MGQ 301 HW2

advertisement
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.
π’Žπ’™ = 𝑬(𝒙) = πŸπŸ‘πŸŽ × πŸπŸ‘πŸŽ = πŸ‘πŸ—πŸŽπŸŽ
𝑺𝒙 = 𝑺𝒏 = πŸ–πŸ“ × πŸπŸ‘πŸŽ = πŸπŸ“πŸ“πŸŽ
b. What are the mean and standard deviation of the mean number of friends per user? (2 points)
π’Žπ’™ = 𝑬(𝒙) = πŸπŸ‘πŸŽ
πŸ–πŸ“
𝑺𝒙 = 𝑺𝒏 =
≈ πŸπŸ“. πŸ“πŸπŸ–πŸ–πŸŽπŸ”
√πŸ‘πŸŽ
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)
πŸπŸ’πŸŽ − πŸπŸ‘πŸŽ
𝑺𝒙 ≈ πŸπŸ“. πŸ“πŸπŸ–πŸ–πŸŽπŸ”, π’ŽπŸ = πŸπŸ‘πŸŽ, 𝑷(𝑴 > πŸπŸ’πŸŽ) = 𝑷 (𝒛 >
)
πŸπŸ“. πŸ“πŸπŸ–πŸ–πŸŽπŸ”
𝟏𝟎
= 𝑷 (𝒛 >
)
≈ 𝑷(𝒛 > 𝟎. πŸ”πŸ’πŸ’πŸ’)
πŸπŸ“. πŸ“πŸπŸ–πŸ–πŸŽπŸ”
𝒛 − 𝒔𝒄𝒐𝒓𝒆 = 𝟎. πŸ‘πŸ• = 𝟎. πŸ”πŸ’πŸ’πŸ‘
𝟏 − 𝟎. πŸ‘πŸ• = 𝟎. πŸ”πŸ‘
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)
𝒔
𝒔
π’Ž ± 𝒛𝒂
, π’Ž = πŸ–πŸ•, 𝒛𝒂 = π‘΄π’‚π’“π’ˆπ’Šπ’ 𝒐𝒇 𝒆𝒓𝒓𝒐𝒓 = 𝟏𝟎
𝟐 √𝒏
πŸπ’
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)
𝒔
𝒛𝒂
= 𝟏𝟎 (πŸ—πŸ“% π’„π’π’π’‡π’Šπ’…π’†π’π’„π’†)
𝟐 √𝒏
𝟏. πŸ—πŸ”
𝒔
√πŸ‘πŸ”
= 𝟏𝟎,
𝒔
√πŸ‘πŸ”
=
𝟏𝟎
𝟏𝟎(√πŸ‘πŸ”)
,𝒔 =
= πŸ‘πŸŽ. πŸ”πŸπŸπŸπŸ’πŸ“
𝟏. πŸ—πŸ”
𝟏. πŸ—πŸ”
𝒛𝒂 = 𝟐. πŸ“πŸ•πŸ“ (πŸ—πŸ—% π’„π’π’π’‡π’Šπ’…π’†π’π’„π’†)
𝟐
𝟐. πŸ“πŸ•πŸ“(πŸ‘πŸŽ. πŸ”πŸπŸπŸπŸ’πŸ“)
√πŸ‘πŸ”
= πŸπŸ‘. πŸπŸ‘πŸ•πŸ•πŸ“πŸ“
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’Ž ± 𝒛𝒂
𝒔
, π’Ž = πŸπŸ—. 𝟎, 𝒔 = πŸ“. πŸ“, 𝒏 = 𝟏𝟐𝟎𝟎, 𝒛𝒂 = 𝟏. πŸ—πŸ” (πŸ—πŸ“% π’„π’π’π’‡π’Šπ’…π’†π’π’„π’†)
𝟐
√𝒏
𝟏. πŸ—πŸ”(πŸ“. πŸ“)
πŸπŸ— ±
= πŸπŸ— ± 𝟎. πŸ‘πŸπŸπŸπŸ—πŸ = (πŸπŸ–. πŸ”πŸ–πŸ–πŸ–πŸŽπŸ–, πŸπŸ—. πŸ‘πŸπŸπŸπŸ—πŸ)
√𝟏𝟐𝟎𝟎
𝟐
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’Ž ± 𝒕𝒂,𝒅𝒇
𝟐
𝒔
√𝒏
, 𝒏 < πŸ‘πŸŽ
π’Ž = πŸ—πŸ–πŸŽ, 𝒏 = 𝟏𝟎, 𝒔 = πŸπŸ—πŸŽ, 𝒅𝒇 = 𝒏 − 𝟏 = 𝟏𝟎 − 𝟏 = πŸ—,
𝟏𝟎𝟎(𝟏 − 𝟎. πŸ—πŸ“)% = 𝟎. πŸŽπŸ“, π’•πŸŽ.πŸŽπŸ“ = 𝟐. πŸπŸ”πŸ
𝟐
πŸ—πŸ–πŸŽ ±
𝟐. πŸπŸ”πŸ(πŸπŸ—πŸŽ)
√𝟏𝟎
,πŸ—
