Q.1 A social psychologist is interested in whether the amount of couple photos (photos featuring the couple together) people post on social media would vary across relationship stages. He administered a questionnaire to a group of couples (one questionnaire for each couple) at the anniversaries of their first year, second year, third year, and fourth year of dating. The questionnaire asked about the total amount of couple photos they posted in social media during the past year. Below are the data, with a higher number denoting more couple photos posted:

a. What are the independent variable and dependent variable of this study?

b. Write down the omnibus null hypothesis and the alternative hypothesis for the overall effect of the independent variable.

c. Conduct a proper statistical test by hand calculation to test the omnibus hypothesis with 5% as the level of significance (α). (For this exercise, the data assumptions of your chosen statistical test can be taken as reasonably met.) Show your calculation formulae and steps. In case you decide to conduct an ANOVA, you are not required to conduct any post-hoc comparisons. Decide whether to reject the null hypothesis or not and state the basis of your decision.

d. Calculate the effect size in terms of eta-squared and omega-squared.

Briefly explain how Levene Test is used. Give an example to illustrate.

Among 16 electrical components exactly 4 are known not to function properly. If 6 components are randomly selected, find the following probabilities: (i) The probability that all selected components function properly. (ii) The probability that exactly 3 are defective. (iii) The probability that at least 1 component is defective.

Listed below are brain volumes (cm3 ) of twins.

Test the claim at the 5% significance level that the brain volume for the first born is different from the second-born twin.

(a) State the null and alternative hypotheses.

(b) Find the critical value and the test statistic.

(c) Should H0 be rejected at the 5% significance level? Make a conclusion.

(d) Construct a 95% confidence interval for the paired difference of the population means

Suppose 70% of all students live in dorms. Of those who live in dorms, 40% of them eat breakfast. Of those who don’t live in dorms, 30% eat breakfast.

5. Make a tree from this information. Label all the probabilities.

6. Make a table from this information. Label everything. Show your work.

7. What percentage of allstudents eat breakfast?

8. What rule/law do we use to answer the previous question?

1. Multiplication rule
2. Law of total probability
3. Bayes rule
4. Addition Rule

9. If a randomly selected student eats breakfast, what is the chance that they live in a dorm?
10. What rule/law do we use to answer the previous question?

1. Multiplication rule
2. Law of total probability
3. Bayes rule
4. Addition Rule

Q.4 A researcher is interested in whether people’s level of loneliness would vary as a function of their relationship status (single vs. in a relationship), and how such difference might depend on whether people own a pet or not. She recruited a group of participants, asking them about their relationship status, pet ownership, and the perceived level of loneliness. The data are as below, with a higher number denoting greater level of loneliness:

a. What are the independent variable(s) and dependent variable of this study?

b. Write down the null hypothesis and the alternative hypothesis for each of the effects in the analysis.

c. Conduct a proper statistical test by hand calculation to test the hypotheses in b., with 5% as the level of significance (α). (For this exercise, the data assumptions of your chosen statistical test can be taken as reasonably met.) Show your calculation formulae and steps. In case you decide to conduct an ANOVA, you are not required to conduct any post-hoc comparisons. Decide whether to reject the null hypothesis or not for each effect and state the basis of your decision

QUESTION 1

1. Assume that the two samples of five cereal boxes (one sample for each of two cereal varieties) listed on the TCCACCTC Web site were collected randomly by organization members. For each sample, assume that the population distribution of individual weight is normally distributed, the average weight is indeed 368 grams and the standard deviation of the process is 15 grams, and obtain the following using Excel and PHStat:

   Weights of Oxford O’s boxes
   360.4
   361.8
   362.3
   364.2
   371.4
   
   Weights of Alpine Frosted Flakes with Vitamins & Minerals boxes
   366.1
   367.2
   365.6
   367.8
   373.5 All answers should be accurate to 2 decimal places.

