null and alternative hypotheses. (b) Calculate the test statistic and p-value. (c) What do we conclude at the \(5 \%\) significance level? At the \(10 \%\) significance level?

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 \(s_{N}=22.8\)

Who Watches More TV: Males or Females? The dataset StudentSurvey has information from males and females on the number of hours spent watching television in a typical week. Computer output of descriptive statistics for the number of hours spent watching TV, broken down by gender, is given: \(\begin{array}{l}\text { Descriptive Statistics: TV } \\ \text { Variable } & \text { Gender } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { TV } & \mathrm{F} & 169 & 5.237 & 4.100 \\ & \mathrm{M} & 192 & 7.620 & 6.427 \\\ \text { Minimum } & \mathrm{Q} 1 & \text { Median } & \mathrm{Q} 3 & \text { Maximum } \\ & 0.000 & 2.500 & 4.000 & 6.000 & 20.000 \\ & 0.000 & 3.000 & 5.000 & 10.000 & 40.000\end{array}\) (a) In the sample, which group watches more TV, on average? By how much? (b) Use the summary statistics to compute a \(99 \%\) confidence interval for the difference in mean number of hours spent watching TV. Be sure to define any parameters you are estimating. (c) Compare the answer from part (c) to the confidence interval given in the following computer output for the same data: \(\begin{array}{l}\text { Two-sample T for TV } \\ \text { Gender } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \mathrm{F} & 169 & 5.24 & 4.10 & 0.32 \\ \mathrm{M} & 192 & 7.62 & 6.43 & 0.46 \\ \text { Difference } & =\mathrm{mu}(\mathrm{F})-\mathrm{mu}(\mathrm{M}) & & \end{array}\) Estimate for for difference: -2.383 r difference: (-3.836,-0.930) \(99 \% \mathrm{Cl}\) for (d) Interpret the confidence interval in context.

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\). A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=3.7, s_{d}=\) 2.1, \(n_{d}=30\)

Number of Fouls in a Season by NBA Players The variable Fouls in the dataset NBAPlayers2015 shows the total number of fouls during the \(2014-2015\) season for all players in the \(\mathrm{NBA}\) (National Basketball Association) who played at least 24 minutes per game that season. We use this group as a sample of all NBA players in all seasons who play regularly. Use this information to test whether there is evidence that NBA players who play regularly have a mean number of fouls in a season less than 160 (or roughly 2 fouls per game).

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.