/*! 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} Q55 E Radon exposure in Egyptian tombs... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Q55 E (page 417)

Radon exposure in Egyptian tombs. Refer to the Radiation Protection Dosimetry (December 2010) study of radon exposure in Egyptian tombs, Exercise 6.30 (p. 349). The radon levels鈥攎easured in becquerels per cubic meter (\({{Bq} \mathord{\left/ {\vphantom {{Bq} {{m^3}}}} \right. \\} {{m^3}}}\) )鈥攊n the inner chambers of a sample of 12 tombs are listed in the table shown below. For the safety of the guards and visitors, the Egypt Tourism Authority (ETA) will temporarily close the tombs if the true mean level of radon exposure in the tombs rises to 6,000\({{Bq} \mathord{\left/ {\vphantom {{Bq} {{m^3}}}} \right. \\} {{m^3}}}\) . Consequently, the ETA wants to conduct a test to determine if the true mean level of radon exposure in the tombs is less than 6,000\({{Bq} \mathord{\left/ {\vphantom {{Bq} {{m^3}}}} \right. \\} {{m^3}}}\) , using a Type I error probability of .10. An SPSS analysis of the data is shown at the bottom of the page. Specify all the elements of the test: \({H_0}\,,{H_a}\) test statistic, p-value,\(\alpha \) , and your conclusion.

50 910 180 580 7800 4000 390 12100 3400 1300 11900 110

Short Answer

Expert verified

The null and alternative hypothesis are given by

\(\begin{aligned}{l}{H_0}:\mu = 6000\\{H_a}:\mu < 6000\end{aligned}\)

Test statistic is -1.766.

We reject the null hypothesis.

Step by step solution

01

Stating the hypothesis

The null and alternative hypothesis are given by

\(\begin{aligned}{l}{H_0}:\mu = 6000\\{H_a}:\mu < 6000\end{aligned}\)

02

Computing mean and standard deviation

The sample size is 12

The mean is given by

\(\begin{aligned}\bar x &= \frac{{50 + 910 + 180 + 580 + 7800 + 4000 + 390 + 12100 + 3400 + 1300 + 11900 + 110}}{{12}}\\ &= \frac{{42720}}{{12}}\\ &= 3560\end{aligned}\)

The standard deviation is given by

\(\begin{aligned}sd &= \sqrt {\frac{\begin{aligned}{l}{\left( { - 3510} \right)^2} + {\left( { - 2650} \right)^2} + {\left( { - 3380} \right)^2} + {\left( { - 2980} \right)^2} + {\left( {4240} \right)^2}{\left( { - 3170} \right)^2}\\ + {\left( {8540} \right)^2} + {\left( { - 160} \right)^2} + {\left( { - 2260} \right)^2} + {\left( {8340} \right)^2} + {\left( { - 3510} \right)^2} + {\left( { - 3450} \right)^2}\end{aligned}}{{11}}} \\ &= \sqrt {\frac{\begin{aligned}{l}12320100 + 7022500 + 11424400 + 8880400 + 17977600 + 10048900 + \\72931600 + 25600 + 5107600 + 69555600 + 12320100 + 11902500\end{aligned}}{{11}}} \\ &= \sqrt {\frac{{251837000}}{{11}}} \\ &= \sqrt {22894272.73} \\ &= 4784.79\end{aligned}\)

The mean and standard deviation are 3560 and 4784.79.

03

Test statistic

The test statistic is computed as

\(\begin{aligned}t &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{3560 - 6000}}{{\frac{{4784.79}}{{\sqrt {12} }}}}\\ &= \frac{{ - 2440}}{{1381.24}}\\ &= - 1.766\end{aligned}\)

Therefore, the test statistic is -1.766.

04

P-value

The Type-1 error probability is given.

Hence,\(\alpha = .10\)

Degrees of freedom are

\(\begin{aligned}df &= n - 1\\ &= 12 - 1\\ &= 11\end{aligned}\)

The tabulated value is -1.363

05

Conclusion

The calculated value is less than the tabulated value.

Therefore, we reject the null hypothesis.

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 91影视!

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

Revenue for a full-service funeral. According to the National Funeral Directors Association (NFDA), the nation's 19,000 funeral homes collected an average of \(7,180 per full-service funeral in 2014 (www.nfda.org). A random sample of 36 funeral homes reported revenue data for the current year. Among other measures, each reported its average fee for a full-service funeral. These data (in thousands of dollars) are shown in the following table.

a. What are the appropriate null and alternative hypotheses to test whether the average full-service fee of U. S. funeral homes this year is less than \)7,180?

b. Conduct the test at\(\alpha = 0.05\). Do the sample data provide sufficient evidence to conclude that the average fee this year is lower than in 2014?

c. In conducting the test, was it necessary to assume that the population of average full-service fees was normally distributed? Justify your answer

Intrusion detection systems. The Journal of Research of the National Institute of Standards and Technology (November鈥� December 2003) published a study of a computer intrusion detection system (IDS). The IDS is designed to provide an alarm whenever unauthorized access (e.g., an intrusion) to a computer system occurs. The probability of the system giving a false alarm (i.e., providing a warning when no intrusion occurs) is defined by the symbol 伪, while the probability of a missed detection (i.e., no warning given when an intrusion occurs) is defined by the symbol 尾. These symbols are used to represent Type I and Type II error rates, respectively, in a hypothesis-testing scenario

a. What is the null hypothesis, H0?

b. What is the alternative hypothesis,Ha?

c. According to actual data collected by the Massachusetts Institute of Technology Lincoln Laboratory, only 1 in 1,000 computer sessions with no intrusions resulted in a false alarm. For the same system, the laboratory found that only 500 of 1,000 intrusions were actually detected. Use this information to estimate the values of 伪and 尾.

Latex allergy in health care workers (cont鈥檇). Refer to Exercise 7.120. Let \({\sigma ^2}\) represent the variance in the number of latex gloves used per week by all hospital employees. Consider testing \({H_0}:{\sigma ^2} = 100\) against\({H_a}:{\sigma ^2} \ne 100.\)

a. Give the rejection region for the test at a significance level of\(\alpha = 0.01.\)

b. Calculate the value of the test statistic.

c. Use the results, parts a and b, to make the appropriate conclusion.

Managers who engage in 鈥渃oopetition.鈥� In business, firms that both cooperate and compete with other firms are described as engaging in 鈥渃oopetition.鈥� A study published in Industrial Marketing Management (February 2016) examined the level of external tension experienced by managers who engage in coopetition. External tension (measured on a 20-point scale) was recorded for each in a sample of 1,532 managers, all from firms that were engaged in coopetition. The sample mean tension was x=10.82 and the sample standard deviation was s=3.04.

Conduct a test (using a=.05) to determine if the true mean external tension level of all managers who engage in coopetition differs from 10.5 points.

A sample of five measurements, randomly selected from a normally distributed population, resulted in the following summary statistics: \(\bar x = 4.8\), \(s = 1.3\) \(\) .

a. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis, 碌<6. Use\(\alpha = .05.\)

b. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis, 碌\( \ne 6\). Use\(\alpha = .05.\)

c. Find the observed significance level for each test.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.