1.The length (in millimetres) of a batch of 9 screws was selected at random from a large consignment and found to have the following information.

8.028.008.018.017.998.007.998.038.01

Construct a 95% confidence interval to estimate the true average length of the screws for the whole consignment.

From a second large consignment, another 16 screws are selected at random and their mean and standard deviation found to be 7.992 mm and 0.01mm. Can you conclude at 5% level of significance that the first batch of screws has greater mean than the second batch?

2.A sample of 8 independent observations provides the following:

3.63.93.84.54.94.24.43.8

Can you conclude at 5% level of significance that the mean is below 5?

3.A house cleaning service claims that they can clean a four bedroom house in less than 2 hours. A sample of n = 16 houses is taken and the sample mean is found to be 1.97 hours and the sample standard deviation is found to be 0.1 hours. (i)Construct a 95% confidence interval for the population mean of cleaning times. (ii)Conduct a hypothesis testing by using 0.05 level of significant to verify the claim.
4.A maker of toothpaste is interested in testing whether the proportion of adults (over age 18) who use their toothpaste and have no cavities within a six-month period is any different than the proportion of children (18 and under) who use the toothpaste and have no cavities within a six-month period. To test this, they have selected a sample of adults and a sample of children randomly from the population of those customers who use their tooth paste. The following results were observed.

AdultsChildren
Sample Size 100 200
Number with 0 cavities 83 165

Based on these sample data and using a significance level of 0.05, what...

...of 1000 flights and proportions of three routes in the sample. He divides them into different sub-groups such as satisfaction, refreshments and departure time and then selects proportionally to highlight specific subgroup within the population. The reasons why Mr Kwok used this sampling method are that the cost per observation in the survey may be reduced and it also enables to increase the accuracy at a given cost.
TABLE 1: Data Summaries of Three Routes
Route 1
Route 2
Route 3
Normal(88.532,5.07943)
Normal(97.1033,5.04488)
Normal(107.15,5.15367)
Summary Statistics
Mean
88.532
Std Dev
5.0794269
Std Err Mean
0.2271589
Upper 95% Mean
88.978306
Lower 95% Mean
88.085694
N
500
Sum
44266
Summary Statistics
Mean
97.103333
Std Dev
5.0448811
Std Err Mean
0.2912663
Upper 95% Mean
97.676525
Lower 95% Mean
96.530142
N
300
Sum
29131
Summary Statistics
Mean
107.15
Std Dev
5.1536687
Std Err Mean
0.3644194
Upper 95% Mean
107.86862
Lower 95% Mean
106.43138
N
200
Sum
21430
From the table above, the total number of passengers for route 1 is 44,266, route 2 is 29,131 and route 3 is 21,430 and the total numbers of passengers for 3 routes are 94,827.
Although route 1 has the highest number of passengers and flights but it has the lowest means of passengers among the 3 routes. From...

...compliments the regular mathematics and therefore both are tested in primary schools. Mathematics is the written application of operation. It teaches students to think clearly, reason well and strategize effectively. Mental Mathematics is the ability to utilise mathematical skills to solve problems mentally. The marks scored by pupils generate statistics which are used by teachers to analyse a student’s performance and development of theories to explain the differences in performance.
The Standard 3 class is where the transition from junior to senior level occurs where teachers expect the transference of concrete to abstract thinking would have occurred.
A common theory by many primary school teachers is ‘Students perform better in Mathematics than Mental math. Mental math is something that has to be developed and involves critical thinking. Mental math requires quick thinking and the student must solve the problem in their minds whereas in regular mathematics, the problem can be solved visually. Therefore, teachers should take these factors into consideration while testing and marking students in these areas.’
In this study, the statistics of 30 students of a standard 3 class of San Fernando Boys’ Government School will be analysed to determine the truth of this theory.
DATA COLLECTION METHODS
Mathematics and mental mathematics marks of term 1 of the class of 2013 were obtained from a Standard 3 teacher of San...

...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)...

...corresponds to the following Confidence Levels:
a. 80%
Confidence level = 80%
or Significancelevel α= 0.2 / =(1-0.80)
No of tails = 2
Z value corresponding to 0.2 significancelevel and 2 tails = 1.2816
=NORMSINV(1-0.2/2)
=ROUND(NORMSINV(1-0.2/2),4)
b. 85%
Confidence level = 85%
or Significancelevel α= 0.15/ =(1-0.85)
No of tails = 2
Z value corresponding to 0.15 significancelevel and 2 tails = 1.4935
=NORMSINV(1-0.15/2)
=ROUND(NORMSINV(1-0.15/2),4)
c. 92%
Confidence level = 92%
or Significancelevel α= 0.08/ =(1-0.92)
No of tails = 2
Z value corresponding to 0.08 significancelevel and 2 tails = 1.7507
=NORMSINV(1-0.08/2)
=ROUND(NORMSINV(1-0.08/0.2),4)
d. 97%
Confidence level = 97%
or Significancelevel α= 0.03/ =(1-0.97)
No of tails = 2
Z value corresponding to 0.03 significancelevel and 2 tails = 2.1701
=NORMSINV(1-0.03/2)
=ROUND(NORMSINV(1-0.03/0.2),4)
2. Using Excel, find the t-score that corresponds to the following Confidence Levels and Sample Sizes:
a. 95% with n = 25
Confidence level = 95%
Or Significancelevel α= 0.05 (1-0.95)
N= 25
degrees of freedom=n-1=...

...Statistics 1
Business Statistics
LaSaundra H. – Lancaster
BUS 308 Statistics for Managers
Instructor Nicole Rodieck
3/2/2014
Statistics 2
When we hear about business statistics, when think about the decisions that a manager makes to help make his/her business successful. But do we really know what it takes to run a business on a statistical level? While some may think that business statistics is too much work because it entails a detailed decision making process that includes calculations, I feel that without educating yourself on the processes first you wouldn’t know how to imply statistics. This is a tool managers will need in order to run a successful business. In this paper I will review types of statistical elements like: Descriptive, Inferential, hypothesis development and testing and the evaluation of the results. Also I will discuss what I have learned from business statistics.
My description of Descriptive statistics is that they are the numerical elements that make up a data that can refer to an amount of a categorized description of an item such as the percentage that asks the question, “How many or how much does it take to “ and the outcome numerical amount. According to “Dr. Ashram’s Statistics site” “The quantities most commonly used to measure the dispersion of the...

...in the preparation of the responses to this quiz. By signing this test, I pledge that I have adhered to the Old Dominion University honor system and the requirements of this quiz.
Student signature
You have two weeks to complete this quiz: it is due prior to class on March 17th. You must email me your test response as ONE Excel file attachment to mcochran@odu.edu.
Please be sure to show all work in order to facilitate partial credit. You may use software in the process of determining the answer. However, if you do so please do not simply provide “the answer”. For example, if you use Excel, be sure all appropriate formula calculations are visible. Note that all graphs, tables and other work must be labeled to a level of professional quality. All work is to be done independently. This quiz contains 7 questions in 4 pages.
1. (13 Points) The time in days between breakdowns of two machines is exponentially distributed with a mean time between failure of 5 for machine A and 6 for machine B.
a. What is the probability that machine A will function for one week before breaking down?
b. What is the probability that both machines will function for a week before breaking down?
c. If machine A has performed satisfactorily for six days, what is the probability that it lasts at least two more days before breaking down?
2. (20 Points) Use Stat Plus and Excel to illustrate the Central Limit Theorem when sampling from...

...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...