/*! 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 41 The article "Americans Seek Spir... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article "Americans Seek Spiritual Guidance on Web" (San Luis Obispo Tribune, October 12,2002 ) reported that \(68 \%\) of the general population belong to a religious community. In a survey on Internet use, \(84 \%\) of "religion surfers" (defined as those who seek spiritual help online or who have used the web to search for prayer and devotional resources) belong to a religious community. Suppose that this result was based on a sample of 512 religion surfers. Is there convincing evidence that the proportion of religion surfers who belong to a religious community is different from \(.68\), the proportion for the general population? Use \(\alpha=.05\).

Short Answer

Expert verified
To provide a short answer, the actual computation of the z score and P-value are necessary. After that, based on the comparison of the P-value to \(\alpha = 0.05\), we would know whether to reject or fail to reject the null hypothesis, and hence could conclude if there is convincing evidence that the proportion of religion surfers who belong to a religious community is different from \(0.68\), as that the proportion for the general population.

Step by step solution

01

State the Hypotheses

We start by stating the null and alternative hypotheses. The null hypothesis assumes no difference between the two proportions, while the alternative hypothesis states otherwise. Let \( p \) denote the proportion of 'religion surfers' who belong to a religious community.\n\nNull hypothesis \( H_0 \): \( p = 0.68 \)\nAlternative hypothesis \( H_a \): \( p ≠ 0.68 \)
02

Calculate the Test Statistic

We'll use the formula for the two-proportion z-test to find the z score:\n\n\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]\n\nwhere:\n\(\hat{p}\) is the sample proportion (0.84),\n\(p_0\) is the stated population proportion (0.68), and\n\(n\) is the sample size (512).\n\nCarrying out the calculation yields: \n\n\[ z = \frac{0.84 - 0.68}{\sqrt{\frac{0.68(1 - 0.68)}{512}}} \]
03

Find the P-value

Once we find the z score, we can determine the P-value, which is the probability under the null hypothesis of obtaining a z score as extreme as the one calculated. The P-value is found using a z-table or statistical software.
04

Draw Conclusion

We compare the P-value to the significance level, \(\alpha = 0.05\). \n\nIf the P-value is less than \(\alpha\), we reject the null hypothesis and conclude that the proportion of religion surfers belonging to a religious community is significantly different from that of the general population.\n\nOtherwise, if the P-value is greater than \(\alpha\), we fail to reject the null hypothesis, implying that the proportion for religion surfers may be equal to 0.68.

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!

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

An automobile manufacturer who wishes to advertise that one of its models achieves \(30 \mathrm{mpg}\) (miles per gallon) decides to carry out a fuel efficiency test. Six nonprofessional drivers are selected, and each one drives a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: \(\begin{array}{llllll}27.2 & 29.3 & 31.2 & 28.4 & 30.3 & 29.6\end{array}\) Assuming that fuel efficiency is normally distributed under these circumstances, do the data contradict the claim that true average fuel efficiency is (at least) \(30 \mathrm{mpg}\) ?

Although arsenic is known to be a poison, it also has some beneficial medicinal uses. In one study of the use of arsenic to treat acute promyelocytic leukemia (APL), a rare type of blood cell cancer, APL patients were given an arsenic compound as part of their treatment. Of those receiving arsenic, \(42 \%\) were in remission and showed no signs of leukemia in a subsequent examination (Washington Post, November 5,1998 ). It is known that \(15 \%\) of APL patients go into remission after the conventional treatment. Suppose that the study had included 100 randomly selected patients (the actual number in the study was much smaller). Is there sufficient evidence to conclude that the proportion in remission for the arsenic treatment is greater than \(.15\), the remission proportion for the conventional treatment? Test the relevant hypotheses using a. 01 significance level.

A well-designed and safe workplace can contribute greatly to increasing productivity. It is especially important that workers not be asked to perform tasks, such as lifting, that exceed their capabilities. The following data on maximum weight of lift (MWOL, in kilograms) for a frequency of 4 lifts per minute were reported in the article "The Effects of Speed, Frequency, and Load on Measured Hand Forces for a Floor-to-Knuckle Lifting Task" (Ergonomics \([1992]: 833-843)\) : \(\begin{array}{lllll}25.8 & 36.6 & 26.3 & 21.8 & 27.2\end{array}\) Suppose that it is reasonable to regard the sample as a random sample from the population of healthy males, age \(18-30\). Do the data suggest that the population mean MWOL exceeds 25 ? Carry out a test of the relevant hypotheses using a \(.05\) significance level.

A hot tub manufacturer advertises that with its heating equipment, a temperature of \(100^{\circ} \mathrm{F}\) can be achieved in at most \(15 \mathrm{~min}\). A random sample of 25 tubs is selected, and the time necessary to achieve a \(100^{\circ} \mathrm{F}\) temperature is determined for each tub. The sample average time and sample standard deviation are \(17.5\) min and \(2.2\) min, respectively. Does this information cast doubt on the company's claim? Carry out a test of hypotheses using significance level \(.05 .\)

Let \(\mu\) denote the true average lifetime for a certain type of pen under controlled laboratory conditions. A test of \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10\) will be based on a sample of size 36. Suppose that \(\sigma\) is known to be \(0.6\), from which \(\sigma_{x}=0.1\). The appropriate test statistic is then $$ z=\frac{\bar{x}-10}{0.1} $$ a. What is \(\alpha\) for the test procedure that rejects \(H_{0}\) if \(z \leq\) \(-1.28 ?\) b. If the test procedure of Part (a) is used, calculate \(\beta\) when \(\mu=9.8\), and interpret this error probability. c. Without doing any calculation, explain how \(\beta\) when \(\mu=9.5\) compares to \(\beta\) when \(\mu=9.8\). Then check your assertion by computing \(\beta\) when \(\mu=9.5\). d. What is the power of the test when \(\mu=9.8 ?\) when \(\mu=9.5 ?\)

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.