A Coca-Cola bottling plant's product line includes 12-ounce cans of Coke products. The

cans are filled by an automated filling process that can be adjusted to any mean filll volume and that

will ll cans according to a normal distribution. However, not all cans will contain the same volume

due to variation in the filling process. Historical records show that regardless of what the mean is set

at, the standard deviation in ll will be 0.035 ounce. Operations managers at the plant know that if

they put too much Coke in a can, the company loses money. If too little is put in the can, customers

are short changed, and the State Department of Weights and Measures may fine the company.

(a) Suppose the industry standards for fill volume call for each 12-ounce can to contain between

11.98 and 12.02 ounces. Assuming that the manager sets the mean fill at 12 ounces, what is the

probability that a can will contain a volume of Coke product that falls in the desired range?

(b) Assume that the manager is focused on an upcoming audit by the Department of Weights and

Measures. She knows the process is to select one Coke can at random and that if it contains less

than 11.97 ounces, the company will be reprimanded and potentially fined. Assuming that the

manager wants at most a 5% chance of this happening, at what level should she set the mean fill

level?

You work as a marine engineer, your job is to design several types of fuels for marine crafts. You do this
by combining three different chemicals in different proportions; chemical x, chemical y, and chemical z.
Your chemicals are stored in vats of equal volume (10 litres).
You are ordering a shipment of new chemicals, and need to work out the density (mass per volume)
of each chemical to calculate freight costs. You look through the documentation that you have on the
chemicals. Although you don’t find a list of their respective densities, you do find a ledger where you
have recorded the number of vats of each chemical required to produced test batches of three different
types of fuel (150 litres of each), and the mass of the fuel produced (in kilograms):
No. vats of chemical x No. vats of chemical y No. vats of chemical z Fuel mass (kg), 150L

(a) By forming an appropriate matrix system and applying Gaussian elimination to an augmented matrix,
show that the mass per vat of chemical x, chemical y and chemical z is 4, 6 and 7 kilograms respectively.
Express your working in fractions to avoid rounding error during to Gaussian elimination.
(b) Assuming that no mass or volume is lost when producing the fuel, determine the densities of each
chemical.
(c) You need to make 150,000 litres of each fuel type. You need to choose between Freight Company
A and Freight Company B; Freight Company A charges $0.08 per kilogram of chemical delivered for
chemicals up to 0.5 kg/L dense, and $0.09 per kilogram for chemicals more dense than 0.5 kg/L whereas
Freight Company B charges a flat rate of $0.05 per litre of chemical delivered.
i) Calculate the total freight cost and use this information to identify which freight company will
be cheaper given that chemicals can not be mixed prior to freighting.
ii) Suppose that you still need to produce 150,000 litres of Fuel A and Fuel C, but now want
to produce more than 150,000 litres of Fuel B. How many litres of Fuel B could you produce before
switching freight companies becomes cheaper?

According to the Behavioral Risk Factor Surveillance System, 58% of all Americans adhere to a sedentary lifestyle. Suppose that you select a sample of 10 individuals and find that 8 of them do not exercise regularly. Assuming that the Surveillance System is correct, what is the probability of finding 8 or more individuals who are sedentary, from a sample of 10?

1.What is the central limit theorem and why is it important in statistics? 2.Explain the differences between the mean, mode and median. Which is the most useful measure of an average and why? 3.Which is a more useful measure of central tendency for stock returns – the arithmetic mean or the geometric mean? Explain your answer.

Open excel In one column enter 8 heights for 10 year old boys (inches) and in the second column enter 8 weights for 10 year old boys (pounds). Ten year old boys height ranges from 42 inches to 62 inches while the weight ranges from 60 to 90 pounds. You can make these up as long as they are reasonable if you do not have access to any data. We want to see if there is a correlation. Move your mouse to put a box around the numbers. From the insert menu select scatter or the chart that says scatter and then select the chart at the top left on the dropdown Click on one of the points Select Add Trendline The default radio button is linear, keep it checked Check display equation on chart and click on the display r-squared value The equation is your least squares line On the least squares line the slope is in front of the x value. Also, you can look at the slope of the scatter plot to see if the slope is positive. This will let you know whether to make the r value positive or negative when you take the square root of r^2. If the r^2 value is say .0382 then you will click on an excel cell and type =.0382^.5 Note the ^ key is above the 6 key This will be your r value and you will make it negative if your slope is negative otherwise leave it positive. If you have a different version of excel then you may want to search to see how to form a trendline. These instructions work with Office 365 and Office 2010. Then follow these instructions: Attach your scatter plot if possible. What did you get for your regression line? What was your r value? What did this tell you? What would be a scenario where you might need to use this application? Explain in a minimum of 250 words

