1.Create a new variable WHISCO that tells us the liking for whiskey and scotch combined. What is the mean of this new variable? 3.3581

2.What is the mean liking for regular beer for the females that

completed the survey? 3.0571

3.Create a new variable HARD that tells us the liking for whiskey, gin, rum and scotch combined. What is the mean of this new variable?3.7703

4.What is the mean liking for whiskey for the males that completed the
survey?4.7105

---------------------------------------------------------------------------------------------------------------------------- The following two statements are Null Hypotheses written in English. For each, rewrite the null using mathematical symbols and notation as shown in the ClassPak. The statement in parentheses refers to the Alternative Hypothesis for each.

5. On average IU students go out 2.5 nights or more per week. (The alternative hypothesis is that IU students go out less than 2.5 times per week.)

Ho: μNights >= 2.5
Ha: μNights < 2.5

6. On average IU students drink exactly 5 1/2 drinks in an average night out. (The alternative hypothesis is that IU students do not drink 5 1/2 drinks per night out.)

Ho: μDrinks = 5.5
Ha: μDrinks ≠ 5.5

Use SPSS to calculate a p-value for each of the null hypotheses you have written in 5 and 6 above:

...I am the Statistical Analyst for Slivers’ Gold Gym (SGG). I was assigned a project to examine the relation of 252 male gym members weight and body fat by conducting a hypothesis test. This report reflects the measures that I examined and how I conducted my hypothesis test to conclude if the male members have an average body fat of 20% as claimed by my boss.
Part I
I analyzed the compiled sample data set that was provided to me for the hypothesis test of the body fat and weight of 252 male members of SGG. I used excel to help me find the mean, median, range and standard deviation. The excel sample data set will is appended to this report. The average body fat of the SGG 252 male gym members I analyzed is 18.9 (SD = 7.8) and the average weight is 178.9lbs (SD = 29.4). The median of the body fat sample data set of the 252 male SGG members is 19.0 and the weight median is 176.5lbs. The range for the body fat is 45.1 and for weight is 244.7lbs. The way I calculated the mean is to simply add the sample data sets of the body fat (4772.5) and weight (45088.95) and then divided these values by the 252 male gym members. The median is simply the middle number of the body fat and weight sample data sets. I calculated the range by finding the difference between the values of the body fat and weight sample data sets, body fat (45.1 - 0.0 = 45.1) weight (363.15 – 118.50 = 244.65). To determine the standard deviation (SD) I calculated how close the sample data sets of...

...so would the representatives from MSF. What would you do if you are representing AFP? But, if you are representing MSF, how would you present your argument? (Hint: Consider your argument based on significance levels.)
8. (8 points) But, wait. What if MSF actually does not know the population standard deviation in this case, would you conduct your hypothesis test differently? Just in case that you are going to perform the hypothesis differently, what would you do instead?
The following information is for Questions 9 and 10.
The tête-à-tête between MSF and AFP broke down, as anyone would have anticipated. They are going to court.
The presiding judge, His Honor Ig Sushi, is rather apprehensive of errors in any statistical testing. Let us assume he picks a 95% significance level, an industrial standard, to be the benchmark of this court case.
I know, I know, everything sounds rather fishy......
9. (14 points) What if the RDA of cobalamine is actually not well established in the medical community? In fact, it may span a range of values in µg, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6. What is the corresponding Type II error and power for each of the given assumed RDA value?
10. (4 points) Plot the corresponding powers against the range of assumed RDA values? What do you observed in this plot, the so-called power plot? Please explain....

...Hypothesis Testing For a Population Mean
The Idea of Hypothesis Testing
Suppose we want to show that only children have an average higher cholesterol level than the national average. It is known that the mean cholesterol level for all Americans is 190. Construct the relevant hypothesis test:
H0: = 190
H1: > 190
We test 100 only children and find that
x = 198
and suppose we know the population standard deviation
= 15.
Do we have evidence to suggest that only children have an average higher cholesterol level than the national average? We have
z is called the test statistic.
Since z is so high, the probability that Ho is true is so small that we decide to reject H0 and accept H1. Therefore, we can conclude that only children have a higher average cholesterol level than the national average.
Rejection Regions
Suppose that = .05. We can draw the appropriate picture and find the z score for -.025 and .025. We call the outside regions the rejection regions.
We call the blue areas the rejection region since if the value of z falls in these regions, we can say that the null hypothesis is very unlikely so we can reject the null hypothesis
Example
50 smokers were questioned about the number of hours they sleep each day. We want to test the hypothesis that the smokers need less sleep than the general public which needs an average of 7.7 hours of sleep. We follow the steps below.
...

...Simple Hypothesis: A statistical hypothesis which specifies the population completely (i.e. the form of probability distribution and all parameters are known) is called a simple hypothesis.
1. Composite Hypothesis: A statistical hypothesis which does not specify the population completely (i.e. either the form of probability distribution or some parameters remain unknown) is called a Composite Hypothesis.
Hypothesis Testing or Test of Hypothesis or Test of Significance
Hypothesis Testing is a process of making a decision on whether to accept or reject an assumption about the population parameter on the basis of sample information at a given level of significance.
Null Hypothesis: Null hypothesis is the assumption which we wish to test and whose validity is tested for possible rejection on the basis of sample information.
It asserts that there is no significant difference between the sample statistic (e.g. Mean, Standard Deviation(S), and Proportion of sample (p)) and population parameter (e.g. Mean(µ), standard deviation (σ), Proportion of Population (P)).
Symbol-It is denoted by Ho
Acceptance- The acceptance of null hypothesis implies that we have no evidence to believe otherwise and indicates that the difference is not significant.
Rejection- The rejection of null hypothesis implies that it is false and indicates that the difference is significant.
Alternative Hypothesis: Alternative hypothesis is the...

