a)
Check
whether
it
can
be
asserted
with
98%
confidence
that
the
underlying
yearly
real
variance
of
the
portfol
Sigma^2=
440
Ho:
sigma^2
=
440
H1:
sigma^2
not
=
440

Statistic

4.9

With
98%
confidence
Lower
limit 3.05348411
Upper
limit 24.7249703
With
98%
confidence
we
cannot
reject
Ho
since
the
statistic
is
inside
the
acceptance
zone
b)
Check
the
same
hypothesis
with
95%
confidence.

With
95%
confidence
Lower
limit 3.81574825
Upper
limit 21.9200493
With
95%
confidence
we
cannot
reject
Ho
since
the
statistic
is
inside
the
acceptance
zone
Question
2
Typical
variance
Number
of
students
New
variance
Normal
population

300
30
480

a)
Check
whether
the
hypothesis
that
the
population
variance
of
their
scores
is
300
can
be
accepted
with
95
%

Ho:
sigma^2
=
300
H1:
sigma^2
not
=
300

Statistic
46.4

Lower
limit 16.0470717
Upper
limit 45.7222858
We
can
reject
Ho
with
95%
confidence
since
it
falls
outside
of
the
interval

b)
Check
with
90%
confidence
the
hypothesis
that
the
population
variance
of
the
scores
is
greater
or
equal
than
Ho:
sigma^2
>=
300
H1:
sigma^2
Question
3
Variance_1
114.09 during
4
years
Variance_2
16.08 following
7
years
Normal
distribution
a)
Test
whether
the
two
population
variances
can
be
considered
equal
with
95
%
confidence.
Ho:
sigma1^2
=
sigma2^2
H1:
sigma1^1
not
equal
to
sigma2^2
Lower
limit
Upper
limit

Statistic

7.09514925

0.06786692
6.59879852

We
can
reject
Ho
with
95%
confidence
since
it
falls
outside
of
the
interval

b)
With
95%
confidence,
test
the
hypothesis
that
the
population
variance
of
the
shares
during
the
years
of
com
Ho:
sigma1^2
>=
sigma2^2
H1:
sigma1^1
<
sigma2^2
Limit

0.11184875

We
cannot
reject
Ho
with
95%
since
it
is
higher
than
the
limit.
Question
4
Percentage
of
forecast
error
Corporation Analyst
A
Analyst
B
1
12.3
7.3
2
15.4
12.1
3
5.3
7.4
4
9.2
8.1
5
8.6
11.3
6
14.2
12.3
7
5.2
3.1

...
MBA 501A – [STATISTICS]
ASSIGNMENT 4
INSTRUCTIONS: You are to work independently on this assignment. The total number of points possible is 50. Please note that point allocation varies per question. Use the Help feature in MINITAB 16 to read descriptions for the data sets so that you can make meaningful comments.
[10 pts] 1. Use the data set OPENHOUSE.MTW in the Student14 folder. Perform the Chi
Square test for independence to determine whether style of home and location are are related. Use α = 0.05. Explain your results.
Pearson Chi-Square = 37.159, DF = 3, P-Value = 0.000
Likelihood Ratio Chi-Square = 40.039, DF = 3, P-Value = 0.000
The P value associated with out chi square is 0.00 and the Alpha level is 0.05 so we reject the null hypothesis. The P- value is less than the alpha level. So, we conclude that style of homes and locations are not related.
[10 pts] 2. Use the data set TEMCO.MTW in the Student14 folder. Perform the Chi
Square test for independence to determine whether department and gender are related. Use α = 0.05. Explain your results.
Pearson Chi-Square = 1.005, DF = 3, P-Value = 0.800
Likelihood Ratio Chi-Square = 1.012, DF = 3, P-Value = 0.798
The P-value associated with out chi square is 0.800 and the Alpha level is 0.05 we can see that we are unable to reject the null hypothesis. The P- value is greater than the alpha level. So, we conclude that departments and gender are related..
[30 pts] 3. Use the data set...

