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Chapter 9: Q. 55. (page 583)

Do you Tweet? The Pew Internet and American Life Project asked a random sample of U.S. adults, 鈥淒o you ever 鈥� use Twitter or another service to share updates about yourself or to see updates about others?鈥� According to Pew, the resulting 95% confidence interval is (0.123, 0.177).11 Based on the confidence interval, is there convincing evidence that the true proportion of U.S. adults who would say they use Twitter or another service to share updates differs from 0.17? Explain your reasoning.

Short Answer

Expert verified

The required answer is:

No, it is not different from0.17

Step by step solution

01

Given information

The confidence interval is(0.123,0.177)

02

The objective is to find whether the true proportion of adults who says that they use Twitter or other series is different from 0.17or not.

The above-mentioned confidence interval clearly shows that 0.17fall within it. As a result, there is insufficient evidence to conclude that the true proportion of adults who use Twitter or other series is greater than 0.17

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Most popular questions from this chapter

No homework? Mr. Tabor believes that less than 75%of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of 50 students at the school.

More lefties?In the population of people in the United States, about 10% are left-handed. After bumping elbows at lunch with several left-handed students, Simon wondered if more than 10%of students at his school are left-handed. To investigate, he selected an SRS of 50students and found 8lefties (p铃�=8/50=0.16).

To determine if these data provide convincing evidence that more than 10%of the students at Simon鈥檚 school are left-handed, 200trials of a simulation were conducted. Each dot in the graph shows the proportion of students that are left-handed in a random sample of 50students, assuming that each student has a 10%chance of being left handed.

a. State appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest.

b. Use the simulation results to estimate the P-value of the test in part (a). Interpret the P-value.

c. What conclusion would you make?

Cell-phone passwords A consumer organization suspects that less than half of parents know their child鈥檚 cell-phone password. The Pew Research Center asked a random sample of parents if they knew their child鈥檚 cell-phone password. Of the 1060parents surveyed, 551reported that they knew the password. Explain why it isn鈥檛 necessary to carry out a significance test in this setting.

Packaging DVDs (6.2,5.3) A manufacturer of digital video discs (DVDs) wants to be sure that the DVDs will fit inside the plastic cases used as packaging. Both the cases and the DVDs are circular. According to the supplier, the diameters of the plastic cases vary Normally with mean 渭=5.3inches and standard deviation 蟽=0.01inch. The DVD manufacturer produces DVDs with mean diameter渭=5.26inches. Their diameters follow a Normal distribution with 蟽=0.02inch.

a. Let X = the diameter of a randomly selected case and Y = the diameter of a randomly selected DVD. Describe the shape, center, and variability of the distribution of the random variable X鈭扽. What is the importance of this random variable to the DVD manufacturer?

b. Calculate the probability that a randomly selected DVD will fit inside a randomly selected case.

c. The production process runs in batches of 100 DVDs. If each of these DVDs is paired with a randomly chosen plastic case, find the probability that all the DVDs fit in their cases.

No homework? Refer to Exercise 1. The math teachers inspect the

homework assignments from a random sample of 50 students at the school. Only 68% of the students completed their math homework. A significance test yields a P-value of 0.1265.

a. Explain what it would mean for the null hypothesis to be true in this setting.

b. Interpret the P-value.
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