1. The mean of all the sample means is _______.

Multiple Choice

• α

• σ

• µ

• X

4. According to the central limit theorem, ____________.

Multiple Choice

• increasing the sample size decreases the dispersion of the sampling distribution

• the sampling distribution of the sample means will be skewed

• the sampling distribution of the sample means is uniform

• sample size is important when the population is not normally distributed

5. The Office of Student Services at a large western state university maintains information on the study habits of its full-time students. Their studies indicate that the mean amount of time undergraduate students study per week is 20 hours. The hours studied follows the normal distribution with a population standard deviation of six hours. Suppose we select a random sample of 144 current students. What is the probability that the mean of this sample is between 19 hours and 20 hours?

Multiple Choice

• Cannot be calculated based on the given information.

• −2.00

• 0.4772

• 2.00

9. Bones Brothers & Associates prepare individual tax returns. Over prior years, Bones Brothers has maintained careful records regarding the time to prepare a return. The mean time to prepare a return is 90 minutes, and the population standard deviation of this distribution is 14 minutes. Suppose 100 returns from this year are selected and analyzed regarding the preparation time. What is the probability that the mean time for the sample of 100 returns is between 88 minutes and 92 minutes?

Multiple Choice

• 0.1664

• 0.8472

• 0.8336

• Approximately 1

11. Suppose we select every fifth invoice in a file. What type of sampling is this?

Multiple Choice

• Cluster

• Systematic

• Stratified

• Random

15. The Office of Student Services at a large western state university maintains information on the study habits of its full-time students. Their studies indicate that the mean amount of time undergraduate students study per week is 20 hours. The hours studied follows the normal distribution with a population standard deviation of six hours. Suppose we select a random sample of 144 current students. What is the standard error of the mean?

Multiple Choice

• 0.50

• 6.00

• 0.25

• 2.00

Analogy: A statistic is to a research sample as is what a _______is to a population.

Group of answer choices

summary

parameter

data set

measurement

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Question 4 1 pts

Which of the following could be a variable to be measured among social work students?

Group of answer choices

income

number of years of school completed

satisfaction with BSW program

all of the above

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Question 5 1 pts

A measurement of a variable is said to be reliable when

Group of answer choices

it shows up on time to class

measures what it is believed to measure

it produces consistent results

none of the above

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Question 61 pts

For the next two questions consider the following research hypothesis: “Younger students will learn to use statistical software more quickly than older students.” What is the dependent variable?

Group of answer choices

age of students

number of computers previously purchased

time required to learn to use statistical software

presence or absence of math phobia

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Question 71 pts

What is the independent variable?

Group of answer choices

age of students

number of computers previously purchased

time required to learn statistical software

presence or absence of math phobia

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Question 8 1 pts

In order for a measurement to be valid it must also be reliable.

Group of answer choices

True

False

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Question 9 1 pts

What type of variable can theoretically assume any numerical value?

Group of answer choices

continuous

discrete

dichotomous

binary

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Question 10 1 pts

What do you call a continuous line that represents the frequency of scores within a class interval?

Group of answer choices

frequency polygon

line graph

histogram

tally marks

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Question 11 1 pts

If you wanted to examine the proportion of students in this class who is male compared to the proportion who is female, which of the following would be best suited to use?

Group of answer choices

ogive

line graph

pie chart

frequency polygon

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A school psychologist is interested in the effect a popular new cognitive therapy on anxiety. The psychologist collects a sample of 13 students and gives them the cognitive therapy once a week for two months. Afterwards the students fill out an anxiety inventory in which their average score was 46.21. Normal individuals in the population have an anxiety inventory average of 49 with a standard deviation of 2.7. What can be concluded with α = 0.10?

a) What is the appropriate test statistic?'

