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 α= 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?

Cell Phone Ownership A Pew Research Center poll of 2076 randomly selected adults showed that 91% of them own cell phones. The following Minitab display results from a test of the claim that 92% of adults own cell phones.

Short Answer

Expert verified

a. The test is two-tailed.

b. The value of the test statistic (z-score) is equal to -1.69.

c. The p-value is equal to 0.091.

d. The null hypothesis is that the proportion of adults who own cell phones is equal to 0.92.The decision is fail to reject the null hypothesis.

e. There is not enough evidence to warrant rejection of the claim that the proportion of adults who own cell phones is equal to 92%.

Step by step solution

01

Given information

It is given that in a survey, out of 2076 adults, 92% of them own cell phones.

02

Tail of the Test

a.

According to the given claim, the proportion of adults who own cell phones is equal to 92%.

The null hypothesis is represented as follows:

H0:p=0.92

The alternative hypothesis is represented as follows:

H1:p0.92

Since there is an unequal 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 -1.69.

04

P-Value

c.

The p-value corresponding to the z-score of -1.69 is given to be equal to 0.091.

05

Null hypothesis and its conclusion

d.

The null hypothesis for this test is as follows:

Null Hypothesis: The proportion of adults who own cell phones is equal to 92%.

Symbolically, H0:p=0.92where

p is the proportion of adults who own cell phones.

Here, the p-value equal to 0.091 is greater than the significance level of 0.05. Thus, the null hypothesis fails to be rejected.

06

Conclusion of the Test

e.

There is not enough evidence to warrant rejection of the claim that the proportion of adults who own cell phones is equal to 92%.

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Most popular questions from this chapter

Technology. In Exercises 9–12, 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.

Airport Data Speeds Data Set 32 “Airport Data Speeds” in Appendix B includes Sprint data speeds (mbps). The accompanying TI-83/84 Plus display results from using those data to test the claim that they are from a population having a mean less than 4.00 Mbps. Conduct the hypothesis test using these results.

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 = 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.

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?

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 .

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.

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 “Holidays, Birthdays, and Postponement of Cancer Death,” 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?

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