...Kruskal-Wallis Tests in SPSS
STAT 314
Three teaching methods were tested on a group of 19 students with homogeneous backgrounds in statistics and comparable aptitudes. Each student was randomly assigned to a method and at the end of a 6-week program was given a standardized exam. Because of classroom space and group size, the students were not equally allocated to each method. The results are shown in the table below. Test for a difference in distributions (medians) of the test scores for the different teaching methods using the Kruskal-Wallis test. Method 1 94 87 90 74 86 97 Method 2 82 85 79 84 61 72 80 Method 3 89 68 72 76 69 65 1. Enter the time values into one variable and the corresponding teaching method number (1 for Method 1, 2 for Method 2, 3 for Method 3) into another variable (see figure, below). Be sure to code your variables appropriately.
2.
Select Graphs Boxplot… (Simple, Summaries for groups of cases) with the variable measured (Test Score) and the category axis variable (Teaching Method) entered (see figures, below). Click “OK”.
3.
Your resulting side-by-side boxplots will appear (see figure, below). As long as the boxes have approximately the same shape, you may continue with the ANOVA procedure.
4.
Select Analyze
Nonparametric Tests
K Independent Samples… (see figure, below).
5.
Select “Test Score” as the test variable, select “Teaching Method” as the grouping factor, and click “Define Range…”. Enter the minimum...

...EXERCISE 36 Questions to be graded
1. The researchers found a significant difference between the two groups (control and treatment) for change in mobility of the women with osteoarthritis (OA) over 12 weeks with the results of F(1, 22) = 9.619,p = 0.005. Discuss each aspect of these results.
Answer
F-value shows significance between the two groups. The null hypothesis should be rejected because the P-Value is 0.005 which would mean that the groups are different.
2. State the null hypothesis for the Baird and Sands (2004) study that focuses on the effect of the GI with PMR treatment on patients’ mobility level. Should the null hypothesis be rejected for the difference between the two groups in change in mobility scores over 12 weeks? Provide a rationale for your answer.
Answer
The null hypothesis would be that the means for each groups the control and the treatment will be equal. The hypothesis will be rejected as the probability of happening is 0.005. Mobility scores are different.
3. The researchers stated that the participants in the intervention group reported a reduction in mobility difficulty at week 12. Was this result statistically significant, and if so at what probability?
Answer
Results are statically significant P=0.005
results are significant at exactly the 99.5% level (very high)
4. If the researchers had set the level of significance or α = 0.01, would the results of p = 0.001 still be statistically significant? Provide a...

...the model of Bernoulli trials is inadequate.
KEY WORDS: Bernoulli trials, the hot-hand, power, simulation study, case study.
1
1 Introduction
In this paper I consider a statistical analysis of basketball shooting in a controlled (practice) setting, with
special interest in the hot-hand. In Section 2, I review and critically examine the two seminal papers on this
topic: Gilovich, Vallone, and Tversky (GVT) [5], and Tversky and Gilovich (TG1) [10]. A simulation study
of power is presented in Section 3. Finally, in Section 4, a case study of 2,000 trials is analyzed.
In GVT and TG1, three additional topics appear which are beyond the scope of this paper:
1. Modeling game free throw shooting,
2. Modeling game shooting, and
3. Opinions and misconceptions of fans.
Readers interested in the first of these topics should refer to Wardrop [12] for a further analysis of the free
throw data from the papers.
Several researchers have considered the second topic; the interested reader is referred to Larkey, Smith,
and Kadane (LSK) [7], Tversky and Gilovich (TG2) [11], Hooke [6] and Forthofer [4]. For related work in
baseball, see Albright [1], Albert [2], Stern and Morris [9] and Stern [8]. Topic 3 is considered briefly in
Section 2 of this paper.
Finally, readers interested in statistical research in sports are referred to Bennett [3]. The chapters on
basketball and baseball should prove to be of special interest to...

...Submit completed tests as word or pdf files via email to paul.kurose@seattlecolleges.edu
Due: Sunday, May 19 (by 8am).
1. a) In Chapter 6, you learned to find interval estimates for two population parameters, a population mean and a population proportion. Explain the meaning of an interval estimate of a population parameter.
An interval estimate for a specified population parameter (such as a mean or proportion) is a range of values in which the parameter is estimated to lie. In Chapter 6, you were assigned to find interval estimates for a population mean and a population proportion.
b) Is finding an interval estimate an example of inferential or descriptive statistics? Explain.
It is an interval estimate is an example of inferential statistics, as an estimate of the value of the population parameter is made based on sample statistics.
c) An interval estimate (23.8, 30.6) is determined for the mean age of NSCC students. Identify the point estimate and the margin of error of the interval estimate.
The point estimate=27.2 and the margin of error=3.4
d) The proportion of registered voters in Washington State in favor of a referendum to lower college tuitions is estimated with a 95% confidence level to be 45% with a margin of error ±3%. Is it possible the referendum will pass? Explain.
Yes, It is possible the referendum will pass. The margin of error is only stated with 95% confidence. In the given case, the estimate that the population proportion will...