1. na 2. z-test 3. one-sample test 4. independent-samples t-test 5. related-samples t-test

b1)
Population: (choose one of the following)

1. normal individuals 2. new hypnosis technique 3. two months 4. students receiving hypnosis

b2)
Sample: (choose one of the following)

1. normal individuals 2. new hypnosis technique 3. two months 4. students receiving hypnosis

c) Obtain/compute the appropriate values to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value = ________ ; test statistic = ___________
Decision: ***(choose one)*** 1. Reject H0 or 2. Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[ ], [ ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and select "na" below.
d = ___________ ;   *(choose one)1. na 2. trivial effect 3. small effect 4. medium effect 5. large effect
r² = ___________ ;   *(choose one)1. na 2. trivial effect 3. small effect 4. medium effect 5. large effect

f) Make an interpretation based on the results. (Choose one)

1.) The population has significantly lower anxiety than students that underwent cognitive therapy.

2.) The population has significantly higher anxiety than students that underwent cognitive therapy.  

3.) The new cognitive therapy technique does not significantly effect anxiety.

A)Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 177 with 10% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

< p <

B) Giving a test to a group of students, the grades and gender are summarized below

Let pp represent the proportion of all male students who would receive a grade of B on this test. Use a 98% confidence interval to estimate pp to three decimal places.

< p <

C) Giving a test to a group of students, the grades and gender are summarized below

Let pp represent the proportion of all male students who would receive a grade of B on this test. Use a 98% confidence interval to estimate p to three decimal places.

Enter your answer as a tri-linear inequality using decimals (not percents).

< p <

State the point estimate for the proportion of all male students who would receive a grade of B on this test:
p≈p≈

State the margin of error for the proportion of all male students who would receive a grade of B on this test:
EBP ≈≈

D) We wish to estimate what percent of adult residents in Ventura County like chocolate. Out of 500 adult residents sampled, 393 like chocolate. Based on this, construct a 95% confidence interval for the proportion (p) of adult residents who like chocolate in Ventura County.

1. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

   Confidence interval =

2. 
   
3. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

   < p <

4. 
   
5. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

   p =  ±±

Assignment #7: One-sample Chi-Square

Directions: Use the Chi-Square option in the Nonparametric Tests menu to answer the questions based on the following scenario. (Assume a level of significance of .05 and use information from the scenario to determine the expected frequencies for each category)

During the analysis of the district data, it was determined that one high school had substantially higher Graduate Exit Test scores than the state average and the averages of high schools in the surrounding districts. To better understand possible reasons for this difference, the superintendent conducted several analyses. One analysis examined the population of students who completed the exam. Specifically, the superintendent wanted to know if the distribution of special education, regular education, and gifted/talented test takers from the local high school differed from the statewide distribution. The obtained data are provided below.

*For purposes of testing, special education includes any student who received accommodations during the test.

1. If the student distribution for the local high school did not differ from the state, what would be the expected percentage of students in each category?

2. What were the actual percentages of local high school students in each category? (Report final answer to two decimal places)

3. State an appropriate null hypothesis for this analysis.

4. What is the value of the chi-square statistic?

5. What are the reported degrees of freedom?

6. What is the reported level of significance?

7. Based on the results of the one-sample chi-square test, was the population of test taking students at the local high school statistically significantly different from the statewide population?

8. Present the results as they might appear in an article. This must include a table and narrative statement that reports and interprets the results of the analysis.

   Note: The table must be created using your word processing program. Tables that are copied and pasted from SPSS are not acceptable.

Modify the program 5-13 from page 279 such that will also compute the class average. This class average is in addition to each individual student score average. To accomplish this additional requirement, you should do the following:

1. Add two more variables of type double: one for accumulating student averages, and one to hold the class average. Don't forget, accumulator variable should be initialized to 0.0.

2. Immediately after computing individual student average, add a statement that will accumulate the newly computed average using the variable from previous step.

3. After the outer loop for the number of students, add a statement that will compute class average as the accumulated averages divided by the number of students.

4. Finally, display the computed class average. Do not use computations inside output statements because the computed value might be needed with later updates.

Save and submit your final code file as Pr5-13_Lab.cpp.

// This program uses a nested loop to average

// a set of test scores for multiple students.