2. (a) For the Oxford's O:

      (i) The value of the sample mean =

      (ii) The proportion of all samples for each process that would have a sample mean less than the value you calculated in step (a)(i) =

      (iii) The proportion of all the individual boxes of cereal that would have a weight less than the value you calculated in step (a)(i) =

      (iv) The probability that an individual box of cereal will weigh less than 368 grams =

      (v) The probability that 4 out of the 5 boxes sampled will weigh less than 368 grams =

      (vi) The lower limit of the 95% confidence interval for the population average weight =

      (vii) The upper limit of the 95% confidence interval for the population average weight =

   (b) For the Alpine Frosted Flakes:

      (i) The value of the sample mean =

      (ii) The proportion of all samples for each process that would have a sample mean less than the value you calculated in step (b)(i) =

      (iii) The proportion of all the individual boxes of cereal that would have a weight less than the value you calculated in step (b)(i) =

      (iv) The probability that an individual box of cereal will weigh less than 368 grams =

      (v) The probability that 4 out of the 5 boxes sampled will weigh less than 368 grams =

      (vi) The lower limit of the 95% confidence interval for the population average weight =

      (vii) The upper limit of the 95% confidence interval for the population average weight =

QUESTION 2

1. Oxford Cereals then conducted a public experiment in which it claimed it had successfully debunked the statements of groups such as the TriCities Consumers Concerned About Cereal Companies That Cheat (TCCACCTC) that claimed that Oxford Cereals was cheating consumers by packaging cereals at less than labeled weights. Review the Oxford Cereals' press release and supporting documents that describe the experiment at the company's Web site and then answer the following assuming that now you have no information about the mean and standard deviation of the population distribution of the weight of all boxes of the cereal produced:

   Weight
   351.8
   360.65
   372.74
   382.96
   375.28
   352.16
   374.15
   361.8
   366.67
   398.86
   384.34
   367.53
   361.59
   364.47
   382.93
   366.88
   368.14
   408.19
   356.03
   379.27
   380.38
   386.44
   378.72
   342.05
   380.29
   361.1
   355.11
   387
   346.86
   391.94
   366.3
   350.52
   397.27
   349
   373.78
   384.04
   392.55
   361.98
   377.07
   390.88
   395.86
   370.21
   380.66
   389.33
   361.15
   386.74
   353
   354.22
   374.24
   363.77
   352.08
   364.11
   359.79
   367.12
   375.84
   343.29
   357.7
   384.75
   380.72
   356.22
   389.72
   375.28
   380.44
   379.14
   364.64
   379.63
   369.29
   337.1
   371.42
   347.63
   363.86
   381.28
   379.21
   366.26
   365.15
   351.33
   375.91
   363.32
   357.96
   375.58

   All answers should be accurate to 2 decimal places.

   (a) For a two-tailed t-test on whether the population mean weight is equal to 368g:

      (i) The value of the t-test statistic is =

      (ii) The p-value of the t-test statistic is =

      (iii) The lower-critical value is =

      (iv) The upper-critical value is =

   (b) For an upper-tailed t-test on whether the population mean weight is more than 368g:

      (i) The value of the t-test statistic is =

      (ii) The p-value of the t-test statistic is =

      (iii) The upper-critical value is =

   (c) For the 95% confidence interval for the population average weight:

      (i) The lower limit =

      (ii) The upper limit =

Researchers conducted a study to investigate whether there is a difference in mean PEF in children with chronic bronchitis as compared to those without. Data on PEF were collected from 100 children with chronic bronchitis and 100 children without chronic bronchitis. The mean PEF for children with chronic bronchitis was 290 with a standard deviation of 64, while the mean PEF for children without chronic bronchitis was 308 with a standard deviation of 77. Based on the data, is there statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without? Run the appropriate test at α=0.05. Assume equal variances. Give each of the following to receive full credit: 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value. You MUST show your work to receive full credit. Partial credit is available.

A class has 40 students.

• Thirty students are prepared for the exam,

• Ten students are unprepared. The professor writes an exam with 10 questions, some are hard and some are easy.

• 7 questions are easy. Based on past experience, the professor knows that: – Prepared students have a 90% chance of answering easy questions correctly – Unprepared students have a 50% chance of answering easy questions correctly.

• 3 questions are hard. Based on past experience, the professor knows that: – Prepared students have a 50% chance of answering hard questions correctly – Unprepared students have a 10% chance of answering hard questions correctly

• Each student’s performance on each question is independent of their performance on other questions.