= πŸ—πŸ–πŸŽ ± πŸπŸŽπŸ•. πŸ’πŸ‘πŸ—πŸŽπŸ— = (πŸ•πŸ•πŸ•πŸ. πŸ“πŸ”πŸŽπŸ—πŸ, πŸπŸπŸ–πŸ•. πŸ’πŸ‘πŸ—πŸŽπŸ—)
b. Find a 98% confidence interval for the mean monthly rent for unfurnished one-bedroom
apartment available for rent in this community. (2 points)
𝒔
π’Ž ± 𝒕𝒂,𝒅𝒇
, 𝒏 < πŸ‘πŸŽ
𝟐
√𝒏
π’Ž = πŸ—πŸ–πŸŽ, 𝒏 = 𝟏𝟎, 𝒔 = πŸπŸ—πŸŽ, 𝒅𝒇 = 𝒏 − 𝟏 = 𝟏𝟎 − 𝟏 = πŸ—,
𝟏𝟎𝟎(𝟏 − 𝟎. πŸ—πŸ–)% = 𝟎. 𝟎𝟐, π’•πŸŽ.𝟎𝟐 = 𝟐. πŸ–πŸπŸ
𝟐
πŸ—πŸ–πŸŽ ±
,πŸ—
𝟐. πŸ–πŸπŸ(πŸπŸ—πŸŽ)
= πŸ—πŸ–πŸŽ ± πŸπŸ“πŸ–. πŸ•πŸŽπŸπŸ•πŸ•πŸ‘ = (πŸ•πŸπŸ. πŸπŸ—πŸ•πŸπŸπŸ•, πŸπŸπŸ‘πŸ–. πŸ•πŸŽπŸπŸ•πŸ•πŸ‘)
√𝟏𝟎
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ŸŽ : 𝝁 = πŸ–πŸ–%, 𝑯𝒂 : 𝝁 > πŸ–πŸ–%
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ŸŽ : 𝝁 = πŸ•πŸ“, 𝑯𝒂 > πŸ•πŸ“
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ŸŽ : 𝝁 = 𝟎, 𝑯𝒂 : 𝝁 ≠ 𝟎
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ŸŽ : 𝝁 = $πŸ’πŸ, πŸ–πŸŽπŸŽ, 𝑯𝒂 : 𝝁 > $πŸ’πŸ, πŸ–πŸŽπŸŽ
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ŸŽ : 𝝁 = 𝟎. πŸ’ 𝒉𝒐𝒖𝒓𝒔, 𝑯𝒂 : 𝝁 ≠ 𝟎. πŸ’ 𝒉𝒐𝒖𝒓𝒔
Q7. A test of the null hypothesis H0: πœ‡= πœ‡0 gives test statistic z = 1.63. (6 points)
a. What is the P-value if the alternative is Ha: πœ‡> πœ‡0? (2 points)
π‘―πŸŽ : 𝝁 = 𝝁 𝟎 , 𝑯𝒂 : 𝝁 > 𝝁 𝟎
𝒛 = 𝟏. πŸ”πŸ‘ = 𝟎. πŸ—πŸ’πŸ–πŸ’, π‘»π’‰π’Šπ’” π’Šπ’” π’“π’Šπ’ˆπ’‰π’• π’•π’‚π’Šπ’π’†π’… 𝒕𝒆𝒔𝒕. 𝟏 − 𝟎. πŸ—πŸ’πŸ–πŸ’ = 𝟎. πŸŽπŸ“πŸπŸ” = 𝑷 − 𝒗𝒂𝒍𝒖𝒆
b. What is the P-value if the alternative is Ha: πœ‡< πœ‡0? (2 points)
π‘―πŸŽ : 𝝁 = 𝝁 𝟎 , 𝑯𝒂 : 𝝁 < 𝝁 𝟎
−𝒛 = −𝟏. πŸ”πŸ‘ = 𝟎. πŸŽπŸ“πŸπŸ”, π‘»π’‰π’Šπ’” π’Šπ’” 𝒍𝒆𝒇𝒕 π’•π’‚π’Šπ’π’†π’… 𝒕𝒆𝒔𝒕. 𝟎. πŸŽπŸ“πŸπŸ” = 𝑷 − 𝒗𝒂𝒍𝒖𝒆
c. What is the P-value if the alternative is Ha: πœ‡≠ πœ‡0? (2 points)
π‘―πŸŽ : 𝝁 = 𝝁 𝟎 , 𝑯𝒂 : 𝝁 ≠ 𝝁 𝟎
𝒛 = 𝟏. πŸ”πŸ‘ = 𝟎. πŸ—πŸ’πŸ–πŸ’, π‘»π’‰π’Šπ’” π’Šπ’” π’•π’˜π’ π’•π’‚π’Šπ’π’†π’… 𝒕𝒆𝒔𝒕.
