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)

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.
𝒎𝒙 = 𝑬(𝒙) = 𝟏𝟑𝟎 × 𝟏𝟑𝟎 = 𝟑𝟗𝟎𝟎
𝑺𝒙 = 𝑺𝒏 = 𝟖𝟓 × 𝟏𝟑𝟎 = 𝟐𝟓𝟓𝟎
b. What are the mean and standard deviation of the mean number of friends per user? (2 points)
𝒎𝒙 = 𝑬(𝒙) = 𝟏𝟑𝟎
𝟖𝟓
𝑺𝒙 = 𝑺𝒏 =
≈ 𝟏𝟓. 𝟓𝟏𝟖𝟖𝟎𝟔
√𝟑𝟎
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)
𝟏𝟒𝟎 − 𝟏𝟑𝟎
𝑺𝒙 ≈ 𝟏𝟓. 𝟓𝟏𝟖𝟖𝟎𝟔, 𝒎𝟏 = 𝟏𝟑𝟎, 𝑷(𝑴 > 𝟏𝟒𝟎) = 𝑷 (𝒛 >
)
𝟏𝟓. 𝟓𝟏𝟖𝟖𝟎𝟔
𝟏𝟎
= 𝑷 (𝒛 >
)
≈ 𝑷(𝒛 > 𝟎. 𝟔𝟒𝟒𝟒)
𝟏𝟓. 𝟓𝟏𝟖𝟖𝟎𝟔
𝒛 − 𝒔𝒄𝒐𝒓𝒆 = 𝟎. 𝟑𝟕 = 𝟎. 𝟔𝟒𝟒𝟑
𝟏 − 𝟎. 𝟑𝟕 = 𝟎. 𝟔𝟑
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
𝒎 ± 𝒛𝒂
, 𝒎 = 𝟖𝟕, 𝒛𝒂 = 𝑴𝒂𝒓𝒈𝒊𝒏 𝒐𝒇 𝒆𝒓𝒓𝒐𝒓 = 𝟏𝟎
𝟐 √𝒏
𝟐𝒏
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 (𝟗𝟓% 𝒄𝒐𝒏𝒇𝒊𝒅𝒆𝒏𝒄𝒆)
𝟐 √𝒏
𝟏. 𝟗𝟔
𝒔
√𝟑𝟔
= 𝟏𝟎,
𝒔
√𝟑𝟔
=
𝟏𝟎
𝟏𝟎(√𝟑𝟔)
,𝒔 =
= 𝟑𝟎. 𝟔𝟏𝟐𝟐𝟒𝟓
𝟏. 𝟗𝟔
𝟏. 𝟗𝟔
𝒛𝒂 = 𝟐. 𝟓𝟕𝟓 (𝟗𝟗% 𝒄𝒐𝒏𝒇𝒊𝒅𝒆𝒏𝒄𝒆)
𝟐
𝟐. 𝟓𝟕𝟓(𝟑𝟎. 𝟔𝟏𝟐𝟐𝟒𝟓)
√𝟑𝟔
= 𝟏𝟑. 𝟏𝟑𝟕𝟕𝟓𝟓
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)
𝒎 ± 𝒛𝒂
𝒔
, 𝒎 = 𝟏𝟗. 𝟎, 𝒔 = 𝟓. 𝟓, 𝒏 = 𝟏𝟐𝟎𝟎, 𝒛𝒂 = 𝟏. 𝟗𝟔 (𝟗𝟓% 𝒄𝒐𝒏𝒇𝒊𝒅𝒆𝒏𝒄𝒆)
𝟐
√𝒏
𝟏. 𝟗𝟔(𝟓. 𝟓)
𝟏𝟗 ±
= 𝟏𝟗 ± 𝟎. 𝟑𝟏𝟏𝟏𝟗𝟐 = (𝟏𝟖. 𝟔𝟖𝟖𝟖𝟎𝟖, 𝟏𝟗. 𝟑𝟏𝟏𝟏𝟗𝟐)
√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)
𝒎 ± 𝒕𝒂,𝒅𝒇
𝟐
𝒔
√𝒏
, 𝒏 < 𝟑𝟎
𝒎 = 𝟗𝟖𝟎, 𝒏 = 𝟏𝟎, 𝒔 = 𝟐𝟗𝟎, 𝒅𝒇 = 𝒏 − 𝟏 = 𝟏𝟎 − 𝟏 = 𝟗,
