The standard deviation alone does not measure relative variation. For example, a standard deviation of $1 would be considered large if it is describing the variability from store to store in the price of an ice cube tray. On the other hand, a standard deviation of $1 would be considered small if it is describing store-to-store variability in the price of a particular brand of freezer. A quantity designed to give a relative measure of variability is the coefficient of variation. Denoted by CV, the coefficient of variation expresses the standard deviation as a percentage of the mean. It is defined by the formula CV = 100(s/ x ). Consider two samples. Sample 1 gives the actual weight (in ounces) of the contents of cans of pet food labeled as having a net weight of 8 oz. Sample 2 gives the actual weight (in pounds) of the contents of bags of dry pet food labeled as having a net weight of 50 lb. There are weights for the two samples.

(a) For each of the given samples, calculate the mean and the standard deviation. (Round all intermediate calculations and answers to five decimal places.)

(b) Compute the coefficient of variation for each sample. (Round all answers to two decimal places.)

1. A company uses three different assembly lines – A1, A2, and A3 – to manufacture a particular component. Of those manufactured by line A1, 5% need rework to remedy a defect, whereas 8% of A2’s components need rework and 10% of A3’s need rework. Suppose that 50% of all components are produced by line A1, 30% are produced by line A2, and 20% come from line A3.

   1. (a) Suppose a component is selected at random, what is the probability that it needs rework?

   2. (b) If a randomly selected component needs rework, what is the probability that it came from line A1?

   3. (c) If a randomly selected component (((((((does not))))))) need rework, what is the proba- bility that it came from line A2?

You are a professor of statistics and have been asked to teach a course in social science statistics off-campus to a class of grad students enrolled in the Continuing Education Program of the University. Since you’ve never taught this program before, you don’t know a great deal about the needs and background of the students in the class. In order to learn more, you hand out a survey to each student asking for information on the following variables: age, undergraduate field, number of stats courses taken, and the level of interest in conducting research (coded as low, medium, high). The results are below:

Analyze your data to give you some useful information about the class. In doing so you need to answer a few things: the level of measurement of the variables, meaningful measures of central tendency for each variable (there can be more than one), the calculated measure of central tendency for each variable (there can be more than one). In order to do that, populate the following table:

Based on the data you collected, calculate the measures of dispersion (specifically the range, variance and standard deviation) for each of the variables that are at the interval level of measurement.

Since all of your data, your measures of central tendency and measures of dispersion. brief paragraph explaining the results of your survey paying special attention to what you, as the instructor, would find useful to bear in mind as you conduct the class. Address each variable, meaningful measures of central tendency for each (providing brief mention as to why you think which are the most useful), and measures of dispersion (where appropriate). You can include graphical representations of the data where it would help to defend the answer.

1. A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is not equal to the mean of WaxCo. In a random sample of 37 floors of WaxWin and 30 of WaxCo. WaxWin had a mean lifetime of 26.2 and WaxCo had a mean lifetime of 21.9. The population standard deviation for WaxWin is assumed to be 9.1 and the population standard deviation for WaxCo is assumed to be 9.2. Perform a hypothesis test using a significance level of 0.10 to help him decide. Let WaxWin be sample 1 and WaxCo be sample 2. The correct hypotheses are: H 0 : μ 1 ≤ μ 2 H 0 : μ 1 ≤ μ 2 H A : μ 1 > μ 2 H A : μ 1 > μ 2 (claim) H 0 : μ 1 ≥ μ 2 H 0 : μ 1 ≥ μ 2 H A : μ 1 < μ 2 H A : μ 1 < μ 2 (claim) H 0 : μ 1 = μ 2 H 0 : μ 1 = μ 2 H A : μ 1 ≠ μ 2 H A : μ 1 ≠ μ 2 (claim) Correct

Since the level of significance is 0.10 the critical value is 1.645 and -1.645

The test statistic is: Incorrect(round to 3 places)

The p-value is: Incorrect(round to 3 places)

A random sample of 30 chemists from Washington state shows an average salary of $42546, the population standard deviation for chemist salaries in Washington state is $868. A random sample of 39 chemists from Florida state shows an average salary of $48395, the population standard deviation for chemist salaries in Florida state is $945. A chemist that has worked in both states believes that chemists in Washington make more than chemists in Florida. At αα=0.05 is this chemist correct?

