1.Assume 20% of all email is spam. A large Internet provider plans on conducting a survey of 900 emails to see what percentage are spam. a.What is the probability they will get a proportion greater than 0.1836?

b.If they get a sample proportion over 24% they are going to shut down their email server. What is the probability this will happen?

2.A survey is done to estimate the proportion of U.S. adults who think that cell phone use while driving should be illegal. In the survey, 54% of a randomly selected sample of 1025 individuals said that cell phone use while driving should be illegal. a.What is the 90% confidence interval for the proportion of adults who think cell phone use should be illegal?

1. A sample of college students was asked whether they would return the money if they found a wallet on the street. Of the 93 women, 84 said “yes,” and of the 75 men, 53 said “yes.” Assume that these students represent all college students (Data source is from UC Davis and can be found in the textbook). a. Is there enough data to calculate a confidence interval for the women?

2. A sample of college students was asked whether they would return the money if they found a wallet on the street. Of the 93 women, 84 said “yes,” and of the 75 men, 53 said “yes.” Assume that these students represent all college students (Data source is from UC Davis and can be found in the textbook). b. Is there enough data to calculate a confidence interval for the women?

3.A CNN/Time poll conducted in the United States October 23-24, 2002, (http://www.pollingreport.com) asked, “Do you favor or oppose the legalization of marijuana?” In the nationwide poll of n = 1007 adults, 34% said that they favored legalization. a.Find the margin of error for a 96% confidence interval.

3. A medical researcher wants to study whether oral contraceptives are correlated with high blood pressure. A sample of 500 women using oral contraceptives showed...

...Z table) to a 3. p-value of 1 – 0.9987 = 0.0013, or 0.13% To be conservative, let us assume a two-sided test. In that case, the p-value would be 0.0026, or 0.26%
a) a) Since p < 0.05, the result is statistically significant, and H0 can be rejected. Therefore, the participants can tell the difference between the two oils and the results are not due to random chance.
b) b) As calculated above, p = 0.0013. This states that there is a 0.13% chance that, should H0 be true, of observing a “correct identification rate” greater than the sample rate observed (57.5% of participants correctly identifying the oils).
c) c) It is safe to say there is a taste difference. Whether or not the difference is large is inconclusive – the study only found that participants can distinguish between the two oils. The analysis of this test helps to answer the question by showing that a taste difference does exist by proving that random chance is not the cause of the test results.
Q. 45 out of 1287 patients took drug and got sick. 25 out of 1299 in control group took placebo and got sick. Cross threshold of statistical significance?
A. H0 = any difference in heart attack or stroke percentage in the groups is due to ordinary chance variation (= we would expect an average difference in death of 0 between the two groups unless we have a variation due to random chance) -- HA = something else beyond random chance is responsible for the difference in attack or stroke...

...Descriptive Statistics
QNT/561
July 29, 2014
Descriptive Statistics
Job Satisfaction
Central Tendency: Mean=8.5 JDI
Dispersion: Standard Deviation=1.16 JDI
Number: 139
Min/Max: 7 to 10 JDI
Confidence Interval: 8.36 to 8.75 JDI
*JDI=Job Descriptive Index
Months of Employment
Central Tendency: Mean= 136.24 Months
Dispersion: Standard Deviation= 117.26 Months
Number: 139
Min/Max: 1 to 359 Months
Confidence Interval: 116.74 to 155.73 Months
Descriptive Statistics Interpretation
Interpretation
Months of Employment
CHJ made a decision to choose 139 random employees. Their months of employment varied between a range of 1 and 359 days the employees estimated months employed with the company was around their months of employment were 136 days, either adding or subtracting 28 days , Nearly half of counted to be 136 days. CHJ strongly believes a 95% population months of employment average is between 116 and 165 days.
Job satisfaction
CHJ carefully selected One hundred employees the range was between 7 and 10. Half of the employees choose 8.75 which meant they were satisfied with their job. The other employees who were surveyed choose 8.6 which mean they were satisfied but CHJ could do a little more improvements.
Appendix A
Raw data used in the analysis
Appendix B
Charts and Tables
Frequency of Male and Female Employees by Department
Legend: Blue=Males Red=Females
Appendix C...

...Convenience Sampling – individuals chosen to be in sample based on convenience or easiness to reach
– Simple Random Sample (SRS) – Every possible sample has same chance (random digits)
– Stratified Random Sample – population is classified into strata, then SRS within strata
– Multistage Sample – take samples at multiple stages (political polling – region then SRS phone numbers)
– Failure to SRS → bias
– Undercoverage, nonresponse, and poorly worded questions cause bias
– Voluntary Response Sample – voluntary participation, super bias, limited access to sample
– Random digit dialing over phone subject to nonresponse and undercoverage to cell-phone only houses
Chapter 9
– Observational Study – observe variables of interest for individuals in samples
– Experimental Study – study impact of explanatory variable(s) on response variable by imposing values of explanatory on the individuals in the sample, then record response.
– Variables are confounded if their effects on a response cannot be distinguished (eliminate by making all same)
– In experiments, treatments are imposed on subjects, variables are factors
– Design of experiment describes choice of treatments and in which manner subjects are assigned treatments
– Randomization uses chance to assign subjects to treatments, to prevent bias (systematic favoritism)
– Double-blind – no one knows what is going on
– Placebo – given to control
– Lurking Variable – can affect...

