QUESTION 1

1.Choose the one alternative that best completes the statement or answers the question. Solve the problem.

Find the installment price of a table bought on the installment plan with a down payment of $30 and 6 payments of $51.12.

$366.72

$30

$336.72

$306.72

10 points   

QUESTION 2

1.Find the monthly interest rate.

What is the monthly interest rate if an annual rate is 17.5%?

1.4%

1.46%

16.5%

14.58%

10 points   

QUESTION 3

1.Solve the problem.

Find the amount financed on a TV with a cash price $430.00 and a down payment of $86.00.

$429.80

$86.00

$344.00 

$430.00 

10 points   

QUESTION 4

1.Solve the problem.

Find the amount financed if a $25 down payment is made on a camera with a cash price of $260.

$235

$25

$260

$315.86 

10 points   

QUESTION 5

1.Solve the problem.

Find the installment price of a laptop computer bought on the installment plan with $90 down and 36 payments of $33.05.

$90

$1279.80

$939.99

$1189.80

10 points   

QUESTION 6

1.Solve the problem.

A mountain bike has a cash price of $690.00. Eric purchases the bike by making a down payment of $69.00 and 24 payments of $38.74. Find the finance charge.

$621.00

$308.76

$69.00

$310.50

10 points   

QUESTION 7

1.Solve the problem.

The installment price of a watch is $310.74 with 6 monthly payments and a down payment of $30. Find the monthly payment.

$280.74

$16.74

$19.26

$46.79 

10 points   

QUESTION 8

1.Solve the problem.

The installment price of a food processor is $379.48 with 18 monthly payments and a down payment of $40. Find the monthly payment.

$18.86

$39.48

$339.48

$0.52

10 points   

QUESTION 9

1.Find the monthly mortgage payment.

$402.39

$481.63

$410.82

$346.19

10 points   

QUESTION 10

1.Solve the problem.

Sarah Fields wants to borrow $121,000 at 6.5% to buy a house. How much interest would she save by going with a 15-year mortgage over a 25-year mortgage?

$55,321.20

$68,752.20

$168,940.20

$85,595.40

10 points   

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1. Direct Materials Purchases Budget

   Marino’s Frozen Pizza Inc. has determined from its production budget the following estimated production volumes for 12” and 16” frozen pizzas for June 2016:

   	Units
   	12″ Pizza		16″ Pizza
   Budgeted production volume	12,900		24,000

   There are three direct materials used in producing the two types of pizza. The quantities of direct materials expected to be used for each pizza are as follows:

   		12″ Pizza		16″ Pizza
   Direct materials:
   	Dough	0.90	lb. per unit	1.50	lbs. per unit
   	Tomato	0.60		1.00
   	Cheese	0.80		1.30

   In addition, Marino’s has determined the following information about each material:

   	Dough		Tomato		Cheese
   Estimated inventory, June 1, 2016	650	lbs.	180	lbs.	360	lbs.
   Desired inventory, June 30, 2016	680	lbs.	170	lbs.	390	lbs.
   Price per pound	$1.1		$2.4		$3.3

   Prepare June’s direct materials purchases budget for Marino’s Frozen Pizza Inc. When required, enter unit prices to the nearest cent. 

   Marino’s Frozen Pizza Inc.
   Direct Materials Purchases Budget
   For the Month Ending June 30, 2016
   	Dough	Tomato	Cheese	Total
   Units required for production:
   12″ pizza	[removed]	[removed]	[removed]
   16″ pizza	[removed]	[removed]	[removed]
   	[removed]	[removed]	[removed]
   Total	[removed]	[removed]	[removed]
   	[removed]	[removed]	[removed]
   Total units to be purchased	[removed]	[removed]	[removed]
   Unit price	x $[removed]	x $[removed]	x $[removed]
   Total direct materials to be purchased	$[removed]	$[removed]	$[removed]	$[removed]

LESSON 08: Hypothesis Testing (continued)

Lesson Objectives:

Testing Hypotheses with a Confidence Interval

Recall that a confidence interval allows the researcher to find a probability with a certain level of confidence.  Remember that the empirical rule states that 68% of the data will fall within 1 standard deviation of the mean in either direction, 95% of the data will fall within 2 standard deviations of the mean, and that 99.7% of the data will fall within 3 standard deviations of the mean as it pertains to the area under the normal distribution curve.  When we are testing a hypothesis we can set the confidence level and find the probability that an event will or will not occur according to the confidence interval.  If we know the confidence level then we can calculate the confidence interval and determine the rejection region.  If the test statistic falls in either of the rejection regions then we can reject the null hypothesis.  When testing a large sample we use the z-score and there are set confidence intervals for each confidence level when using z-scores and they are as follows:

Confidence Level                     Corresponding z-score                         Corresponding z-score rejection region
90%                                                   1.645                                                             z < -1.645    or z > 1.645
95%                                                   1.96                                                               z < -1.96      or z < 1.96
99%                                                   2.575                                                             z < 2.575     or z > 2.575

Recall that the confidence interval formula is the sample mean +/- z(standard deviation/√n)

For t-scores, it is a little different since the t-score depends on n – 1 degrees of freedom.  The first row of the t-table gives alpha (a) which is 1 – confidence level.  The 2nd column has .40 which is equal to the amount of area left in the two tails of the normal distribution if the confidence level is set at 60%.  The 3rd column represents 75% and so on.  Below is a list of the confidence levels and the corresponding alpha for each one so that you will know which column holds the correct t-score based on the given confidence level.  Remember that the table gives you the total area left after you have considered the confidence level and it will need to be divided by two to come up with the area in each tail of the normal distribution.

Confidence level                     Corresponding a-value for t-table                      
60%                                                    .40
75%                                                    .25
90%                                                    .10
95%                                                    .05
97.5%                                                 .025
99%                                                    .01
99.5%                                                 .005
99.95%                                               .0005

Recall that the confidence interval formula is the sample mean +/- t(standard deviation/√n)

Let’s practice: 

The average yearly income of families in a particular state is $40,000.  However a sample of 100 people show that the average income is $45,000 with a population standard deviation of $500.  Conduct a hypothesis test at confidence level 95% to see if the true average income of families reported by the state is correct.

Step 1:  State the null and alternative hypothesis
H₀ = sample mean = population mean
H₁ = sample mean ≠ population mean

Step 2:  Determine the confidence interval.  (we will use a z-score since this is a large sample)
Since the confidence level is 95%, the confidence interval is -1.96< z <1.96

Step 3:  Determine the z-score

z = 45000-40000/500=10

Step 3: Compare this z-score to the confidence interval and make a conclusion.
This is outside of the confidence interval which means that we must reject the null hypothesis and conclude that the average income reported by the state is incorrect.

A particular school gives annual standardized tests at the end of the year and last year’s average score was 70 with a standard deviation of 5. A sample of 10 students’ tests were pulled and the average score was 85.  Considering a 99% confidence level, did the school report the correct overall average for the students’ test scores?

State the null and alternative hypotheses.
H₀ = sample mean = population mean
H₁ = sample mean ≠ population mean

Determine the confidence level(we will use a t-score since the sample size is small)
10 -1 = 9 degrees of freedom
99% confidence level = t_{.005 with 9 degrees of freedom}= 3.24 or 3.24/2 = 1.62
So the confidence interval is -1.62 > t > 1.62 and the rejection region is t < -1.62 or t > 1.62

Determine the t-score

t = 85 – 70 / 5 = 3

Compare the t-score to the confidence interval and make a conclusion.
Since the t-score is beyond the confidence interval and within the rejection region we must reject the null hypothesis.

Using the Chi-Squared Distribution 

The chi-squared statistic (x²) is used to compare observed values to the expected values in an experiment.  The formula for the chi-squared statistic is the sum of all  (observed values – expected value)² / expected value. The chi-squared table is much  like the t-table meaning that you have to calculate the degrees of freedom but this time we use the number of categories -1 rather than n -1.  The chi-squared table has p-values listed in the first row which represent the area left under the chi-squared distribution as it relates to the confidence interval.  P=.05 represents a 95% confidence interval, P= .01 represents a 90% confidence interval, and P= .001 represents a 99% interval.  Once you have found the degrees of freedom then you look over to the corresponding column for the p-value.  The number where that row and that column meet is the chi-squared statistic.  The data must fill two conditions in order to use the chi-squared distribution: 1)  the total observed values must exceed 20 and 2) the expected value must exceed 4 for each category.  Watch this video to see what the chi-squared distribution looks like and how it works.

Let’s Practice:  

A high school principal gave a questionnaire to 25 boys and 25 girls to see if gender played a role in the students’ responses to the following statement:

What does this mean?