#include <iostream>

using namespace std;

int main()

{

   int numStudents, // Number of students

   numTests; // Number of tests per student

   double average; // Average test score for a student

   // Get the number of students

   cout << "This program averages test scores.\n";

   cout << "How many students are there? ";

   cin >> numStudents;

   // Get the number of test scores per student

   cout << "How many test scores does each student have? ";

   cin >> numTests;

   cout << endl;

   // Read each student's scores and compute their average

   for (int snum = 1; snum <= numStudents; snum++) //Outer loop

   { double total = 0.0; // Initialize accumulator

for (int test = 1; test <= numTests; test++) //Inner loop

{ int score;

   // Read a score and add it to the accumulator

cout << "Enter score " << test << " for ";

   cout << "student " << snum << ": ";

   cin >> score;

   total += score;

} //End inner loop

// Compute and display the student's average

average = total / numTests;

cout << "The average score for student " << snum;

cout << " is " << average << "\n\n";

   } //End outer loop

   return 0;

}

1)A university financial aid office polled a random sample of 824 male undergraduate students and 731 female undergraduate students. Each of the students was asked whether or not they were employed during the previous summer. 568 of the male students and 391 of the female students said that they had worked during the previous summer. Give a 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer.

Step 1 of 3:

Find the point estimate that should be used in constructing the confidence interval. Round your answer to three decimal places

Step 2 of 3:

Find the margin of error. Round your answer to six decimal places.

Step 3 of 3:

Construct the 90%confidence interval. Round your answers to three decimal places.

2)The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day. They would like the estimate to have a maximum error of 0.14 gallons. A previous study found that for an average family the standard deviation is 2.3 gallons and the mean is 15 gallons per day. If they are using a 99% level of confidence, how large of a sample is required to estimate the mean usage of water? Round your answer up to the next integer

3)Given two independent random samples with the following results:

n1        7          n2        11

x1bar   143      x2bar   162

s1        12        s2        33

data: n1=7 x‾1=143    s1=12   n2=11 x‾2=162 s2=33

Use this data to find the 90% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 3:

Find the point estimate that should be used in constructing the confidence interval.

Step 2 of 3:

Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Step 3 of 3:

Construct the 90% confidence interval. Round your answers to the nearest whole number

// CSE240 Fall 2020 HW 7 & 8
// Write your name here
// Write the compiler used: Visual studio or gcc

// READ BEFORE YOU START:
// You are given a partially completed program that creates a linked list of employee information.
// The global linked list 'list' is a list of employees with each node being struct 'employeeList'.
// 'employeeList' consists of struct 'employee' which has: employee name, room number, and a linked list of 'supervisors'.
// The linked list of supervisors has each node containing simply the name of the supervisor.
// HW7 ignores the 'supervisors' linked list since there is no operation/manipulation to be done on 'supervisors' list in HW7.
// HW8 has operations/manipulations to be done with 'supervisors' list like add a supervisor, display last added supervisor.

// To begin, you should trace through the given code and understand how it works.
// Please read the instructions above each required function and follow the directions carefully.
// If you modify any of the given code, the return types, or the parameters, you risk getting compile error.
// You are not allowed to modify main ().
// You can use string library functions.

// ***** WRITE COMMENTS FOR IMPORANT STEPS OF YOUR CODE. *****
// ***** GIVE MEANINGFUL NAMES TO VARIABLES. *****
// ***** Before implementing any function, see how it is called in executeAction() *****

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#pragma warning(disable: 4996) // for Visual Studio

#define MAX_NAME 30

// global linked list 'list' contains the list of employees
struct employeeList {
   struct employee* employee;
   struct employeeList* next;
} *list = NULL;               // currently empty list

// structure "employee" contains the employee's name, room number and linked list of supervisors
struct employee {
   char name[MAX_NAME];
   unsigned int roomNumber;
   struct supervisor* supervisors;       // linked list 'supervisors' contains names of supervisors
};

// structure 'supervisor' contains supervisor's name
struct supervisor {
   char name[MAX_NAME];
   struct supervisor* next;
};

// forward declaration of functions (already implmented)
void flushStdIn();
void executeAction(char);