𝟏𝟎𝟎(𝟏 − 𝟎. 𝟗𝟓)% = 𝟎. 𝟎𝟓, 𝒕𝟎.𝟎𝟓 = 𝟐. 𝟐𝟔𝟐
𝟐
𝟗𝟖𝟎 ±
𝟐. 𝟐𝟔𝟐(𝟐𝟗𝟎)
√𝟏𝟎
,𝟗
= 𝟗𝟖𝟎 ± 𝟐𝟎𝟕. 𝟒𝟑𝟗𝟎𝟗 = (𝟕𝟕𝟕𝟐. 𝟓𝟔𝟎𝟗𝟏, 𝟏𝟏𝟖𝟕. 𝟒𝟑𝟗𝟎𝟗)
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
𝒎 ± 𝒕𝒂,𝒅𝒇
, 𝒏 < 𝟑𝟎
𝟐
√𝒏
𝒎 = 𝟗𝟖𝟎, 𝒏 = 𝟏𝟎, 𝒔 = 𝟐𝟗𝟎, 𝒅𝒇 = 𝒏 − 𝟏 = 𝟏𝟎 − 𝟏 = 𝟗,
𝟏𝟎𝟎(𝟏 − 𝟎. 𝟗𝟖)% = 𝟎. 𝟎𝟐, 𝒕𝟎.𝟎𝟐 = 𝟐. 𝟖𝟐𝟏
𝟐
𝟗𝟖𝟎 ±
,𝟗
𝟐. 𝟖𝟐𝟏(𝟐𝟗𝟎)
= 𝟗𝟖𝟎 ± 𝟐𝟓𝟖. 𝟕𝟎𝟐𝟕𝟕𝟑 = (𝟕𝟐𝟏. 𝟐𝟗𝟕𝟐𝟐𝟕, 𝟏𝟐𝟑𝟖. 𝟕𝟎𝟐𝟕𝟕𝟑)
√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)
𝑯𝟎 : 𝝁 = 𝟖𝟖%, 𝑯𝒂 : 𝝁 > 𝟖𝟖%
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)
𝑯𝟎 : 𝝁 = 𝟕𝟓, 𝑯𝒂 > 𝟕𝟓
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)
𝑯𝟎 : 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)
𝑯𝟎 : 𝝁 = $𝟒𝟐, 𝟖𝟎𝟎, 𝑯𝒂 : 𝝁 > $𝟒𝟐, 𝟖𝟎𝟎
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)
𝑯𝟎 : 𝝁 = 𝟎. 𝟒 𝒉𝒐𝒖𝒓𝒔, 𝑯𝒂 : 𝝁 ≠ 𝟎. 𝟒 𝒉𝒐𝒖𝒓𝒔
Q7. A test of the null hypothesis H0: 𝜇= 𝜇0 gives test statistic z = 1.63. (6 points)
a. What is the P-value if the alternative is Ha: 𝜇> 𝜇0? (2 points)
𝑯𝟎 : 𝝁 = 𝝁 𝟎 , 𝑯𝒂 : 𝝁 > 𝝁 𝟎
𝒛 = 𝟏. 𝟔𝟑 = 𝟎. 𝟗𝟒𝟖𝟒, 𝑻𝒉𝒊𝒔 𝒊𝒔 𝒓𝒊𝒈𝒉𝒕 𝒕𝒂𝒊𝒍𝒆𝒅 𝒕𝒆𝒔𝒕. 𝟏 − 𝟎. 𝟗𝟒𝟖𝟒 = 𝟎. 𝟎𝟓𝟏𝟔 = 𝑷 − 𝒗𝒂𝒍𝒖𝒆
b. What is the P-value if the alternative is Ha: 𝜇< 𝜇0? (2 points)
𝑯𝟎 : 𝝁 = 𝝁 𝟎 , 𝑯𝒂 : 𝝁 < 𝝁 𝟎
−𝒛 = −𝟏. 𝟔𝟑 = 𝟎. 𝟎𝟓𝟏𝟔, 𝑻𝒉𝒊𝒔 𝒊𝒔 𝒍𝒆𝒇𝒕 𝒕𝒂𝒊𝒍𝒆𝒅 𝒕𝒆𝒔𝒕. 𝟎. 𝟎𝟓𝟏𝟔 = 𝑷 − 𝒗𝒂𝒍𝒖𝒆
