/*! 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} Q11BSC Testing Claims About Proportions... [FREE SOLUTION] | 91影视

91影视

Testing Claims About Proportions. In Exercises 9鈥32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Stem Cell Survey Adults were randomly selected for a Newsweek poll. They were asked if they 鈥渇avor or oppose using federal tax dollars to fund medical research using stem cells obtained from human embryos.鈥 Of those polled, 481 were in favor, 401 were opposed, and 120 were unsure. A politician claims that people don鈥檛 really understand the stem cell issue and their responses to such questions are random responses equivalent to a coin toss. Exclude the 120 subjects who said that they were unsure, and use a 0.01 significance level to test the claim that the proportion of subjects who respond in favor is equal to 0.5. What does the result suggest about the politician鈥檚 claim?

Short Answer

Expert verified

Nullhypothesis: The proportion of adults who responded in favor is equal to 0.5.

Alternativehypothesis: The proportion of adults who responded in favor is not equal to 0.5.

Teststatistic: 2.694

Criticalvalue: 1.96

P-value: 0.0071

The null hypothesis is rejected.

There isenough evidence to reject the claim that the proportion of adults who responded in favor is equal to 0.5.

Because it appears that more than half of individuals support utilizing federal tax dollars to fund medical research using stem cells derived from human embryos, the politician's claim that people's reactions are random and have no significant meaning is completely false.

Step by step solution

01

Given information

The number of adults who responded in favor is equal to 481, and the number of adults opposed is equal to 401.

02

Hypotheses

The null hypothesis is written as follows:

The proportion of adults who responded in favor is equal to 0.5.

\({H_0}:p = 0.5\)

The alternative hypothesis is written as follows:

The proportion of adults who responded in favor is not equal to 0.5.

\({H_1}:p \ne 0.5\)

The test is two-tailed.

03

Sample size, sample proportion, and population proportion

The sample size is computed below:

\[\begin{array}{c}n = 481 + 401\\ = 882\end{array}\]

The sample proportion of adults who responded in favor isas follows:

\[\begin{array}{c}\hat p = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{adults}}\;{\rm{who}}\;{\rm{responded}}\;{\rm{in}}\;{\rm{favour}}}}{{{\rm{Sample}}\;{\rm{Size}}}}\\ = \frac{{481}}{{882}}\\ = 0.545\end{array}\]

The population proportion of adults who responded in favor is equal to 0.5.

04

Test statistic

The value of the test statistic is computed below:

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.545 - 0.5}}{{\sqrt {\frac{{0.5\left( {1 - 0.5} \right)}}{{882}}} }}\\ = 2.694\end{array}\)

Thus, z=2.694.

05

Critical value and P-value

Referring to the standard normal distribution table, the critical value of z at\(\alpha = 0.05\)for a two-tailed test is equal to 1.96.

Referring to the standard normal distribution table, the p-value for the test statistic value of 2.694 is equal to 0.0071.

Since the p-value is less than 0.05, the null hypothesis is rejected.

06

Conclusion of the test

There is enough evidence to reject the claim that the proportion of adults who responded in favor is equal to 0.5.

Since it appears that more than 50% of the adults favor using federal tax dollars to fund medical research using stem cells obtained from human embryos, the politicians claim that the responses of the people are random and do not hold any substantial meaning incorrect.

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

Critical Values. In Exercises 21鈥24, refer to the information in the given exercise and do the following.

a. Find the critical value(s).

b. Using a significance level of = 0.05, should we reject H0or should we fail to reject H0?

Exercise 20

Testing Hypotheses. In Exercises 13鈥24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Heights of Supermodels Listed below are the heights (cm) for the simple random sample of female supermodels Lima, Bundchen, Ambrosio, Ebanks, Iman, Rubik, Kurkova, Kerr,Kroes, Swanepoel, Prinsloo, Hosk, Kloss, Robinson, Heatherton, and Refaeli. Use a 0.01 significance level to test the claim that supermodels have heights with a mean that is greater than the mean height of 162 cm for women in the general population. Given that there are only 16 heights represented, can we really conclude that supermodels are taller than the typical woman?

178 177 176 174 175 178 175 178 178 177 180 176 180 178 180 176

Technology. In Exercises 9鈥12, test the given claim by using the display provided from technology. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Body Temperatures Data Set 3 鈥淏ody Temperatures鈥 in Appendix B includes 93 body temperatures measured at 12 虏鲁 on day 1 of a study, and the accompanying XLSTAT display results from using those data to test the claim that the mean body temperature is equal to 98.6掳F. Conduct the hypothesis test using these results.

Critical Values. In Exercises 21鈥24, refer to the information in the given exercise and do the following.

a. Find the critical value(s).

b. Using a significance level of = 0.05, should we reject H0or should we fail to reject H0?

Exercise 17

Testing Claims About Proportions. In Exercises 9鈥32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Medication Usage In a survey of 3005 adults aged 57 through 85 years, it was found that 87.1% of them used at least one prescription medication (based on data from 鈥淯se of Prescription Over-the-Counter Medications and Dietary SupplementsAmong Older Adultsin the United States,鈥 by Qato et al., Journal of the American Medical Association,Vol. 300,No. 24). Use a 0.01 significance level to test the claim that more than 3/4 of adults use at least one prescription medication. Does the rate of prescription use among adults appear to be high?

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.