...Statistics 5371 Final Exam Review Fall 2012.
1. Suppose a researcher wants to design a new study with a power of 0.8 and a significance of 0.05 to test whether the caffeine content for a brand of coffee is really 100mg. A previous study gave a mean caffeine level for this brand of 110 mg and a standard deviation of 7 mg. Use PROC POWER to determine how many cups of coffee need testing.
2. A company did a study to estimate the effect of different promotional strategies on the market share of one of their products. Over a period of 36 months they varied their promotional strategy. There are four strategies: Ordinary (standard pricing and advertising); Discount (price discount with standard advertising); Promotion (standard pricing, enhanced advertising); and Both (price discount with enhanced advertising). Each month one of the strategies was employed and the market share (the percentage of purchasers of the given product type that selected the company's product). Higher market share is better for the company. The table below records the number of months in which each strategy was used (n) and a summary of the market share results in those months.
|Strategy |Pricing |Advertising |n |mean |SD |
|Ordinary |standard |standard |8 |2.40 |0.12 |...

...Tutorial Questions for Exploratory Data Analysis – Summary Statistics & Graphs
1. Customers of a particular bank rated the service provided by the bank on a scale of one to ten, correct to one decimal point. The bank categorised their customers as either (1) Private Account holders or (2) Business Account holders. The information below summarises customer attitudes towards the quality of service provided by the bank. Use the output to answer the questions below.
a) Briefly describe and compare the distributions of results for the two groups of customers. You should mention appropriate measures of centre and spread, and any other points you feel may be of interest.
b) The survey results were described as being symmetrically distributed. What does this mean, and what evidence is there below to support this claim?
c) The standard deviation of the ratings for the Private Account holders is 1.336. What does this value mean?
d) Verify the value of the standard error of the mean (SE Mean) for the Business Account holders.
e) Determine a 90% confidence interval for the true mean of the Business Account holders and interpret the result in the context of the situation.
Descriptive Statistics: Quality
Variable Use N N* Mean SE Mean StDev Minimum Q1 Median Q3
Quality 1 45 0 5.971 0.199 1.336 3.700 4.800 6.000 7.050
2 30 0 8.323 0.172 0.941 6.200 7.675 8.400 9.025
Variable Use...

...Chapter 7
Student Lecture Notes
7-1
Business Statistics: A Decision-Making Approach
6th Edition
Chapter 7 Estimating Population Values
Fundamentals of Business Statistics – Murali Shanker
Chap 7-1
Confidence Intervals
Content of this chapter Confidence Intervals for the Population Mean, μ
when Population Standard Deviation σ is Known when Population Standard Deviation σ is Unknown
Determining the Required Sample Size
Fundamentals of Business Statistics – Murali Shanker
Chap 7-2
Fundamentals of Business Statistics – Murali Shanker
Chapter 7
Student Lecture Notes
7-2
Confidence Interval Estimation for μ
Suppose you are interested in estimating the average amount of money a Kent State Student (population) carries. How would you find out?
Fundamentals of Business Statistics – Murali Shanker
Chap 7-3
Point and Interval Estimates
A point estimate is a single number, a confidence interval provides additional information about variability
Lower Confidence Limit
Point Estimate Width of confidence interval
Upper Confidence Limit
Fundamentals of Business Statistics – Murali Shanker
Chap 7-4
Fundamentals of Business Statistics – Murali Shanker
Chapter 7
Student Lecture Notes
7-3
Estimation Methods
Point Estimation
Provides single value Based on observations from 1 sample Gives no information on how close value is to the population parameter
Interval Estimation
Provides range of values...

...STATISTIC
(1) A study of the number of cars sold looked at the number of cars sold at 500
Dealers The smallest dealer had 11 cars sold and the largest had 154 cars sold. If you were going to create a frequency distribution of the number of cars sold using six classes in your frequency distribution, which of the following might be a
reasonable first class?
(a) 0 to 25
_
(2) Give that a sample of 25 had x = 75, and (x-x)² = 48 the mean and standard
deviation are
_
(c) X = 3.0 and S = 48/24
(3) Which of the following is always true for a test of a hypothesis?
(a) The type ll error is more important than type l error (b) is larger then
(c ) is the probability of a type ll error (d) A type l error occurs when you
accept the null hypothesis when it is not true (e) None of the above is true.
(4) Suppose that, in attesting situation with a population that is normally distributed,
The null hypothesis is that the population mean is equal to 41. A sample of size 36 had a sample mean of 41 and sample standard deviation of 2. Which of the
following is correct?
(b) You cannot reject the null hypothesis
(5) If you take a sample of size 24 from a normal...

...Chapter 7: Intro to Sampling Distributions
Sampling Error = x̄ - μ
Z-Values for a sampling distribution of x̄ :
Z =
Z-Values adjusted with Finite Population Correction
Applied if: the sample is large relative to the population (n is greater than 5% of N) and sampling Is without replacement
Z =
Using the Sampling Distribution for Means
Compute the Sample Mean
Define the sampling distribution
μx̄ =
Define the probability statement of interest
P(z30 will give sampling distribution that is nearly normal
fairly symmetric, n>15
The sampling distribution with have:
Sampling Distribution of a Proportion
Sampling Proportion Error = p –
Sampling Distribution of P is approximated by a normal distribution if
Where this applies:
Mean =
Standard Error =
Z-Values for Proportions
Using the sample distribution for Proportions
Determine the population proportion,
Calculate the sample proportion, p
Derive the mean and standard deviation of the sampling distribution
Define the event of interest
If n and n(1- are both > 5, the convert p to z-value
Use the standard normal table to determine probability
Chapter 8: Estimating Single Population Parameters
Point and Confidence Interval Estimates for a Pop Mean
Point Estimate ± (Critical Value)(Standard Error)
Confidence interval for μ ( known)
Steps:
Define the population of interest and select a simple random sample of size n
Specify the confidence level
Compute the sample mean
Determine...