Answer ASAP please!!!

1. You are the researcher and you develop a hypothesis. It states that by providing relaxation therapy to a group of 10 smokers who average 20 cigarettes a day, you hypothesize that they will reduce the number of cigarettes they smoke after eight weeks of therapy. You will take measurements on the group you have provided relaxation therapy to after eight weeks and the mean is 16 cigarettes. You compare the mean to two other groups who did not receive the relaxation therapy. One group, n=10 received no treatment and averaged 19.7 cigarettes at the end of eight weeks, and the other group, n=10 who was given literature on the negative effects of smoking averaged 18.1 cigarettes after eight weeks. You calculate the means of all groups using the appropriate hypothesis statistic. You hypothesize that there will be a difference in the means after eight weeks. You measure smoking levels after you terminate the therapy to see if there are effects from the independent variable. The obtained test statistic is 12.396 and a p value of 0.000152. You used the alpha level of 0.05.

1. How many groups (K) are involved in this study________
2. How many participants do you have in the study N=_____
3. How many degrees of freedom are there _BG_____   _WG_____Total_____
4. What is the critical value/region of rejection base on an alpha level of .05_______
5. What is the correct test statistic_________
6. What is the independent variable_________
7. What is the dependent variable __________
8. Based on the obtained value, do you reject or fail to reject H0:______
1. State the reason you made your decision__________

A manufacturing plant bought a new machine to package its product. After some testing, the plant supervisor decided that the packages needed about 300 packing peanuts to keep the product safe. She set the new machine to dispense the peanuts into each box but wanted to test to make sure the machine was operating correctly. She ran the machine 20 times and counted the number of packing peanuts dispensed.

The plant conducts a one-mean hypothesis at the 5% significance level, to test if the the mean number of peanuts dispensed is different from 300.

(a) H0:μ=300; Ha:μ≠300, which is a two-tailed test.

(b) Student test scores are given below.

Use Excel to test whether the mean number of peanuts dispensed is different from 300. Identify the test statistic, t, from the output, rounding to two decimal places.

286
291
268
305
321
311
295
341
325
280
295
329
296
340
335
307
325
310
331
291

test statistic =         p-value =

1. How does your estimate of beta compare with the beta estimate provided (1.07)? Why might your estimate differ from estimated beta of 1.07?

2. How much of the variability of your security’s return is “explained” by the variability of returns in the “market”? (Note: In your case, the market is represented by the S&P 500 Index.) Do you think that a different market index might be a better representation of the market for your particular security? Why/Why not?

3. What is the correlation of returns for your security with the market for the selected time period? Might this relationship change over time, and if so, how and why?

4. Does the relationship between your security and the market appear to be statistically significantly different than zero? What evidence from the regression supports your conclusion?

5. Review the standardized residuals and comment about the importance of individual data points (if any) that may have influenced your estimation of beta. (observation 2 is the only skewed one)

1. Choose the t test (which mean pick ONE) you think is appropriate out of the three t tests listed below to determine the t-value for the data (you can copy the results from your SPSS output and past them here:
   1. One sample (Mean) test
   2. Two samples (independent t-test) test
   3. Paired/related sample t tests
2. From the t test you ran above, tell me the confidence intervals (look under 95% CI, lower and upper) for the data and interpret what they mean.

Quality of Marriage             Quality of the Parent–Child Relationship

76                                                                    43

81                                                                    33

78                                                                    23

76                                                                    34

76                                                                    31

78                                                                    51

76                                                                    56

78                                                                    43

98                                                                    44

88                                                                    45

76                                                                    32

66                                                                    33

44                                                                    28

67                                                                    39

65                                                                    31

59                                                                    38

87                                                                    21

77                                                                    27

79                                                                    43

85                                                                    46

68                                                                    41

76                                                                    41

77                                                                    48

98                                                                    56

98                                                                    56

99                                                                   55

98                                                                    45

87                                                                   68

67                                                                   54

78                                                                   33

Learning Objectives

1. Understand when and how Chi-square testing is used.
2. Become proficient with the Chi-square analysis in SPSS.
3. Gain more experience with customizing statistical calculations in Excel.
4. Perform a basic interpretation of a Chi-Square analysis.