// functions that need implementation:
// HW 7
void addEmployee(char* employeeNameInput, unsigned int roomNumInput); // 20 points
void displayEmployeeList(struct employeeList* tempList);   // 15 points
struct employee* searchEmployee(char* employeeNameInput);   // 15 points
//HW 8
void addSupervisor(char* employeeNameInput, char* supervisorNameInput);   // 15 points
void displayEmployeeSupervisorList(struct employeeList* tempList);   // 15 points
void removeEmployee(char* employeeNameInput);           // 20 points

int main()
{
   char selection = 'a';       // initialized to a dummy value
   do
   {
       printf("\nCSE240 HW 7,8\n");
       printf("Please enter your selection:\n");
       printf("HW7:\n");
       printf("\t a: add a new employee to the list\n");
       printf("\t d: display employee list (no supervisors)\n");
       printf("\t b: search for an employee on the list\n");
       printf("\t q: quit\n");
       printf("HW8:\n");
       printf("\t c: add a supervisor of a employee\n");
       printf("\t l: display employees who report to a specific supervisor\n");
       printf("\t r: remove an employee\n");
       printf("\t q: quit\n");

       selection = getchar();
       flushStdIn();
       executeAction(selection);
   } while (selection != 'q');

   return 0;
}

// flush out leftover '\n' characters
void flushStdIn()
{
   char c;
   do c = getchar();
   while (c != '\n' && c != EOF);
}

// Ask for details from user for the given selection and perform that action
// Read the function case by case
void executeAction(char c)
{
   char employeeNameInput[MAX_NAME], supervisorNameInput[MAX_NAME];
   unsigned int roomNumInput;
   struct employee* searchResult = NULL;

   switch (c)
   {
   case 'a':   // add employee
               // input employee details from user
       printf("\nPlease enter employee's name: ");
       fgets(employeeNameInput, sizeof(employeeNameInput), stdin);
       employeeNameInput[strlen(employeeNameInput) - 1] = '\0';   // discard the trailing '\n' char
       printf("Please enter room number: ");
       scanf("%d", &roomNumInput);
       flushStdIn();

       if (searchEmployee(employeeNameInput) == NULL)   // un-comment this line after implementing searchEmployee()                  
       //if (1)                                   // comment out this line after implementing searchEmployee()
       {
           addEmployee(employeeNameInput, roomNumInput);
           printf("\nEmployee successfully added to the list!\n");
       }
       else
           printf("\nThat employee is already on the list!\n");
       break;

   case 'd':       // display the list
       displayEmployeeList(list);
       break;

   case 'b':       // search for an employee on the list
       printf("\nPlease enter employee's name: ");
       fgets(employeeNameInput, sizeof(employeeNameInput), stdin);
       employeeNameInput[strlen(employeeNameInput) - 1] = '\0';   // discard the trailing '\n' char

       if (searchEmployee(employeeNameInput) == NULL)   // un-comment this line after implementing searchEmployee()                  
       //if (0)                                   // comment out this line after implementing searchEmployee()
           printf("\nEmployee name does not exist or the list is empty! \n\n");
       else
       {
           printf("\nEmployee name exists on the list! \n\n");
       }
       break;

   case 'r':       // remove employee
       printf("\nPlease enter employee's name: ");
       fgets(employeeNameInput, sizeof(employeeNameInput), stdin);
       employeeNameInput[strlen(employeeNameInput) - 1] = '\0';   // discard the trailing '\n' char

       if (searchEmployee(employeeNameInput) == NULL)   // un-comment this line after implementing searchEmployee()                  
       //if (0)                                   // comment out this line after implementing searchEmployee()
           printf("\nEmployee name does not exist or the list is empty! \n\n");
       else
       {
           removeEmployee(employeeNameInput);
           printf("\nEmployee successfully removed from the list! \n\n");
       }
       break;

   case 'c':       // add supervisor
       printf("\nPlease enter employee's name: ");
       fgets(employeeNameInput, sizeof(employeeNameInput), stdin);
       employeeNameInput[strlen(employeeNameInput) - 1] = '\0';   // discard the trailing '\n' char