...1. A radio station that plays classical music has a “By Request” program each Saturday night. The percentage of requests for composers on a particular night are listed below:
Composers Percentage of Requests
Bach 5
Beethoven 26
Brahms 9
Dvorak 2
Mendelssohn 3
Mozart 21
Schubert 12
Schumann 7
Tchaikovsky 14
Wagner 1
a. Does the data listed above comprise a valid probability distribution? Explain.
The individual probabilities are all between 0 & 1 and the sum = 100%
b. What is the probability that a randomly selected request is for one of the three B’s?
P(one of the B’s) = P(Bach) + P(Beethoven) + P(Brahms) = 5 + 26 + 9 = 40%
c. What is the probability that a randomly selected request is for a Mozart piece?
P(Mozart) = 21%
d. What is the probability that a randomly selected request is not for one of the two S’s?
P(not Schubert or Schumann) = 1 – P(Schubert or Schumann)
= 1 – (12 + 7)
= 81%
e. Neither Bach nor Wagner wrote any symphonies. What is the probability that a randomly selected request is for a composer who wrote at least one symphony?
P(Symphony) = 1 – P(Bach or Wagner)
= 1 – (5 + 1)
= 94%
f. What is the probability that a randomly selected request is for a composer other...

...1. Introduction
This report is about the case study of PAR, INC. From the following book: Statistics for Business an Economics, 8th edition by D.R. Anderson, D.J. Sweeney and Th.A. Williams, publisher: Dave Shaut. The case is described at page 416, chapter 10.
2. Problem statement
Par, Inc. has produced a new type of golf ball. The company wants to know if this new type of golf ball is comparable to the old ones. Therefore they did a test, which consists out of 40 trials with the current and 40 trials with the new golf balls. The testing was performed with a mechanical fitting machine so that any difference between the mean distances for the two models could be attributed to a difference in the design. The outcomes are given in the table of appendix 1.
3. Hypothesis testing
The first thing to do is to formulate and present the rationale for a hypothesis test that Par, Inc. could use to compare the driving distance of the current and new golf balls. By formulation of these hypothesis there is assumed that the new and current golf balls show no significant difference to each other. The hypothesis and alternative hypothesis are formulated as follow:
Question 1
H0 : µ1 - µ2 = 0 (they are the same)
Ha : µ1 - µ2 ≠ 0 (the are not the same)
4. P-value
Secondly; analyze the data to provide the hypothesis testing conclusion. The p-value for the test is:
Question 2
Note: the statistical data is provide in § 5.
-one machine
-two...

...typically have? You take a random sample of 51 reduced-fat cookies and test them in a lab, finding a mean fat content of 4.2 grams. You calculate a 95% confidence interval and find that the margin of error is ±0.8 grams. A) You are 95% confident that the mean fat in reduced fat cookies is between 3.4 and 5 grams of fat. B) We are 95% confident that the mean fat in all cookies is between 3.4 and 5 grams. C) We are 95% sure that the average amount of fat in the cookies in this study was between 3.4 and 5 grams. D) 95% of reduced fat cookies have between 3.4 and 5 grams of fat. E) 95% of the cookies in the sample had between 3.4 and 5 grams of fat. Determine the margin of error in estimating the population parameter. 12) How tall is your average statistics classmate? To determine this, you measure the height of a random sample of 15 of your 100 fellow students, finding a 95% confidence interval for the mean height of 67.25 to 69.75 inches. A) 1.5 inches B) 0.25 inches C) 1.06 inches D) 1.25 inches E) Not enough information is given. 12) 11) 10)
3
Construct the indicated confidence interval for the difference between the two population means. Assume that the assumptions and conditions for inference have been met. 13) The table below gives information concerning the gasoline mileage for random samples of trucks of two different types. Find a 95% confidence interval for the difference in the means m X - m Y. Brand X Brand Y 50 50 20.1 24.3 2.3 1.8 13)...

...
Final Project: Nyke Shoe Company
Barbara Greczyn
STA 201 - Principles of Statistics
Instructor Alok Dihtal
April 26, 2015
Introduction
Nyke Shoe Company has been in business for over 50 years. Over the last five years, the company has been undergoing some financial hardship due to an erratic market and an inability to understand what the consumer actually needs. In a last ditch effort to avoid bankruptcy, they have adopted a new business model which entails the development of only one shoe size. In order to achieve this goal, statistical data must be utilized and applied to make the best choice. The data used will be explained to the fullest and a conclusion will be then obtained.
Methodology
A sample group of 35 participants was gathered, 18 females and 17 males. Their heights and shoe sizes were gathered and their data was processed in three categories: shoe size, height, gender. Descriptive statistics was applied to three separate data sets, one with all participants included, one sets with just female participants, and one with just male participants. Then a two sample t-test was conducted with the assumption that there were unequal variances amongst both male and female data sets.
Results
There is a normal distribution of the data with ranges in size from size 5 to size 14 amongst the participants. With these ranges, the mean is 9.142, with a standard deviation of 2.583 and a variance of 6.670....

