MGQ 301 – Spring 2015 Homework 2 Seungmin Lee 50093307 Statistical Decisions in Management Q1. Facebook provides a variety of statistic on their Web site that detail the growth and popularity of the site. One such statistic is that the average user has 130 friends. This distribution only takes integer values, so it is certainly not Normal. We will also assume it is skewed to the right with a standard deviation σ = 85. Consider a SRS of 30 Facebook users. You must show your work in answering the following questions. (5 points) a. What are the mean and standard deviation of the total number of friends in this sample? (2 points) ο· The population is not normal distribution. It skewed right side. But, the sample has large random sample (n≥30). The mean of the sample equals to the population mean. ππ = π¬(π) = πππ × πππ = ππππ πΊπ = πΊπ = ππ × πππ = ππππ b. What are the mean and standard deviation of the mean number of friends per user? (2 points) ππ = π¬(π) = πππ ππ πΊπ = πΊπ = ≈ ππ. ππππππ √ππ c. Use the central limit theorem to find the probability that the average number of friends in 30 Facebook users is greater than 140. (1 point) πππ − πππ πΊπ ≈ ππ. ππππππ, ππ = πππ, π·(π΄ > πππ) = π· (π > ) ππ. ππππππ ππ = π· (π > ) ≈ π·(π > π. ππππ) ππ. ππππππ π − πππππ = π. ππ = π. ππππ π − π. ππ = π. ππ 1-0.37=0.63 Q2. A study based on a sample size of 36 reported a mean of 87 with a margin of error of 10 for 95% confidence. (3 points) a. Give the 95% confidence interval. (2 points) π π π ± ππ , π = ππ, ππ = π΄πππππ ππ πππππ = ππ π √π ππ 87±10=(77,97) b. If you wanted 99% confidence for the same study, would your margin of error be greater than, equal to, or less than 10? (1 point) π ππ = ππ (ππ% ππππππ ππππ) π √π π. ππ π √ππ = ππ, π √ππ = ππ ππ(√ππ) ,π = = ππ. ππππππ π. ππ π. ππ ππ = π. πππ (ππ% ππππππ ππππ) π π. πππ(ππ. ππππππ) √ππ = ππ. ππππππ Q3. The Student Monitor surveys 1200 undergraduates from 100 colleges semiannually to understand trends among college students. Recently, the Student Monitor reported that the average amount of time spent per week on the Internet was 19.0 hours. Assume that the standard deviation is 5.5 hours. Give a 95% confidence interval for the mean time spent per week on the Internet. (4 points) π ± ππ π , π = ππ. π, π = π. π, π = ππππ, ππ = π. ππ (ππ% ππππππ ππππ) π √π π. ππ(π. π) ππ ± = ππ ± π. ππππππ = (ππ. ππππππ, ππ. ππππππ) √ππππ π Q4. You want to rent an unfurnished one-bedroom apartment in Dallas next year. The mean monthly rent for a random sample of 10 apartments advertised in the local newspaper is $980. Assume the monthly rents in Dallas follow a Normal distribution with a standard deviation of $290. (6 points) a. Find a 95% confidence interval for the mean monthly rent for unfurnished one-bedroom apartment available for rent in this community. (2 points) π ± ππ,π π π π √π , π < ππ π = πππ, π = ππ, π = πππ, π π = π − π = ππ − π = π, πππ(π − π. ππ)% = π. ππ, ππ.ππ = π. πππ π πππ ± π. πππ(πππ) √ππ ,π = πππ ± πππ. πππππ = (ππππ. πππππ, ππππ. πππππ) b. Find a 98% confidence interval for the mean monthly rent for unfurnished one-bedroom apartment available for rent in this community. (2 points) π π ± ππ,π π , π < ππ π √π π = πππ, π = ππ, π = πππ, π π = π − π = ππ − π = π, πππ(π − π. ππ)% = π. ππ, ππ.ππ = π. πππ π πππ ± ,π π. πππ(πππ) = πππ ± πππ. ππππππ = (πππ. ππππππ, ππππ. ππππππ) √ππ c. What can you say about the relationship between range of the confidence interval and the level of confidence? (2 points) - When increased level of confidence, the range of the confidence interval changes to wide. - When decreased level of confidence, the range of the confidence interval changes to narrow. Q5. State the appropriate null hypothesis is H0 and alternative hypothesis Ha in each of the following cases. (6 points) a. A 2008 study reported that 88% of students owned a cell phone. You plan to take an SRS of students to see if the percentage has increased. (2 points) π―π : π = ππ%, π―π : π > ππ% b. The examinations in a large freshman chemistry class are scaled after grading so that the mean score is 75. The professor thinks that students who attend early morning recitation sections will have a higher mean score than the class as a whole. Her students this semester can be considered a sample from the population of all students she might teach, so she compares their mean score with 75. (2 points) π―π : π = ππ, π―π > ππ c. The student newspaper at your college recently changed the format of their opinion page. You take a random sample of students and select those who regularly read the newspaper. They are asked to indicate their opinions on the changes using a five-point scale: -2 if the new format is much worse than the old, -1 if the new format is somewhat worse than the old, 0 if the new format is the same as the old, +1 fi the new format is somewhat better than the old, and +2 if the