c. What is the P-value if the alternative is Ha: 𝜇≠ 𝜇0? (2 points)
𝑯𝟎 : 𝝁 = 𝝁 𝟎 , 𝑯𝒂 : 𝝁 ≠ 𝝁 𝟎
𝒛 = 𝟏. 𝟔𝟑 = 𝟎. 𝟗𝟒𝟖𝟒, 𝑻𝒉𝒊𝒔 𝒊𝒔 𝒕𝒘𝒐 𝒕𝒂𝒊𝒍𝒆𝒅 𝒕𝒆𝒔𝒕.
𝟏 − 𝟎. 𝟗𝟒𝟖𝟒 = 𝟎. 𝟎𝟓𝟏𝟔, 𝟎. 𝟎𝟓𝟏𝟔 × 𝟐 = 𝟎. 𝟏𝟎𝟑𝟐 = 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 = 𝟏𝟔 − 𝟏 = 𝟏𝟓, 𝒕 = 𝟐. 𝟏𝟎
𝟎. 𝟎𝟐𝟓 < 𝑪𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒗𝒂𝒍𝒖𝒆 < 𝟎. 𝟎𝟓
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)
𝑯𝟎 : 𝝁 = 𝟖, 𝑯𝒂 : 𝝁 > 𝟖
𝒅𝒇 = 𝒏 − 𝟏 = 𝟏𝟔 − 𝟏 = 𝟏𝟓, 𝒕 = 𝟐. 𝟏𝟎
𝟎. 𝟎𝟐𝟓(𝒕𝟎.𝟎𝟐𝟓 = 𝟐. 𝟏𝟑𝟏) < 𝑪𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒗𝒂𝒍𝒖𝒆 < 𝟎. 𝟎𝟓(𝒕𝟎.𝟎𝟓 = 𝟏. 𝟕𝟓𝟑)
𝜶 = 𝟎. 𝟎𝟓, 𝒕𝟎.𝟎𝟓 = 𝟏. 𝟕𝟓𝟑
𝒄𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒗𝒂𝒍𝒖𝒆 > 𝜶, 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎
𝜶 = 𝟎. 𝟎𝟏, 𝒕𝟎.𝟎𝟏 = 𝟐. 𝟔𝟎𝟐
𝜶 > 𝒄𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒗𝒂𝒍𝒖𝒆, 𝑭𝒂𝒊𝒍 𝒕𝒐 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎
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)
𝑯𝟎 : 𝝁 = 𝟏𝟎, 𝑯𝒂 : 𝝁 < 𝟐𝟎
𝒏 = 𝟏𝟒, 𝒅𝒇 = 𝟏𝟑, 𝒕 = −𝟐. 𝟓𝟓
𝟎. 𝟎𝟐𝟓(𝒕𝟎.𝟎𝟐𝟓 = 𝟐. 𝟏𝟔𝟎) < 𝒄𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒗𝒂𝒍𝒖𝒆𝒔 < 𝟎. 𝟏(𝒕𝟎.𝟏 = 𝟏. 𝟑𝟓𝟎)
𝑰𝒇 𝒕𝟎.𝟏 > 𝒕𝜶 , 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎
𝑰𝒇 𝒕𝟎.𝟎𝟐𝟓 < 𝒕𝜶 , 𝑭𝒂𝒊𝒍 𝒕𝒐 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎
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
𝑥̅
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)
𝒎𝟏 = 𝑪𝒐𝒎𝒑𝒖𝒕𝒆𝒓 𝑷𝒓𝒐𝒈𝒓𝒂𝒎, 𝒎𝟐 = 𝑫𝑽𝑫 𝑷𝒓𝒐𝒈𝒓𝒂𝒎
𝑯𝟎 : 𝒎𝟏 − 𝒎𝟐 = 𝟎, 𝑯𝒂 : 𝒎𝟏 − 𝒎𝟐 ≠ 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
( 𝒏𝟏 + 𝒏𝟐 )
𝟏
𝟐
𝟐
𝟐
𝟏
𝒔 𝟐
𝟏
𝒔 𝟐
( 𝟏 ) +
( 𝟐 )
𝒏𝟏 − 𝟏 𝒏𝟏
𝒏𝟐 − 𝟏 𝒏𝟐
𝟐
𝒅𝒇 =
𝟖. 𝟐𝟖𝟐 𝟖. 𝟑𝟑𝟐
( 𝟐𝟒𝟐 +
)
𝟐𝟒𝟓
𝟐
𝟐
𝟏
𝟖. 𝟐𝟖𝟐
𝟏
𝟖. 𝟑𝟑𝟐
(
)
+
(
)
𝟐𝟒𝟐 − 𝟏 𝟐𝟒𝟐
𝟐𝟒𝟓 − 𝟏 𝟐𝟒𝟓
𝒅𝒇 =
(𝟎. 𝟐𝟖𝟑𝟐𝟗𝟗 + 𝟎. 𝟐𝟖𝟑𝟐𝟐)𝟐
𝟎. 𝟎𝟖𝟎𝟐𝟓𝟖 𝟎. 𝟎𝟖𝟎𝟐𝟏𝟒
+
𝟐𝟒𝟏
𝟐𝟒𝟒
𝒅𝒇 =
𝟎. 𝟑𝟐𝟎𝟗𝟒𝟒
= 𝟒𝟖𝟒. 𝟗𝟖𝟎𝟔𝟓𝟔
𝟎. 𝟎𝟎𝟎𝟔𝟔𝟏𝟕𝟔𝟔𝟔
𝒂 = 𝟎. 𝟎𝟓, 𝒕𝟎.𝟎𝟓 = 𝟏. 𝟗𝟖𝟒 [𝒕𝒘𝒐 − 𝒕𝒂𝒊𝒍]
𝒕=
𝒕=
(𝒙
̅̅̅𝟏 − ̅̅̅)
𝒙𝟐 − (𝒎𝟏 − 𝒎𝟐 )
(𝟑. 𝟗 − 𝟏. 𝟖) − 𝟎
√𝟔𝟖. 𝟓𝟓𝟖𝟒 + 𝟔𝟗. 𝟑𝟖𝟖𝟗
𝟐𝟒𝟐
𝟐𝟒𝟓
=
𝟐
𝟐
√𝒔𝟏 + 𝒔𝟐
𝒏𝟏
𝒏𝟐
=
𝟐. 𝟏
√𝟎. 𝟐𝟖𝟑𝟐𝟗𝟗 + 𝟎. 𝟐𝟖𝟑𝟐𝟐
=
𝟐. 𝟏
√𝟎. 𝟓𝟔𝟔𝟓𝟏𝟗
𝟐. 𝟏
= 𝟐. 𝟕𝟗𝟎𝟎𝟒𝟗
𝟎. 𝟕𝟓𝟐𝟔𝟕𝟓
𝒕𝟎.𝟎𝟎𝟐 = 𝟑. 𝟏𝟕𝟒 < 𝑪𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒗𝒂𝒍𝒖𝒆 < 𝒕𝟎.𝟎𝟏 = 𝟐. 𝟔𝟐𝟔
𝒕𝟎.𝟎𝟓 = 𝟏. 𝟗𝟖𝟒 < 𝒕𝟎.𝟎𝟏 = 𝟐. 𝟔𝟐𝟔, 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎
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)
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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.