(a) Find the probability that a prepared student answers all 10 questions correctly.

(b) What is the probability that at least one of the 30 prepared students answers all 10 questions correctly. Assume that each student’s score is independent of every other student.

(c) Let P be the number of questions answered correctly by a randomly chosen prepared student, and let U be the number answered correctly by a randomly chosen unprepared student. Find E[P] and E[U]

(d) Find Var(P) and Var(U)

In a survey of 2 comma 418 ​adults, 1 comma 886 reported that​ e-mails are easy to​ misinterpret, but only 1 comma 228 reported that telephone conversations are easy to misinterpret. Complete parts​ (a) through​ (c) below.

a. Construct a​ 95% confidence interval estimate for the population proportion of adults who report that​ e-mails are easy to misinterpret. less than or equalspiless than or equals ​(Round to four decimal places as​ needed.)

b. Construct a​ 95% confidence interval estimate for the population proportion of adults who report that telephone conversations are easy to misinterpret. less than or equalspiless than or equals ​(Round to four decimal places as​ needed.)

c. Compare the results of​ (a) and​ (b). Which statement below regarding the implications of the information found in​ (a) and​ (b) is​ correct?

A. More adults believe that​ e-mails are easy to misinterpret than believe that telephone conversations are easy to misinterpret.

B. The number of adults that believe that​ e-mails are easy to misinterpret and the number of adults that believe that telephone conversations are easy to misinterpret are roughly the same.

C. More adults believe that telephone conversations are easy to misinterpret than believe that​ e-mails are easy to misinterpret.

D. The information cannot be compared because it is derived from two different opinions.

The scientific productivity of major world cities was the subject of a recent study. The study determined the number of scientific papers published between 1994 and 1997 by researchers from each of the 20 world cities, and is shown below.

Construct a 99 % confidence interval for the average number of papers published in major world cities.

<μ<

7.46 A poll had a sample of 66 people choose their favorite Skittles flavor by color (green, orange, purple, red or yellow). A separate poll sampled 91 people, again asking them their favorite Skittles flavor, but rather than by color they asked by the actual flavor (lime, orange, grape, strawberry, and lemon, respectively). The table below shows the results from both polls. Does the way people choose their favorite Skittles type, by color or flavor, appear to be related to which type is chosen?

1. State the null and alternative hypotheses.
2. Give a table with the expected counts for each of the 10 cells.
3. Are the expected counts large enough for a chi-square test?
4. How many degrees of freedom do we have for this test?
5. Calculate the chi-square test statistic.
6. At the α=0.05 level, do we find evidence that method of choice affects which is chosen?

Table: Skittles popularity

Please answer using your own words (i.e., don't quote directly from the text).: Describe and contrast the three ways of describing results: comparing group percentages, correlating scores, and comparing group means. For each, give two examples (other than those given in the text) of their appropriate use.

What is a z-test, when is it used, and what does it tell us?

A criminologist conducted a survey to determine whether the incidence of certain types of crime varied from one district of a large city to another. The particular crimes of interest were assault, burglary, larceny, and homicide. The following table shows the number of crimes committed in four districts of the city during the past year.

Type of Crime

Can we conclude from the data at the 0.01 significance level that the occurrence of these types of crime is dependent on the city district?

Can we conclude from the data at the 0.01 significance level that the occurrence of these types of crime is dependent on the city district?

       Note:   The confidence level is equivalent to 1 – α. So, if your significance

                    level is 0.05, the corresponding confidence level is 95%.

Note: You can use the functions qchisq() in R to help you in solving the following.

Why we are using qchisq() function in R?

The qchisq() function in R allows us to specify a desired area in a tail and the number of degrees of freedom. From that information, qchisq() computes the required x-value to get the specified area in the specified tail with the specified number of degrees of freedom.

1. (1 point) State the two hypothesis of interest.

2. (2 points) Calculate an appropriate test statistic for (a) by hand. Use the table below to do the calculations of the expected frequencies.

Note: Round the numbers to two decimals.

C. (2 points) Write your conclusion using the rejection region method “critical value method” include both statistical and related to the topic of the question (practical) interpretation use the function qchisq() in R