Using Chi Square

Using a statistical test without having a good idea of what it can and cannot do means that you may misuse the test. This also means that you won't have a clear grasp of what your results really mean. We know that there are basically two types of random variables and they yield two types of data: numerical and categorical. Up to this point in class, we've focused mostly on numerical data. A Chi-square (X²) analysis is used to investigate whether distributions of categorical variables differ from one another. We do this by comparing observed frequencies in different categories of one or more independent variables. Then, we compare these observed frequencies with expected frequencies and determine if there is a significant, proportional difference in the frequency counts of the different categories.

About Your Data

A Psychologist is interested in looking at the personality traits of Business majors. She administers the Big Five personality inventory to 258 Business majors at her university. For each participant, their strongest personality trait was determined and tallied in the corresponding category. The results she obtained are reported below. The psychologist has brought you in to analyze her data. You will need to perform a Chi-square test of goodness of fit.

Instructions

In this lab, you will be completing a Chi-square test in both SPSS and Excel. You will interpret your results. You will submit screenshots and your interpretations.

Chi-Square In SPSS

1. Open SPSS.

2. Enter the personality data in SPSS as per the instructions in the video tutorial.
3. Perform a Chi-square, goodness of fit analysis, as per the instructions in the video tutorial.
4. Assess your output from the analysis. You should see three tables: Descriptives, Frequencies, and Test Statistics. By this point in the semester, you should be able to quickly determine whether you will reject or fail to reject the null hypothesis for this study.
5. Take a screenshot showing your output. I'm most interested in seeing the Test Statistics table.
6. Near the end of the SPSS tutorial, you will see Todd offer conclusions that look a lot like the ones we have practiced in other labs. Use these conclusions as a template to write up your own. Please make sure that your conclusions are clear, correct and free of grammatical mistakes.

Chi-Square In Excel

1. Open Microsoft Excel.

2. Open the Chi-Square in Excel video above. Using the personality data listed in the previous section, follow the steps in the video exactly. Perform all of the same steps. This will be more like a computer-assisted walkthrough of Chi-square. The only thing you don't need to worry about is highlighting rows and columns with color, though I would encourage you to do so. If you run into problems, you might try searching YouTube for other videos providing similar tutoring.

3. When you have completed the Excel analysis, check your Chi-square statistic with the one calculated in SPSS. Do they match? They should.
4. Take a Screenshot of your Excel spreadsheet with the completed work. It should look like the one in the Excel video.

Mother's age 18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,51

Female 1, 0, 2, 2, 3, 4, 7, 3, 2, 4, 7, 1, 6, 4, 5, 3, 1, 4, 0, 1, 1, 1, 0, 1, 0

Use the stem and leaf plots that you previously created to help you draw and label histograms on your scratch paper with bin width of 2 for mothers's age at birth of female students and for mother's age at birth of male students. Make the lower bound of your first bin 16.

Comment: Bin width of 2 is not a typo. Yes, your stem and leaf plot has bins of 5 so some thinking is required, but at least your stem and leaf plot has the values in order for you.

• Stem
  1
  1
  2
  2
  3
  3
  4
  4
  5

  Stem	Leaf
  1
  1	8
  2	0	0	1	1	2	2	3	3	3	3	4	4	4	4	4	4	4
  2	5	5	5	6	6	7	7	7	7	8	8	8	8	8	8	8	9
  3	0	0	0	0	0	0	1	1	1	1	2	2	2	2	2	3	3	3	4
  3	5	5	5	5	7	8	9
  4	1
  4
  5

A presidential candidate's aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at least 70%. If 158 out of a random sample of 235 college students expressed an intent to vote, can we reject the aide's estimate at the 0.1 level of significance?Perform a one-tailed test. Then fill in the table below.Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)

Salmon Weights: Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh 18 such salmon. The mean weight from your sample is 22.2 pounds with a standard deviation of 4.7 pounds. You want to construct a 99% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River.

(a) What is the point estimate for the mean weight of all spawning Chinook salmon in the Columbia River?
pounds =

(b) Construct the 99% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River. Round your answers to 1 decimal place.
? < μ < ?

(c) Are you 99% confident that the mean weight of all spawning Chinook salmon in the Columbia River is greater than 20 pounds and why?

No, because 20 is above the lower limit of the confidence interval.

Yes, because 20 is below the lower limit of the confidence interval.     

No, because 20 is below the lower limit of the confidence interval.

Yes, because 20 is above the lower limit of the confidence interval.