       if (searchEmployee(employeeNameInput) == NULL)   // un-comment this line after implementing searchEmployee()                  
       //if (0)                                       // comment out this line after implementing searchEmployee()
           printf("\nEmployee name does not exist or the list is empty! \n\n");
       else
       {
           printf("\nPlease enter supervisor's name: ");
           fgets(supervisorNameInput, sizeof(supervisorNameInput), stdin);
           supervisorNameInput[strlen(supervisorNameInput) - 1] = '\0';   // discard the trailing '\n' char

           addSupervisor(employeeNameInput, supervisorNameInput);
           printf("\nSupervisor added! \n\n");
       }
       break;

   case 'l':       // list supervisor's employees
       displayEmployeeSupervisorList(list);
       break;

   case 'q':       // quit
       break;

   default: printf("%c is invalid input!\n", c);
   }
}

// HW7 Q1: addEmployee (20 points)
// This function is used to insert a new employee in the linked list.
// You must insert the new employee to the head of linked list 'list'.
// You need NOT check if the employee already exists in the list because that is taken care by searchEmployee() called in executeAction(). Look at how this function is used in executeAction().
// Don't bother to check how to implement searchEmployee() while implementing this function. Simply assume that employee does not exist in the list while implementing this function.
// NOTE: The function needs to add the employee to the head of the list.
// NOTE: This function does not add supervisors to the employee info. There is another function addSupervisor() in HW8 for that.
// Hint: In this question, no supervisors means NULL supervisors.

void addEmployee(char* employeeNameInput, unsigned int roomNumInput)
{
   // YOUR CODE HERE
}

// HW7 Q2: displayEmployeeList (15 points)
// This function displays the employee details (struct elements) of each employee.
// Parse through the linked list 'list' and print the employee details ( name and room number) one after the other. See expected output screenshots in homework question file.
// You should not display supervisor names (because they are not added in HW7).
// You MUST use recursion in the function to get full points. Notice that 'list' is passed to the function argument. Use recursion to keep calling this function till end of list.

void displayEmployeeList(struct employeeList* tempList)
{
   // YOUR CODE HERE

}

// HW7 Q3: searchEmployee (15 points)
// This function searches the 'list' to check if the given employee exists in the list. Search by employee name.
// If it exists then return that 'employee' node of the list. Notice the return type of this function.
// If the employee does not exist in the list, then return NULL.
// NOTE: After implementing this fucntion, go to executeAction() to comment and un-comment the lines mentioned there which use searchEmployee()
// in case 'a', case 'r', case 'l' (total 3 places)
struct employee* searchEmployee(char* employeeNameInput)
{

   struct employeeList* tempList = list;           // work on a copy of 'list'
  
   // YOUR CODE HERE

   return NULL;
}

Help me complete the code thank you.

The table below represents 2 attempts that students had to complete the same statistics exam in a course.

1) The professor believes that, on average, students will do better on the second attempt than on the first.

a) Choose an appropriate test to determine if students improved on the second attempt compared to their first. Draw appropriate conclusions.

b) Calculate the size/magnitude of this effect.

c) Identify the 95% confidence interval around our measurement and explain what this result tells us about our data.

2) I modify the table so that the column labelled 1st attempt now represents the results from students in Prof A’s statistics class, and the column labelled 2^nd attempt represents the results of a completely different group of students taking Prof B’s class.

a) Choose the appropriate test to demonstrate if there is a significant difference in the results in Professor A’s class compared to Professor B.

b) Is my approach to this problem the same as in Question 1). Why or why not?

3) I keep the table changes mentioned in question 2. The 1st attempt column still represents the results from students in Professor A's statistics class and the 2nd attempt column represents different students from Professor B.'s class. Professor A discovers that 3 students in his class cheated so he eliminates their grades from the group. If I now wanted to compare the performance of class A and class B, should the statistical approach change compared to Question 2? Why or why not? (Note: You do not need to do the calculations; you just need to provide an explanation.)

**********************
I'm going to modify the table a bit. Data now represent the marks obtained by students in the statistics course at the mid-term exam and in the final exam in 2018.

