Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

We examine the effect of different inputs on determining the sample size needed. Find the sample size needed to give a margin of error within ±3 with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. Assume that we use \(\tilde{\sigma}=30\) as our estimate of the standard deviation in each case. 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 levels are 237, 139, and 101 respectively. There exists an inverse relationship between the confidence level and the required sample size.

Step by step solution

01

Calculate for 99% confidence level

Firstly, one must find the Z-value corresponding to a 99% confidence interval, which is 2.576. Substituting the values into the sample size formula results in \( n =\left(\frac{2.576*30}{3}\right)^2= 236.25\). However, since we cannot have a fraction of a sample, round up to the nearest whole number, giving a sample size of 237.
02

Calculate for 95% confidence level

Now, find the Z-value corresponding to a 95% confidence interval, which is 1.96. Substituting into the formula gives \( n = \left(\frac{1.96*30}{3}\right)^2 = 138.04 \). Again, since we cannot have a fraction of a sample, round up to the nearest whole number, giving a sample size of 139.
03

Calculate for 90% confidence level

Lastly, find the Z-value corresponding to a 90% confidence interval, which is 1.645. Substituting into the formula gives \( n = \left(\frac{1.645*30}{3}\right)^2 = 100.23 \). We round up to the nearest whole number, giving a sample size of 101.
04

Comment on relationship

As observed, as the confidence level decreases, the required sample size decreases as well. This implies an inverse relationship between confidence level and 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 Vaia!

Key Concepts

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

Margin of Error
Understanding the margin of error is crucial when conducting surveys or studies. Essentially, it is an expression of the amount of random sampling error in a survey's results. Think of it as a measurement of the potential discrepancy between the sample's estimate of a population parameter and the actual population parameter. It is represented by the letter 'E' in many mathematical formulas.

For example, if you're seeing a margin of error of ±3, this means that the true population parameter is expected to lie within 3 units above or below the survey's reported value. In our exercise, the goal was to determine the sample size needed to achieve this specified margin of error, which requires other statistical components like the confidence level and the standard deviation to be factored in. The smaller the margin of error, the closer we can expect our sample's estimate to reflect the true population parameter, but this often requires a larger sample size.
Confidence Level
A confidence level signifies the degree of certainty with which we can expect the results from our sample to reflect the true population parameters. Expressed as a percentage, common confidence levels include 90%, 95%, and 99%.

Choosing a higher confidence level implies greater assurance in our results, but it also necessitates a larger sample size to maintain a constant margin of error. In practical terms, if you choose a 99% confidence level, as in the given exercise, you are saying that if the same population were sampled multiple times, 99 out of 100 sample results would fall within the margin of error of the true population parameter. The confidence level is directly tied to the Z-value, which adjusts the sample size calculation accordingly.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of data values. A lower standard deviation indicates that the data points are closer to the mean or average value, while a higher standard deviation indicates a wider spread of values.

In the context of sample size determination, standard deviation (often estimated as \( \tilde{\sigma} \) in the absence of true population statistics) helps to capture the variability we expect within our target population. The greater the variability or standard deviation, the larger the sample size required to achieve a specific margin of error at a given confidence level. This relationship is pivotal in calculating the right sample size for reliable statistical inference.
Z-value
The Z-value, also known as the Z-score, is a numerical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations. In the realm of determining sample size, the Z-value corresponds to the chosen confidence level and can be obtained from a standard normal distribution table or a Z-table.

As the Z-value increases, which happens with higher confidence levels, the sample size needs to be larger to maintain the same margin of error. In our exercise, varying Z-values for different confidence levels (99%, 95%, and 90%) were provided, and each critical Z-value resulted in different sample size requirements: the higher the Z-value, the larger the sample size that's needed. This showcases how the Z-value influences the sample size calculation to ensure the study's accuracy and reliability.

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

As part of the same study described in Exercise 6.254 , the researchers also were interested in whether babies preferred singing or speech. Forty-eight of the original fifty infants were exposed to both singing and speech by the same woman. Interest was again measured by the amount of time the baby looked at the woman while she made noise. In this case the mean time while speaking was 66.97 with a standard deviation of \(43.42,\) and the mean for singing was 56.58 with a standard deviation of 31.57 seconds. The mean of the differences was 10.39 more seconds for the speaking treatment with a standard deviation of 55.37 seconds. Perform the appropriate test to determine if this is sufficient evidence to conclude that babies have a preference (either way) between speaking and singing.

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}<\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllll} \hline \text { Treatment } 1 & 16 & 12 & 18 & 21 & 15 & 11 & 14 & 22 \\ \text { Treatment } 2 & 18 & 20 & 25 & 21 & 19 & 8 & 15 & 20 \\ \hline \end{array} $$

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. To study the effect of sitting with a laptop computer on one's lap on scrotal temperature, 29 men have their scrotal temperature tested before and then after sitting with a laptop for one hour.

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.

A survey of 1000 adults in the US conducted in March 2011 asked "Do you favor or oppose 'sin taxes' on soda and junk food?" The proportion in favor of taxing these foods was \(32 \% .10\) (a) Find a \(95 \%\) confidence interval for the proportion of US adults favoring taxes on soda and junk food. (b) What is the margin of error? (c) If we want a margin of error of only \(1 \%\) (with \(95 \%\) confidence \()\), what sample size is needed?

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.

Sign-up for free