The observed values are the actual number of males and females who answered disagree to the question.  The expected values is the average of the observed values(total expected value/# of categories).  Now that we have calculated the chi-squared or critical value we must compare it to the actual chi-squared statistic.  We find the degrees of freedom by subtracting 1 from the number of categories.  Since we have two categories (male and female) the degrees of freedom = 2 – 1 = 1.  Remember that the confidence level is 95% so that means that there is a = .05 so we are looking for the number where row 1(degrees of freedom) and column p = .05 (alpha level) intersect.  Find this number on the chi-squared table.  The value is 3.84.  To conclude we must determine whether x²  is greater than or equal to that value or less than that value.  If the chi squared that we calculated from the set of data is ≥ the p-value (value from chi-squared table) then we must reject the null hypothesis and if it is less than the p-value then we can not reject the null hypothesis.  Since x² = 3.2 and 3.2 < 3.84  we can not reject the null hypothesis.  So the principal can not rule out the fact that the genders did not play a role in the students’ responses to the statement.

Additional Resources

chi-squared worked examples

NEXT TEACHER OFFICE HOURS ARE: 

Grading Rubric: 

Assignment:

For questions 1- 5 use confidence intervals to test the hypothesis.

1)  A light bulb producing company states that its lights will last an average of 1200 hours with a standard deviation of 200 hours.  A sample of 100 light bulbs from the company were tested and the researcher found that the average life of each light bulb was 1050 hours.  At a 95% confidence level, determine whether these light bulbs are in compliance with the company’s claim.

2)  A company’s human resource department claims that all employees are present on the average 4 days out of the work week with a standard deviation of 1.  They hired an outside company to do an audit of their employees’ absences.  The company took a sample a 10 people and found that on the average the employees were present 3 days per week.  With a 95% confidence level, determine whether the company’s claim is true based on the data from the sample.

3)  A teacher claims that all of her students pass the state mandated test with an average of 90 with a standard deviation of  10.  The principal gave the test to 20 of her students to see if the teacher’s claim was true.  He found that the average score was 75.  With a 95% confidence level, determine whether the teacher is making the correct claim about all of her students.

4)  The lifeguard’s at a local pool have to be able to respond to a distressed swimmer at an average of 10 seconds with a standard deviation of 4 in order to be considered for employment.  If a sample of 100 lifeguards showed that their average response time is 15 seconds, with a confidence level of 95% determine whether this group may be considered for employment.

5)  It is believed that an average of  20 mg of iodine is in each antibiotic cream produced by a certain company with a standard deviation of 5 mg.  The company pulled 150 of its antibiotic creams and found that on the average each cream contained 29 mg of iodine.  Determine with a 95% confidence level whether or not these creams are in compliance with the company’s belief?

For questions 6 – 10 use the chi-squared distribution to test the hypothesis.

6)  A restaurant owner wants to see if the business is good enough for him to purchase a restaurant.  He asks the present owner for a breakdown of how many customers that come in for lunch each day and the results are as follows:  Monday – 20, Tuesday – 30, Wednesday – 25, Thursday – 40 and Friday – 55.  The prospective owner observes the restaurant and finds the following number of customers coming for lunch each day:  Monday- 30, Tuesday – 15, Wednesday- 7, Thursday 40, and Friday – 33.  At a 95% confidence level determine whether the present owner reported the correct number of customers for lunch each day.

7)  An employer polled its employers to see if they agree with the proposed new store hours and whether or not their present shift made a difference in their answers.  The customers answered 1 for agree, 2 for don’t know, and 3 for disagree.  Nine first shift employees answered “agree”, 15 second shift employees answered “agree”, and 20 third shift employees answered agree.  With a 95% confidence level determine whether or not the employees’ present shift played a role in their responses to the poll.

8)  A politician surveyed 100 citizens to determine if their job title had anything to do with the way they responded to the following statement:  “A city-wide curfew will be put into place.  Select the time that you think it should be put into place.  8pm, 9pm, or 10pm”.  He is mostly concerned with the 10 pm responses.  25 teachers chose 10pm, 40 doctors chose 10pm, and 35 police responded 10pm.  With a 95% confidence level, determine whether job title plays a role in how the citizens responded to the statement.

9)  A meter reader did an experiment to see if there is a relationship between the number of tickets she writes and the number of blocks she is away from the park that is considered the heart of the city.  At 0 blocks from the park she writes 35 tickets, at 1 block away from the park she writes 25 tickets, at 2 blocks from the park she writes 20 tickets and at 3 blocks from the park she writes 25 tickets.  Use a 95% confidence level.

10)  A high school principal asks his students to respond to the following statement:  “School should start at 9:00am rather than 7:00am.  Answer 1 for agree, 2 for don’t know, and 3 for disagree.”  There were 90 seniors who answered agree, 35 juniors, 30 sophomores, and 25 freshmen.  Help the principal decide with a 95% confidence level that the students’ status played a role in how they responded to the question.

To upload a file for the teacher to see, click here.