...that we are accepting the alternative hypothesis and this statement works vice versa. In this case that means that the null hypothesis can be rejected or disproving. For the data set that was given the null hypothesis also known as H-nought was µ1=µ2, while the alternative hypothesis is µ1<µ2. Null hypothesis states that the amount of rural nurse homes was equal to the average amount of beds used. Alternative hypothesis states that rural area nursing homes uses fewer amounts of beds. The claim indicated to what kind of test was going to be used and since I claimed that the rural area were going to have a lower average number of beds it states that the shaded area on the critical value test will be less than zero.
Table 1. Descriptive statistics for the given null and alternative hypothesis that includes the sample, mean, median, standard deviation, maximum values, and minimum values.
Sample Size
Mean
Median
Standard Deviation
Maximum Value
Minimum Value
Rural Area
34
0.6538
1.0000
0.4803845
1
0
Bed
4850
93.27
88.00
40.85273
244.00
25.00
Figure 1. This figure illustrates the critical value test for the left-tailed test. The critical value that was needed for the test was -1.692 according to the t-table since our sample size was 34. Used the degree of freedom formula to find the critical value.
Figure 2. This figure reflects to the p-value. When we figured out the p-value we used pt(t,33). Since pt(t,33) equaled 0.0137855 that indicated...

...
1 - Produce the relevant descriptive graph and table to summarise the MODE variable (labelled Control: Completes by phone/mail/web). The MODE variable summarises the different ways that each institution completed the survey. Write a paragraph explaining the key features of the data observed through the output.
The distribution of the surveyance mode which was conducted for a sample of 2000. The mode is displayed in the above chart. The most common primary mode was the web (53.9%) it was above the average. The second and third in common was the Mail and the Phone – Full mode at 24% and 16.3% respectively, very few people uses the Phone – Short method (5.9%)
2 - Produce the relevant descriptive graph and table to summarise the OPSEXTOTM variable (labelled QB5b_num_mal. Male outpatients – Total). This variable measures the total number of male outpatients at each institution. Write a paragraph explaining the key features of the data observed through the output
The distribution of male outpatients at each institution in a sample of 2000 is displayed in the above histogram. The distribution is positively skewed with 50% below 96 cases or less. Typically male outpatient cases were between 30 and 223 with mean at 191.62 and standard deviation of 349.62. However there were quite a number of outliers with 3 significant outliers above 4000 cases in ID 2926, 7251 and 6751 and there were a number of institutions with 0 cases as...

...Charismatic Condition
Mean 4.204081633
Standard Error 0.097501055
Median 4.2
Mode 4.8
Standard Deviation 0.682507382
Sample Variance 0.465816327
Kurtosis 5.335286065
Skewness -1.916441174
Range 3.5
Minimum 1.5
Maximum 5
Sum 206
Count 49
Confidence Level(95.0%) 0.196039006
In both the Charismatic and the punitive condition data sets there were 49 people surveyed. We know this because we were able to use descriptive statistics to show the count and that shows the number of people surveyed. The average or the mean of the charismatic condition is 4.20. The standard error is saying that 0.0975 % is the error that will normally occur if two different people are comparing results. The middle value or the median of the data set is 4.2 and the most frequently occurring value or mode is 4.8. The charismatic condition is skewed to the left because we are getting a negative number for the skewness data. The skewness is at -1.916441174. The difference between the largest and the smallest value which is also called the range is 3.5. The minimum score is 1.5 and the maximum score was a 5 given by the students for the charismatic condition.
The Frequency for the charismatic condition is telling us the summary of the data that is presented is the form of class intervals and frequency. This bar graph will show you the values of the scores starting at 1.5 and going up to 5. The frequency shows us the number of students who picked the...