The director of library services at a college did a survey of types of books (by subject) in the circulation library. Then she used library records to take a random sample of 888 books checked out last term and classified the books in the sample by subject. The results are shown below.

Using a 5% level of significance, test the claim that the subject distribution of books in the library fits the distribution of books checked out by students.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: The distributions are the same.
H₁: The distributions are the same.H₀: The distributions are different.
H₁: The distributions are the same.    H₀: The distributions are the same.
H₁: The distributions are different.H₀: The distributions are different.
H₁: The distributions are different.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

chi-squarenormal    uniformbinomialStudent's t

What are the degrees of freedom?


(c) Estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, the evidence is sufficient to conclude that the subject distribution of books in the library is different from that of books checked out by students.At the 5% level of significance, the evidence is insufficient to conclude that the subject distribution of books in the library is different from that of books checked out by studentsThe director of library services at a college did a survey of types of books (by subject) in the circulation library. Then she used library records to take a random sample of 888 books checked out last term and classified the books in the sample by subject. The results are shown below.

Using a 5% level of significance, test the claim that the subject distribution of books in the library fits the distribution of books checked out by students.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: The distributions are the same.
H₁: The distributions are the same.H₀: The distributions are different.
H₁: The distributions are the same.    H₀: The distributions are the same.
H₁: The distributions are different.H₀: The distributions are different.
H₁: The distributions are different.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

chi-squarenormal    uniformbinomialStudent's t

What are the degrees of freedom?


(c) Estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, the evidence is sufficient to conclude that the subject distribution of books in the library is different from that of books checked out by students.At the 5% level of significance, the evidence is insufficient to conclude that the subject distribution of books in the library is different from that of books checked out by students

6.27  Average hours per week listening to the radio. The Student Monitor surveys 1200 undergraduates from four-year colleges and universities throughout the United States semiannually to understand trends among college students.¹¹ Recently, the Student Monitor reported that the average amount of time listening to the radio per week was 11.5 hours. Of the 1200 students surveyed, 83% said that they listened to the radio, so this collection of listening times has around 204 (17% × 1200) zeros. Assume that the standard deviation is 8.3 hours.

1. (a) Give a 95% confidence interval for the mean time spent per week listening to the radio.

2. (b) Is it true that 95% of the 1200 students reported weekly times that lie in the interval you found in part (a)? Explain your answer.

3. (c) It appears that the population distribution has many zeros and is skewed to the right. Explain why the confidence interval based on the Normal distribution should nevertheless be a good approximation.

The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students. In the table, IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and ES refers to extrovert, sensing.

Use a chi-square test to determine if Myers-Briggs preference type is independent of area of study at the 0.05 level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: Myers-Briggs type and area of study are independent.
H₁: Myers-Briggs type and area of study are independent.H₀: Myers-Briggs type and area of study are independent.
H₁: Myers-Briggs type and area of study are not independent.    H₀: Myers-Briggs type and area of study are not independent.
H₁: Myers-Briggs type and area of study are independent.H₀: Myers-Briggs type and area of study are not independent.
H₁: Myers-Briggs type and area of study are not independent.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

normalchi-square    Student's tbinomialuniform

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)

p-value > 0.1000.050 < p-value < 0.100    0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs type and area of study are not independent.At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs type and area of study are not independent.    