MAT540

Week 1 Homework

Chapter 1 

2.    The Retread Tire Company recaps tires. The fixed annual cost of the recapping operation is $60,000.The variable cost of recapping a tire is $9.The company charges $25 to recap a tire.

a. For an annual volume of 12,000 tires, determine the total cost, total revenue, and profit.

b. Determine the annual break-even volume for the Retread Tire Company operation.

4.    Evergreen Fertilizer Company produces fertilizer. The company’s fixed monthly cost is $25,000, and its variable cost per pound of fertilizer is $0.15. Evergreen sells the fertilizer for $0.40 per pound. Determine the monthly break-even volume for the company.

12. If Evergreen Fertilizer Company in Problem 4 changes the price of its fertilizer from $0.40 per pound to $0.60 per pound, what effect will the change have on the break-even volume?

14. If Evergreen Fertilizer Company increases its advertising expenditures by $14,000 per year, what effect will the increase have on the break-even volume computed in Problem 13?

Reference Problem 13:  If Evergreen Fertilizer Company changes its production process to add a weed killer to the fertilizer in order to increase sales, the variable cost per pound will increase from $0.15 to $0.22. What effect will this change have on the break-even volume computed in Problem 12?

20. Annie McCoy, a student at Tech, plans to open a hot dog stand inside Tech’s football stadium during home games. There are seven home games scheduled for the upcoming season. She must pay the Tech athletic department a vendor’s fee of $3,000 for the season. Her stand and other equipment will cost her $4,500 for the season. She estimates that each hot dog she sells will cost her $0.35. She has talked to friends at other universities who sell hot dogs at games. Based on their information and the athletic department’s forecast that each game will sell out, she anticipates that she will sell approximately 2,000 hot dogs during each game.

a. What price should she charge for a hot dog in order to break even?

b. What factors might occur during the season that would alter the volume sold and thus the break-even price Annie might charge?

22. The College of Business at Tech is planning to begin an online MBA program. The initial start-up cost for computing equipment, facilities, course development, and staff recruitment and development is $350,000.The college plans to charge tuition of $18,000 per student per year. However, the university administration will charge the college $12,000 per student for the first 100 students enrolled each year for administrative costs and its share of the tuition payments.

a. How many students does the college need to enroll in the first year to break even?

b. If the college can enroll 75 students the first year, how much profit will it make?

c. The college believes it can increase tuition to $24,000, but doing so would reduce enrollment to 35.  Should the college consider doing this?

Chapter 11  

18.  The following probabilities for grades in management science have been determined based on past records:

The grades are assigned on a 4.0 scale, where an A is a 4.0, a B a 3.0, and so on. Determine the expected grade and variance for the course.

20. An investment firm is considering two alternative investments, A and B, under two possible future sets of economic conditions, good and poor. There is a .60 probability of good economic conditions occurring and a .40 probability of poor economic conditions occurring. The expected gains and losses under each economic type of conditions are shown in the following table:

Using the expected value of each investment alternative, determine which should be selected.

26. The weight of bags of fertilizer is normally distributed, with a mean of 50 pounds and a standard deviation of 6 pounds. What is the probability that a bag of fertilizer will weigh between 45 and 55 pounds?

28. The Polo Development Firm is building a shopping center. It has informed renters that their rental spaces will be ready for occupancy in 19 months.  If the expected time until the shopping center is completed is estimated to be 14 months, with a standard deviation of 4 months, what is the probability that the renters will not be able to occupy in 19 months?

30. The manager of the local National Video Store sells videocassette recorders at discount prices. If the store does not have a video recorder in stock when a customer wants to buy one, it will lose the sale because the customer will purchase a recorder from one of the many local competitors. The problem is that the cost of renting warehouse space to keep enough recorders in inventory to meet all demand is excessively high. The manager has determined that if 90% of customer demand for recorders can be met, then the combined cost of lost sales and inventory will be minimized. The manager has estimated that monthly demand for recorders is normally distributed, with a mean of 180 recorders and a standard deviation of 60. Determine the number of recorders the manager should order each month to meet 90% of customer demand.

To test the effectiveness of a business school preparation course, 8 students took a general business test before and after the course. The results are given below.

                                                         Exam Score                   Exam Score

                            Student                Before Course             After Course

                                 1                            530                                       670

                                 2                            690                                       770

                                 3                            910                                       1,000

                                 4                            700                                       710

                                 5                            450                                       550

                                 6                            820                                       870

                                 7                            820                                       770

                                 8                            630                                       610

Construct the 95% confidence interval for the mean difference between the exam scores before the course and exam scores after the course.