4) I would like to know if there is a relationship or link between the grades that the students obtained in the midterm exam and in the final exam.

a) Make an appropriate graphical representation to illustrate these data.

b) What conclusions can we draw only by looking at this graph? Are there any data points that seem problematic?

5) a) What is the strength of the relationship between these two variables?

b) What part of the variance could be explained by the relationship that exists between these variables?

c) Is this relationship statistically significant?

6) In the 2019 winter semester, a student obtained a grade of 64 in her midterm exam. What grade could we predict that she will get in the final exam?

7) a) If I wanted to test the relationship between the midterm exam performance and the final exam using a chi-square test, how would the above data table need be rearranged/modified?

b) Despite my suggestion in 7a) to use a chi-square test, it would actually be a bad idea to use the chi-square test with this type of problem. Why might that be the case? What problem (s) would it cause? (Think of the rules we discussed for using chi-square.)

A sample of 12 of bags of Calbie Chips were weighed (to the nearest gram), and listed here as follows.

219, 226, 217, 224, 223, 216, 221, 228, 215, 229, 225, 229 Find a 95% confidence interval for the mean mass of bags of Calbie Chips.

Find a 95% confidence interval for the mean mass of bags of Calbie Chips

2. (b) Professor GeniusAtCalculus has two lecture sections (A and B) of the same 4th year Advanced Calculus (AMA 4301) course in Semester 2. She wants to investigate whether section A students maybe ”smarter” than section B students by comparing their perfor- mances in the midterm test. A random sample of 12 students were taken from section A, with mean midterm test score of 78.8 and standard deviation 8.5; and a random sample of 9 students were taken from section B, with mean midterm test score of 86 and standard deviation 9.3. Assume the population standard deviations of midterm test scores for both sections are the same. Construct the 90% confidence interval for the difference in midterm test scores of the two sections. Based on the sample midterm test scores from the two sections, can Professor GeniusAtCalculus conclude that there is any evidence that one section of students are ”smarter” than the other section? Justify your conclusions.

   [8 marks]

3. (c) The COVID-19 (coronavirus) mortality rate of a country is defined as the ratio of the number of deaths due to COVID-19 divided by the number of (confirmed) cases of COVID-19 in that country. Suppose we want to investigate if there is any difference between the COVID-19 mortality rate in the US and the UK. On April 18, 2020, out of a sample of 671,493 cases of COVID-19 in the US, there was 33,288 deaths; and out of a sample of 109,754 cases of COVID-19 in the UK, there was 14,606 deaths. What is the 92% confidence interval in the true difference in the mortality rates between the two countries? What can you conclude about the difference in the mortality rates between the US and the UK? Justify your conclusions.

fidence interval for the mean mass of bags of Calbie Chips. [9 marks] (b) Professor GeniusAtCalculus has two lecture sections (A and B) of the same 4th year Advanced Calculus (AMA 4301) course in Semester 2. She wants to investigate whether section A students maybe ”smarter” than section B students by comparing their perfor- mances in the midterm test. A random sample of 12 students were taken from section A, with mean midterm test score of 78.8 and standard deviation 8.5; and a random sample of 9 students were taken from section B, with mean midterm test score of 86 and standard deviation 9.3. Assume the population standard deviations of midterm test scores for both sections are the same. Construct the 90% confidence interval for the difference in midterm test scores of the two sections. Based on the sample midterm test scores from the two sections, can Professor GeniusAtCalculus conclude that there is any evidence that one section of students are ”smarter” than the other section? Justify your conclusions. [8 marks] (c) The COVID-19 (coronavirus) mortality rate of a country is defined as the ratio of the number of deaths due to COVID-19 divided by the number of (confirmed) cases of COVID-19 in that country. Suppose we want to investigate if there is any difference between the COVID-19 mortality rate in the US and the UK. On April 18, 2020, out of a sample of 671,493 cases of COVID-19 in the US, there was 33,288 deaths; and out of a sample of 109,754 cases of COVID-19 in the UK, there was 14,606 deaths. What is the 92% confidence interval in the true difference in the mortality rates between the two countries? What can you conclude about the difference in the mortality rates between the US and the UK? Justify your conclusions.