𝟏 − 𝟎. πŸ—πŸ’πŸ–πŸ’ = 𝟎. πŸŽπŸ“πŸπŸ”, 𝟎. πŸŽπŸ“πŸπŸ” × πŸ = 𝟎. πŸπŸŽπŸ‘πŸ = 𝑷
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)
𝒅𝒇 = 𝒏 − 𝟏 = πŸπŸ” − 𝟏 = πŸπŸ“, 𝒕 = 𝟐. 𝟏𝟎
𝟎. πŸŽπŸπŸ“ < π‘ͺπ’“π’Šπ’•π’Šπ’„π’‚π’ 𝒗𝒂𝒍𝒖𝒆 < 𝟎. πŸŽπŸ“
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ŸŽ : 𝝁 = πŸ–, 𝑯𝒂 : 𝝁 > πŸ–
𝒅𝒇 = 𝒏 − 𝟏 = πŸπŸ” − 𝟏 = πŸπŸ“, 𝒕 = 𝟐. 𝟏𝟎
𝟎. πŸŽπŸπŸ“(π’•πŸŽ.πŸŽπŸπŸ“ = 𝟐. πŸπŸ‘πŸ) < π‘ͺπ’“π’Šπ’•π’Šπ’„π’‚π’ 𝒗𝒂𝒍𝒖𝒆 < 𝟎. πŸŽπŸ“(π’•πŸŽ.πŸŽπŸ“ = 𝟏. πŸ•πŸ“πŸ‘)
𝜢 = 𝟎. πŸŽπŸ“, π’•πŸŽ.πŸŽπŸ“ = 𝟏. πŸ•πŸ“πŸ‘
π’„π’“π’Šπ’•π’Šπ’„π’‚π’ 𝒗𝒂𝒍𝒖𝒆 > 𝜢, 𝒓𝒆𝒋𝒆𝒄𝒕 π‘―πŸŽ
𝜢 = 𝟎. 𝟎𝟏, π’•πŸŽ.𝟎𝟏 = 𝟐. πŸ”πŸŽπŸ
𝜢 > π’„π’“π’Šπ’•π’Šπ’„π’‚π’ 𝒗𝒂𝒍𝒖𝒆, π‘­π’‚π’Šπ’ 𝒕𝒐 𝒓𝒆𝒋𝒆𝒄𝒕 π‘―πŸŽ
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ŸŽ : 𝝁 = 𝟏𝟎, 𝑯𝒂 : 𝝁 < 𝟐𝟎
𝒏 = πŸπŸ’, 𝒅𝒇 = πŸπŸ‘, 𝒕 = −𝟐. πŸ“πŸ“
𝟎. πŸŽπŸπŸ“(π’•πŸŽ.πŸŽπŸπŸ“ = 𝟐. πŸπŸ”πŸŽ) < π’„π’“π’Šπ’•π’Šπ’„π’‚π’ 𝒗𝒂𝒍𝒖𝒆𝒔 < 𝟎. 𝟏(π’•πŸŽ.𝟏 = 𝟏. πŸ‘πŸ“πŸŽ)
𝑰𝒇 π’•πŸŽ.𝟏 > π’•πœΆ , 𝒓𝒆𝒋𝒆𝒄𝒕 π‘―πŸŽ
𝑰𝒇 π’•πŸŽ.πŸŽπŸπŸ“ < π’•πœΆ , π‘­π’‚π’Šπ’ 𝒕𝒐 𝒓𝒆𝒋𝒆𝒄𝒕 π‘―πŸŽ
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)
π’ŽπŸ = π‘ͺπ’π’Žπ’‘π’–π’•π’†π’“ π‘·π’“π’π’ˆπ’“π’‚π’Ž, π’ŽπŸ = 𝑫𝑽𝑫 π‘·π’“π’π’ˆπ’“π’‚π’Ž
π‘―πŸŽ : π’ŽπŸ − π’ŽπŸ = 𝟎, 𝑯𝒂 : π’ŽπŸ − π’ŽπŸ ≠ 𝟎
