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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.

Touch Therapy Repeat the preceding exercise using a 0.01 significance level. Does the conclusion change?

Short Answer

Expert verified

Null hypothesis: The proportion of correct guesses is equal to 0.5.

Alternative hypothesis: The proportion of correct guesses is not equal to 0.5.

Test statistic: -2.032

Critical value: 2.5758

P-value: 0.0422

The null hypothesis is failed to reject.

There is not enough evidence to reject the claim that the touch therapists randomly guess the correct answer.

Step by step solution

01

Given information

Out of 280 trials, 123 guesses were correct by the touch therapists. It is claimed that that touch therapists randomly guess the answer.

02

Hypotheses

The null hypothesis is written as follows.

The proportion of correct guesses is equal to 0.5.

H0:p=0.5

The alternative hypothesis is written as follows.

The proportion of correct guesses is not equal to 0.5.

H1:p0.5

The test is two-tailed.

03

Sample size, sample proportion, and population proportion

The sample size is n=280.

The sample proportion of correct guesses is computed below.

p^=NumberofcorrectguessesSampleSize=123280=0.439

The population proportion of correct guesses is equal to 0.5.

04

Test statistic

The value of the test statistic is computed below.

z=p^-ppqn=0.439-0.50.51-0.5280=-2.032

Thus, z=-2.032.

05

Critical value and p-value

Referring to the standard normal table, the critical value of z at for a two-tailed test is equal to 2.5758.

Referring to the standard normal table, the p-value for the test statistic value of -2.032 is equal to 0.0422.

As the p-value is greater than 0.01, the null hypothesis is failed to reject.

06

Conclusion of the test

There is not enough evidence to reject the claim that the touch therapists randomly guess the correct answer.

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Most popular questions from this chapter

P-Values. In Exercises 17–20, do the following:

a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.

b. Find the P-value. (See Figure 8-3 on page 364.)

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

The test statistic of z = -2.50 is obtained when testing the claim that p<0.75

Finding Critical t Values When finding critical values, we often need significance levels other than those available in Table A-3. Some computer programs approximate critical t values by calculating t=df×eA2/df-1where df = n-1, e = 2.718, A=z8×df+3/8×df+1, and z is the critical z score. Use this approximation to find the critical t score for Exercise 12 “Tornadoes,” using a significance level of 0.05. Compare the results to the critical t score of 1.648 found from technology. Does this approximation appear to work reasonably well?

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.

Touch Therapy Repeat the preceding exercise using a 0.01 significance level. Does the conclusion change?

Interpreting Power Chantix (varenicline) tablets are used as an aid to help people stop smoking. In a clinical trial, 129 subjects were treated with Chantix twice a day for 12 weeks, and 16 subjects experienced abdominal pain (based on data from Pfizer, Inc.). If someone claims that more than 8% of Chantix users experience abdominal pain, that claim is supported with a hypothesis test conducted with a 0.05 significance level. Using 0.18 as an alternative value of p, the power of the test is 0.96. Interpret this value of the power of the test.

P-Values. In Exercises 17–20, do the following:

a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.

b. Find the P-value. (See Figure 8-3 on page 364.)

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

The test statistic of z = -1.94 is obtained when testing the claim that p=38 .

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