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

Exercise 8- 5. Perception and Reality In a presidential election, 308 out of 611 voters surveyed said that they voted for the candidate who won (based on data from ICR Survey Research Group). Use a 0.05 significance level to test the claim that among all voters, the percentage who believe that they voted for the winning candidate is equal to 43%, which is the actual percentage of votes for the winning candidate. What does the result suggest about voter perceptions?

Short Answer

Expert verified

There is not enough evidence to support the claim that 43% of voters favored the candidate. This suggests that the voters did not reveal the true votes.

Step by step solution

01

Given information

The summary of the results obtained from the survey is as follows.

Sample size \(n = 611\).

Votes in favor of winning candidate \(x = 308\).

Level of significance \(\alpha = 0.05\).

It is claimed that the proportion of voters in favor of the winning candidate is 43% or 0.43.

02

State the hypotheses

Null hypothesis: The votes for the winning candidate are equal to 43%.

Alternative hypothesis: The votes for the winning candidate are not equal to 43%

Let p be the true proportion of voters who vote for the candidate who won.

Mathematically,

\({H_0}:p = 0.43\)

\({H_1}:p \ne 0.43\)

03

Compute the test statistic

In testing claims for the population proportion, the z-test is used.

The sample proportion is given as follows.

\(\begin{array}{c}\hat p = \frac{{308}}{{611}}\\ = 0.5041\end{array}\).

From the given information,

\(\begin{array}{c}p = 0.43\\q = 1 - p\\ = 1 - 0.43\\ = 0.57\end{array}\).

The test statistic is computed as follows.

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.5041 - 0.43}}{{\sqrt {\frac{{0.43 \times 0.57}}{{611}}} }}\\ = 3.70\end{array}\).

04

Compute the critical value

From the standard normal table, the critical values for a two-tailed test with a 0.05 significance level are obtained as follows.

The left-tailed critical value will have an area of 0.025, corresponding to row -1.9 and column 0.06.

The right-tailed critical value will have an area of 0.975, corresponding to row 1.9 and column 0.06.

Thus, the two critical values are -1.96 and 1.96.

05

State the decision

If the test statistic lies between the critical values, the null hypothesis is failed to be rejected. Otherwise, the null hypothesis is rejected.

Here, 3.70 does not fall between the critical values -1.96 and 1.96.

Thus, the null hypothesis is rejected.

06

Conclusion

The result suggests that there is not sufficient evidence to support the claim that the actual percentage of voters who favored the winning candidate is 43%.

As the proportion of voters in the survey is quite high, it suggests that the voters did not reveal the true votes in the survey.

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

Test Statistics. In Exercises 13โ€“16, refer to the exercise identified and find the value of the test statistic. (Refer to Table 8-2 on page 362 to select the correct expression for evaluating the test statistic.)

Exercise 7 โ€œPulse Ratesโ€

Testing Claims About Proportions. In Exercises 9โ€“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.

Postponing Death An interesting and popular hypothesis is that individuals can temporarily postpone death to survive a major holiday or important event such as a birthday. In a study, it was found that there were 6062 deaths in the week before Thanksgiving, and 5938 deaths the week after Thanksgiving (based on data from โ€œHolidays, Birthdays, and Postponement of Cancer Death,โ€ by Young and Hade, Journal of the American Medical Association, Vol. 292, No. 24). If people can postpone death until after Thanksgiving, then the proportion of deaths in the week before should be less than 0.5. Use a 0.05 significance level to test the claim that the proportion of deaths in the week before Thanksgiving is less than 0.5. Based on the result, does there appear to be any indication that people can temporarily postpone death to survive the Thanksgiving holiday?

Using Technology. In Exercises 5โ€“8, identify the indicated values or interpret the given display. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section. Use = 0.05 significance level and answer the following:

a. Is the test two-tailed, left-tailed, or right-tailed?

b. What is the test statistic?

c. What is the P-value?

d. What is the null hypothesis, and what do you conclude about it?

e. What is the final conclusion?

Biometric Security In a USA Today survey of 510 people, 53% said that we should replace passwords with biometric security, such as fingerprints. The accompanying Statdisk display results from a test of the claim that half of us say that we should replace passwords with biometric security.

t Test Exercise 2 refers to a t test. What is the t test? Why is the letter t used? What is unrealistic about the z test methods in Part 2 of this section?

Calculating Power Consider a hypothesis test of the claim that the Ericsson method of gender selection is effective in increasing the likelihood of having a baby girl, so that the claim is p>0.5. Assume that a significance level of ฮฑ= 0.05 is used, and the sample is a simple random sample of size n = 64.

a. Assuming that the true population proportion is 0.65, find the power of the test, which is the probability of rejecting the null hypothesis when it is false. (Hint: With a 0.05 significance level, the critical value is z = 1.645, so any test statistic in the right tail of the accompanying top graph is in the rejection region where the claim is supported. Find the sample proportion in the top graph, and use it to find the power shown in the bottom graph.)

b. Explain why the green-shaded region of the bottom graph represents the power of the test.

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