(d) Recognizing the sample size is less than 30, why could we use the above method to find the confidence interval?

Because the sample size is less than 100.

Because the sample size is greater than 10.  

   Because the parent population is assumed to be normally distributed.

Because we do not know the distribution of the parent population.

Suppose that 3 balls are chosen from an urn which contains 5 red, 6 white and 10 blue balls. Assume X and Y represents, respectively, the number of red balls chosen and summation of chosen red and white balls.
(a) Determine the outcomes, and then represent all possible outcomes with a chart of X and Y.
(b) Determine pXY [x,y], and then from pXY [x,y] determine, pX[x] and pY [y].
(c) Determine Cov(X, Y ) and ρXY .
(d) Determine the joint CDF.

A group of 10 adults is asked to type a passage of text. Here are their times in seconds: 28.9, 27.3, 29.1, 31.5, 27.7, 29.3, 28.3, 30.1, 30.7, 30.9 where ¯x=29.38 and s=1.4. Typing time for the same text passage is normally distributed with unknown mean μ, and known standard deviation σ=1.6. At significance level α=0.005, is the sample showing strong evidence that mean typing time of this text passage is other than 30? Accurate to 4 decimal places, which of the following is σ¯x used for this testing hypothesis problem?

The price of a share of stock divided by the company's estimated future earnings per share is called the P/E ratio. High P/E ratios usually indicate "growth" stocks, or maybe stocks that are simply overpriced. Low P/E ratios indicate "value" stocks or bargain stocks. A random sample of 51 of the largest companies in the United States gave the following P/E ratios†.

(a) Use a calculator with mean and sample standard deviation keys to find the sample mean x and sample standard deviation s. (Round your answers to one decimal place.)

(b) Find a 90% confidence interval for the P/E population mean μ of all large U.S. companies. (Round your answers to one decimal place.)

(c) Find a 99% confidence interval for the P/E population mean μ of all large U.S. companies. (Round your answers to one decimal place.)

Q: The mars company claims that 13 percent of M&Ms plain candies distributed into bags are brown. Investigate this claim with an appropriate hypothesis test. Use a significance level of a= 0.05

1. The p-value for this test statistic is: _______________.

2. Null Hypothesis:

3. Alternative Hypothesis:

4. Conclusion: We REJECT/DO NOT REJECT the null hypothesis. (Circle the correct answer) State what this conclusion means in terms of the problem.

5. Would it be more likely the null hypothesis is rejected for an individual bag of M&M’s, or when we poolthe class results together? Explain your answer.

Discuss how more sophisticated simulation and Crystal Ball can be used to address complex business issues.

The objective of the question is to test the Hypothesis If the Mean travel time in minutes between Point A to Point B is equal to the mean of the travel time in minutes from Point B to your A. First you must find the mean and standard deviations. Then perform and list the complete required steps for the TWO required Hypothesis tests and ALSO USE THE P-VALUE AS A REJECTION RULE FOR BOTH TESTS.

One Hypothesis test is an F test for the equality of the variances of travel Times and the second test is a T test for the equality of the means of travel times in minutes. The F test must be performed first in order to select either Case1 or Case 2 for the T-test. PLEASE SHOW HOW YOU OBTAINED ALL ANSWERS

Recorded Time values in minutes from point A to point B: 32, 34, 51, 30, 29, 35, 36, 29, 32, 29, 33, 32, 29, 30, 33, 30, 30, 33, 30, 31, 35, 35, 34, 32, 33, 33, 31, 33, 34, 30, 30, 29, 34, 32, 36, 29, 30, 32, 30, 33, 31

Recorded Time values in minutes from point B to point A: 36, 28, 48, 28, 27, 54, 34, 29, 26, 34, 33, 42, 29, 34, 31, 48, 27, 42, 28, 45, 26, 43, 32, 41, 30, 36, 27, 44, 29, 29, 35, 26, 31, 28, 27, 28, 32, 41, 34, 28, 31

For a sample of 10 individuals, a researcher calculates residuals for the relationship between “number of delinquent friends” and “number of prior arrests” and finds that the positive residuals = 125. The researcher then collects a second sample of 10 individuals and calculates the residuals on the same two variables and discovers the sum of the positive residuals = 75. What can you conclude about the strength of the relationship between “number of delinquent peers” and “number of prior arrests” across these two random samples? How are they similar/different?