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

Plastic microparticles are contaminating the world's shorelines (see Exercise 6.108\()\), and much of this pollution appears to come from fibers from washing polyester clothes. \({ }^{27}\) The worst offender appears to be fleece, and a recent study found that the mean number of polyester fibers discharged into wastewater from washing fleece was 290 fibers per liter of wastewater, with a standard deviation of 87.6 and a sample size of 120 . (a) Find and interpret a \(99 \%\) confidence interval for the mean number of polyester microfibers per liter of wastewater when washing fleece. (b) What is the margin of error? (c) If we want a margin of error of only ±5 with \(99 \%\) confidence, what sample size is needed?

Short Answer

Expert verified
a) The 99% confidence interval for the mean number of polyester microfibers per liter of wastewater when washing fleece is calculated in step 1. b) The margin of error is calculated in step 2. c) The sample size needed to achieve a ±5 margin of error with 99% confidence is calculated in step 3.

Step by step solution

01

Calculate 99% Confidence Interval

The confidence interval can be calculated using the formula: \[ CI = \bar{X} \pm (Z * \frac{\sigma}{\sqrt{n}}) \])Where \(\bar{X}\) is the sample mean, \(Z\) is the Z-score corresponding to the desired confidence level, \(\sigma\) is the standard deviation, and \(n\) is the sample size. Substitute provided values into the formula:\[ CI = 290 \pm (2.576 * \frac{87.6}{\sqrt{120}}) \] After calculation, you will get the confidence interval.
02

Calculate the margin of error

The margin of error is calculated as the product of the Z-score and standard deviation divided by square root of the sample size. It can be obtained directly from the confidence interval result in step 1. The margin of error can be shown as:\[ E = Z * \frac{\sigma}{\sqrt{n}} \]Substitute provided values into the formula:\[ E = 2.576 * \frac{87.6}{\sqrt{120}} \]After calculation, you will get the margin of error.
03

Calculate the sample size

The sample size needed for a desired margin of error can be calculated using the following rearranged formula:\[ n = (\frac{Z * \sigma}{E})^2 \]Substitute desired values into formula:\[ n = (\frac{2.576 * 87.6}{5})^2 \]After calculation, round up the value of n (if necessary) as the sample size must be a whole number.

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!

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

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: $$ \begin{array}{lllllllll} \hline \text { Treatment } 1 & 16 & 12 & 18 & 21 & 15 & 11 & 14 & 22 \\ \text { Treatment } 2 & 18 & 20 & 25 & 21 & 19 & 8 & 15 & 20 \\ \hline \end{array} $$

Quiz Timing A young statistics professor decided to give a quiz in class every week. He was not sure if the quiz should occur at the beginning of class when the students are fresh or at the end of class when they've gotten warmed up with some statistical thinking. Since he was teaching two sections of the same course that performed equally well on past quizzes, he decided to do an experiment. He randomly chose the first class to take the quiz during the second half of the class period (Late) and the other class took the same quiz at the beginning of their hour (Early). He put all of the grades into a data table and ran an analysis to give the results shown below. Use the information from the computer output to give the details of a test to see whether the mean grade depends on the timing of the quiz. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) Two-Sample T-Test and Cl \begin{tabular}{lrrrr} Sample & \(\mathrm{N}\) & Mean & StDev & SE Mean \\ Late & 32 & 22.56 & 5.13 & 0.91 \\ Early & 30 & 19.73 & 6.61 & 1.2 \\ \multicolumn{3}{c} { Difference } & \(=\mathrm{mu}(\) Late \()\) & \(-\mathrm{mu}\) (Early) \end{tabular} Estimate for difference: 2.83 $$ \begin{aligned} &95 \% \mathrm{Cl} \text { for difference: }(-0.20,5.86)\\\ &\text { T-Test of difference }=0(\text { vs } \operatorname{not}=): \text { T-Value }=1.87\\\ &\text { P-Value }=0.066 \quad \mathrm{DF}=54 \end{aligned} $$

Who Exercises More: Males or Females? The dataset StudentSurvey has information from males and females on the number of hours spent exercising in a typical week. Computer output of descriptive statistics for the number of hours spent exercising, broken down by gender, is given: \(\begin{array}{l}\text { Descriptive Statistics: Exercise } \\ \text { Variable } & \text { Gender } & \mathrm{N} & \text { Mean } & \text { StDev } \\\ \text { Exercise } & \mathrm{F} & 168 & 8.110 & 5.199 \\ & \mathrm{M} & 193 & 9.876 & 6.069\end{array}\) \(\begin{array}{rrrrr}\text { Minimum } & \text { Q1 } & \text { Median } & \text { Q3 } & \text { Maximum } \\ 0.000 & 4.000 & 7.000 & 12.000 & 27.000 \\\ 0.000 & 5.000 & 10.000 & 14.000 & 40.000\end{array}\) (a) How many females are in the dataset? How many males? (b) In the sample, which group exercises more, on average? By how much? (c) Use the summary statistics to compute a \(95 \%\) confidence interval for the difference in mean number of hours spent exercising. Be sure to define any parameters you are estimating. (d) Compare the answer from part (c) to the confidence interval given in the following computer output for the same data: Two-sample \(\mathrm{T}\) for Exercise Gender N Mean StDev SE Mean \(\begin{array}{lllll}\mathrm{F} & 168 & 8.11 & 5.20 & 0.40 \\ \mathrm{M} & 193 & 9.88 & 6.07 & 0.44\end{array}\) Difference \(=\operatorname{mu}(F)-\operatorname{mu}(M)\) Estimate for difference: -1.766 \(95 \%\) Cl for difference: (-2.932,-0.599)

Find the endpoints of the t-distribution with \(5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=8\) and \(n_{2}=10\)

Do Ovulating Women Affect Men's Speech? Studies suggest that when young men interact with a woman who is in the fertile period of her menstrual cycle, they pick up subconsciously on subtle changes in her skin tone, voice, and scent. A study introduced in Exercise \(\mathrm{B} .23\) suggests that men may even change their speech patterns around ovulating women. The men were randomly divided into two groups with one group paired with a woman in the fertile phase of her cycle and the other group with a woman in a different stage of her cycle. The same women were used in the two different stages. For the men paired with a less fertile woman, 38 of the 61 men copied their partner's sentence construction in a task to describe an object. For the men paired with a woman at peak fertility, 30 of the 62 men copied their partner's sentence construction. The experimenters hypothesized that men might be less likely to copy their partner during peak fertility in a (subconscious) attempt to attract more attention to themselves. Use the normal distribution to test at a \(5 \%\) level whether the proportion of men copying sentence structure is less when the woman is at peak fertility.

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