new format is much better than the old. (2 points) π―π : π = π, π―π : π ≠ π Q6. Translate each of the following research questions into appropriate H0 and Ha. (4 points) a. Census Bureau data show that the mean household income in the area served by a shopping mall is $42,800 per year. A market research firm questions shoppers at the mall to find out whether the mean household income of mall shoppers is higher than that of the general population. (2 points) π―π : π = $ππ, πππ, π―π : π > $ππ, πππ b. Last year, your online registration technicians took an average of 0.4 hours to respond to trouble calls from students trying to register. Do this year’s data show a different average response time? (2 points) π―π : π = π. π πππππ, π―π : π ≠ π. π πππππ Q7. A test of the null hypothesis H0: π= π0 gives test statistic z = 1.63. (6 points) a. What is the P-value if the alternative is Ha: π> π0? (2 points) π―π : π = π π , π―π : π > π π π = π. ππ = π. ππππ, π»πππ ππ πππππ ππππππ ππππ. π − π. ππππ = π. ππππ = π· − πππππ b. What is the P-value if the alternative is Ha: π< π0? (2 points) π―π : π = π π , π―π : π < π π −π = −π. ππ = π. ππππ, π»πππ ππ ππππ ππππππ ππππ. π. ππππ = π· − πππππ c. What is the P-value if the alternative is Ha: π≠ π0? (2 points) π―π : π = π π , π―π : π ≠ π π π = π. ππ = π. ππππ, π»πππ ππ πππ ππππππ ππππ. π − π. ππππ = π. ππππ, π. ππππ × π = π. ππππ = π· Q8. The one-sample t statistic for testing H0: μ = 8 versus Ha: μ > 8 from a sample of n =16 observations has the value t = 2.10. (4 points) a. What are the degrees of freedom for this statistic? Give the two critical values t* from Table D that bracket t. (2 points) π π = π − π = ππ − π = ππ, π = π. ππ π. πππ < πͺπππππππ πππππ < π. ππ b. Between what two values does the P-value of the test fall? Is the value t = 2.10 significant at the 5% level? Is it significant at the 1% level? (2 points) π―π : π = π, π―π : π > π π π = π − π = ππ − π = ππ, π = π. ππ π. πππ(ππ.πππ = π. πππ) < πͺπππππππ πππππ < π. ππ(ππ.ππ = π. πππ) πΆ = π. ππ, ππ.ππ = π. πππ ππππππππ πππππ > πΆ, ππππππ π―π πΆ = π. ππ, ππ.ππ = π. πππ πΆ > ππππππππ πππππ, ππππ ππ ππππππ π―π Q9. The one-sample t statistic for testing H0: μ = 10 versus Ha: μ < 20 from a sample of n =14 observations has the value t = - 2.55. What are the degrees of freedom for this statistic? Between what two values does the P-value of the test fall? (3 points) π―π : π = ππ, π―π : π < ππ π = ππ, π π = ππ, π = −π. ππ π. πππ(ππ.πππ = π. πππ) < ππππππππ ππππππ < π. π(ππ.π = π. πππ) π°π ππ.π > ππΆ , ππππππ π―π π°π ππ.πππ < ππΆ , ππππ ππ ππππππ π―π Q10. The assessment of computerized brain-training programs is a rapidly growing area of research. Researchers are now focusing on whom does this training benefit most, what brain functions can be best improved, and which products are most effective. A recent study looked at 487 communitydwelling adults aged 65 and older, each randomly assigned to one of two training groups. In one group, the participants used a computerized program 1 hour per day. In the other, DVD-based educational programs were shown with quizzes following each video. The training period lasted 8 weeks. The response was the improvement in a composite score obtained from an auditory memory/attention survey given before and after the 8 weeks. The results are summarized in the following table. (9 points) Group n s π₯Μ Computer Program 242 3.9 8.28 DVD Program 245 1.8 8.33 a. Given that there are other studies showing a benefit of computerized brain training, state the null and alternative hypotheses. (3 points) ππ = πͺπππππππ π·ππππππ, ππ = π«π½π« π·ππππππ π―π : ππ − ππ = π, π―π : ππ − ππ ≠ π b. Report the test statistic, its degrees of freedom, and the P-value. What is your conclusion using significance level πΌ = 0.05? (3 points) π π π = π π π π ( ππ + ππ ) π π π π π π π π π π ( π ) + ( π ) ππ − π ππ ππ − π ππ π π π = π. πππ π. πππ ( πππ + ) πππ π π π π. πππ π π. πππ ( ) + ( ) πππ − π πππ πππ − π πππ π π = (π. ππππππ + π. πππππ)π π. ππππππ π. ππππππ + πππ πππ π π = π. ππππππ = πππ. ππππππ π. ππππππππππ π = π. ππ, ππ.ππ = π. πππ [πππ − ππππ] π= π= (π Μ Μ Μ π − Μ Μ Μ ) ππ − (ππ − ππ ) (π. π − π. π) − π √ππ. ππππ + ππ. ππππ πππ πππ = π π √ππ + ππ ππ ππ = π. π √π. ππππππ + π. πππππ = π. π √π. ππππππ π. π = π. ππππππ π. ππππππ ππ.πππ = π. πππ < πͺπππππππ πππππ < ππ.ππ = π. πππ ππ.ππ = π. πππ < ππ.ππ = π. πππ, ππππππ π―π c. Can you conclude that this computerized brain training always improves a person’s auditory memory better than the DVD program? If not, explain why in one sentence. (3 points) - At the 0.05 level of significance, there is difference with DVD program and computer program. But, we can’t conclude that this computerized brain training always improves a person’s auditory memory better than the DVD program. Because this decision is based on a sample, there is the possibility of making the wrong decision.