Using Technology. In Exercises 5–8, 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\(\alpha \)= 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.

Short Answer

Expert verified

a.The test is two-tailed.

b.The value of the test statistic (z-score) is equal to 1.3284.

c.The p-value is equal to 0.1840.

d. The null hypothesis is that the proportion of people who say that passwords should be replaced with biometric security is equal to 50%.The null hypothesis is failed to reject.

e. There is not enough evidence to conclude that the proportion of people who say that passwords should be replaced with biometric security is not equal to 50%.

Step by step solution

01

Given information

It is given that out of 510 respondents, 53% say that passwords should be replaced with biometric security.

A claim is made that exactly half of the people say that passwords should be replaced with biometric security.

02

Tail of the test

a.

According to the given claim, the proportion of people who say that passwords should be replaced with biometric security is equal to 50%.

This implies that the alternative hypothesis will be as follows.

Alternative hypothesis:\({H_1}:p \ne 0.5\).

Asthere is a not equal sign in the alternative hypothesis, the test is two-tailed.

03

Test statistic

b.

The test statistic to test the given claim is the z-score.

Here, the value of the test statistic (z-score) is equal to 1.3284.

04

P-value

c.

The p-value corresponding to the z-score of 1.3284 is equal to 0.1840.

05

Null hypothesis and conclusion of the test

d.

The null hypothesis for this test is as follows.

Null hypothesis: The proportion of people who say that passwords should be replaced withbiometric security is equal to 50% or 0.5.

Symbolically,\({H_0}:p = 0.5\),where

p is the proportion of people who say that passwords should be replaced withbiometric security.

Here, the p-value equal to 0.1840 is greater than the significance level of 0.05. Thus, the null hypothesis is failed to reject.

06

Conclusion of the test

e.

There is not enough evidence to conclude that the proportion of people who say that passwords should be replaced withbiometric security is not equal to 50% or 0.5.

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Most popular questions from this chapter

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?

Testing Hypotheses. In Exercises 13–24, 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.

How Many English Words? A simple random sample of 10 pages from Merriam-Webster’s Collegiate Dictionary is obtained. The numbers of words defined on those pages are found, with these results: n = 10, x = 53.3 words, s = 15.7 words. Given that this dictionary has 1459 pages with defined words, the claim that there are more than 70,000 defined words is equivalent to the claim that the mean number of words per page is greater than 48.0 words. Assume a normally distributed population. Use a 0.01 significance level to test the claim that the mean number of words per page is greater than 48.0 words. What does the result suggest about the claim that there are more than 70,000 defined words?

Type I and Type II Errors. In Exercises 29–32, provide statements that identify the type I error and the type II error that correspond to the given claim. (Although conclusions are usually expressed in verbal form, the answers here can be expressed with statements that include symbolic expressions such as p = 0.1.).

The proportion of people who write with their left hand is equal to 0.1.

PowerFor a hypothesis test with a specified significance level , the probability of a type I error is, whereas the probability of a type II error depends on the particular value ofpthat is used as an alternative to the null hypothesis.

a.Using an alternative hypothesis ofp< 0.4, using a sample size ofn= 50, and assumingthat the true value ofpis 0.25, find the power of the test. See Exercise 34 “Calculating Power”in Section 8-1. [Hint:Use the valuesp= 0.25 andpq/n= (0.25)(0.75)/50.]

b.Find the value of , the probability of making a type II error.

c.Given the conditions cited in part (a), find the power of the test. What does the power tell us about the effectiveness of the test?

Testing Claims About Proportions. In Exercises 9–32, 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.

Drug Screening The company Drug Test Success provides a “1-Panel-THC” test for marijuana usage. Among 300 tested subjects, results from 27 subjects were wrong (either a false positive or a false negative). Use a 0.05 significance level to test the claim that less than 10% of the test results are wrong. Does the test appear to be good for most purposes?

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