b. Report the test statistic, its degrees of freedom, and the P-value. What is your conclusion using
significance level 𝛼 = 0.05? (3 points)
𝟐
𝒅𝒇 =
𝒔 𝟐 𝒔 𝟐
( π’πŸ + π’πŸ )
𝟏
𝟐
𝟐
𝟐
𝟏
𝒔 𝟐
𝟏
𝒔 𝟐
( 𝟏 ) +
( 𝟐 )
π’πŸ − 𝟏 π’πŸ
π’πŸ − 𝟏 π’πŸ
𝟐
𝒅𝒇 =
πŸ–. πŸπŸ–πŸ πŸ–. πŸ‘πŸ‘πŸ
( πŸπŸ’πŸ +
)
πŸπŸ’πŸ“
𝟐
𝟐
𝟏
πŸ–. πŸπŸ–πŸ
𝟏
πŸ–. πŸ‘πŸ‘πŸ
(
)
+
(
)
πŸπŸ’πŸ − 𝟏 πŸπŸ’πŸ
πŸπŸ’πŸ“ − 𝟏 πŸπŸ’πŸ“
𝒅𝒇 =
(𝟎. πŸπŸ–πŸ‘πŸπŸ—πŸ— + 𝟎. πŸπŸ–πŸ‘πŸπŸ)𝟐
𝟎. πŸŽπŸ–πŸŽπŸπŸ“πŸ– 𝟎. πŸŽπŸ–πŸŽπŸπŸπŸ’
+
πŸπŸ’πŸ
πŸπŸ’πŸ’
𝒅𝒇 =
𝟎. πŸ‘πŸπŸŽπŸ—πŸ’πŸ’
= πŸ’πŸ–πŸ’. πŸ—πŸ–πŸŽπŸ”πŸ“πŸ”
𝟎. πŸŽπŸŽπŸŽπŸ”πŸ”πŸπŸ•πŸ”πŸ”πŸ”
𝒂 = 𝟎. πŸŽπŸ“, π’•πŸŽ.πŸŽπŸ“ = 𝟏. πŸ—πŸ–πŸ’ [π’•π’˜π’ − π’•π’‚π’Šπ’]
𝒕=
𝒕=
(𝒙
Μ…Μ…Μ…πŸ − Μ…Μ…Μ…)
π’™πŸ − (π’ŽπŸ − π’ŽπŸ )
(πŸ‘. πŸ— − 𝟏. πŸ–) − 𝟎
√πŸ”πŸ–. πŸ“πŸ“πŸ–πŸ’ + πŸ”πŸ—. πŸ‘πŸ–πŸ–πŸ—
πŸπŸ’πŸ
πŸπŸ’πŸ“
=
𝟐
𝟐
√π’”πŸ + π’”πŸ
π’πŸ
π’πŸ
=
𝟐. 𝟏
√𝟎. πŸπŸ–πŸ‘πŸπŸ—πŸ— + 𝟎. πŸπŸ–πŸ‘πŸπŸ
=
𝟐. 𝟏
√𝟎. πŸ“πŸ”πŸ”πŸ“πŸπŸ—
𝟐. 𝟏
= 𝟐. πŸ•πŸ—πŸŽπŸŽπŸ’πŸ—
𝟎. πŸ•πŸ“πŸπŸ”πŸ•πŸ“
π’•πŸŽ.𝟎𝟎𝟐 = πŸ‘. πŸπŸ•πŸ’ < π‘ͺπ’“π’Šπ’•π’Šπ’„π’‚π’ 𝒗𝒂𝒍𝒖𝒆 < π’•πŸŽ.𝟎𝟏 = 𝟐. πŸ”πŸπŸ”
π’•πŸŽ.πŸŽπŸ“ = 𝟏. πŸ—πŸ–πŸ’ < π’•πŸŽ.𝟎𝟏 = 𝟐. πŸ”πŸπŸ”, 𝒓𝒆𝒋𝒆𝒄𝒕 π‘―πŸŽ
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.
Download