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let X denote the number of hoses being used on the self-service island at a particular time, and let Y denote the number of hoses on the full-service island in use at that time. The joint pmf of X and Y appears in the accompanying tabulation. y p(x, y) 0 1 2 x 0 0.10 0.05 0.02 1 0.07 0.20 0.08 2 0.06 0.14 0.28 (a) Given that X = 1, determine the conditional pmf of Y—i.e., pY|X(0|1), pY|X(1|1), pY|X(2|1). (Round your answers to four decimal places.) y 0 1 2 pY|X(y|1) (b) Given that two hoses are in use at the self-service island, what is the conditional pmf of the number of hoses in use on the full-service island? (Round your answers to four decimal places.) y 0 1 2 pY|X(y|2) (c) Use the result of part (b) to calculate the conditional probability P(Y ≤ 1 | X = 2). (Round your answer to four decimal places.) P(Y ≤ 1 | X = 2) = (d) Given that two hoses are in use at the full-service island, what is the conditional pmf of the number in use at the self-service island? (Round your answers to four decimal places.) x 0 1 2 pX|Y(x|2)

The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below.

Use a 5% level of significance to test the claim that the distribution of U.S. households fits the Dove Creek distribution.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: The distributions are different.
H₁: The distributions are different.H₀: The distributions are the same.
H₁: The distributions are different.    H₀: The distributions are different.
H₁: The distributions are the same.H₀: The distributions are the same.
H₁: The distributions are the same.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to two decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

binomialchi-square    normaluniformStudent's t

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)


(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, the evidence is sufficient to conclude that the community household distribution does not fit the general U.S. household distribution.At the 5% level of significance, the evidence is insufficient to conclude that the community household distribution does not fit the general U.S. household distribution.    

A telephone company claims that the service calls which they receive are equally distributed among the five working days of the week. A survey of 80 randomly selected service calls was conducted. Is there enough evidence to refute the telephone company's claim that the number of service calls does not change from day-to-day?

Step 1 of 10:

H0: Service calls are not equally distributed over the five working days.

Ha: Service calls are equally distributed over the five working days.

Step 2 of 10: What does the null hypothesis indicate about the proportions of service calls received each day?

The proportions of service calls received each day are all thought to be equal.

The proportions of service calls received each day are different for each category (and equal to the previously accepted values).

Step 3 of 10: State the null and alternative hypothesis in terms of the expected proportions for each category.

Ho:Pi

Ha: There is some difference amongst the proportions.

Step 4 of 10: Find the expected value for the number of service calls received on Monday. Round your answer to two decimal places.

Step 5 of 10: Find the expected value for the number of service calls received on Tuesday. Round your answer to two decimal places.

Step 6 of 10: Find the value of the test statistic. Round your answer to three decimal places.

Step 7 of 10: Find the degrees of freedom associated with the test statistic for this problem.

Step 8 of 10: Find the critical value of the test at the 0.0250.025 level of significance. Round your answer to three decimal places.

Step 9 of 10: Make the decision to reject or fail to reject the null hypothesis at the 0.0250.025 level of significance.

Fail to Reject Null Hypothesis

Reject Null Hypothesis

Step 10 of 10: State the conclusion of the hypothesis test at the 0.0250.025 level of significance.

There is not enough evidence to refute the claim that the service calls are distributed evenly among the days.

There is enough evidence to refute the claim that the service calls are distributed evenly among the days.

Two different pain reliever drugs, “Drug A” and “Drug B,” are developed to relieve arthritis pain. A clinical study recruits 15 individuals with arthritis. The individuals are given each drug on different days and the number of hours of pain relief for each drug is recorded. This data is in the “arthritis.csv” data file. a) State the appropriate hypotheses to test if there is a difference in the hours of pain relief for Drug A versus Drug B, using “Drug A” as sample 1. b) Use RStudio and report a 93% confidence interval for the difference. Based on this confidence interval, what conclusion can be reached with respect to the hypotheses in a)?

Data

Suppose we have the following information on GMAT scores for business and non-business majors: Business Majors Non-Business Majors n1 = 8 n2 = 5 _ _ X1 = 545 X2 = 525 s1 = 120 s2 = 60

a. Using a 0.05 level of significance, test to see whether the population variances are equal.

b. Using a 0.05 level of significance, test the clam that average GMAT scores for business majors is above the average GMAT scores for non-business majors in the population. Assume unequal population variances.

Just before Quebec referendum on sovereignty, a local newspaper polls

400 voters in an attempt to predict whether it would succeed. Suppose that

sovereignty has the support of 48% of the voters. What is the probability that the

news paper’s sample will lead them to predict victory? Interpret the results in the

context of this problem. Be sure to verify that the assumptions and conditions

necessary for your analysis are met