Let Washington be sample 1 and Florida be sample 2.

The correct hypotheses are:

• H0:μ1≤μ2H0:μ1≤μ2
  HA:μ1>μ2HA:μ1>μ2(claim)
• H0:μ1≥μ2H0:μ1≥μ2
  HA:μ1<μ2HA:μ1<μ2(claim)
• H0:μ1=μ2H0:μ1=μ2
  HA:μ1≠μ2HA:μ1≠μ2(claim)

Since the level of significance is 0.05 the critical value is 1.645
The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

A researcher is interested in seeing if the average income of rural families is greater than that of urban families. To see if his claim is correct he randomly selects 45 families from a rural area and finds that they have an average income of $66299 with a population standard deviation of $668. He then selects 31 families from a urban area and finds that they have an average income of $67979 with a population standard deviation of $534. Perform a hypothesis test using a significance level of 0.01 to test his claim. Let rural families be sample 1 and urban familis be sample 2.

The correct hypotheses are:

• H0:μ1≤μ2H0:μ1≤μ2
  HA:μ1>μ2HA:μ1>μ2(claim)
• H0:μ1≥μ2H0:μ1≥μ2
  HA:μ1<μ2HA:μ1<μ2(claim)
• H0:μ1=μ2H0:μ1=μ2
  HA:μ1≠μ2HA:μ1≠μ2(claim)

Since the level of significance is 0.01 the critical value is 2.326
The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

A researcher is interested in seeing if the average income of rural families is greater than that of urban families. To see if his claim is correct he randomly selects 45 families from a rural area and finds that they have an average income of $66299 with a population standard deviation of $668. He then selects 31 families from a urban area and finds that they have an average income of $67979 with a population standard deviation of $534. Perform a hypothesis test using a significance level of 0.01 to test his claim. Let rural families be sample 1 and urban familis be sample 2.

The correct hypotheses are:

• H0:μ1≤μ2H0:μ1≤μ2
  HA:μ1>μ2HA:μ1>μ2(claim)
• H0:μ1≥μ2H0:μ1≥μ2
  HA:μ1<μ2HA:μ1<μ2(claim)
• H0:μ1=μ2H0:μ1=μ2
  HA:μ1≠μ2HA:μ1≠μ2(claim)

Since the level of significance is 0.01 the critical value is 2.326
The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

Question:

The survey is conducted to calculate average number of sick days.The employees are categorised into nurses, doctors and administrators. Stratified random sampling, is used with each group forming a separate strata in order to conduct survey. The no. of employees in the strata are given in table below:

Table-1:

Number of employees in each category

Number of Sick days for 9 doctors, 7 nurses and 4 administrators

Table-2:

a.  Calculate the sample size from each strata if proportional allocation is done for selecting total 25 employees from table-1

b.  Using the data from table-2 calculate the mean number of sick days for that year.

c.  Place a 95% confidence bound on the mean.

d. Calculate the total number of sick days for the year.

e.  Place a bound on the total number of sick days and interpret this bound.

Consider the following information about travelers on vacation: 40% check work email, 30% use a cell phone to stay connected to work, 25% bring a laptop with them, 21% both check work email and use a cell phone to stay connected, and 59% neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition, 88 out of every 100 who bring a laptop also check work email, and 70 out of every 100 who use a cell phone to stay connected also bring a laptop.

(a)

What is the probability that a randomly selected traveler who checks work email also uses a cell phone to stay connected?

(b)

What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected?

(c)

If the randomly selected traveler checked work email and brought a laptop, what is the probability that he/she uses a cell phone to stay connected? (Round your answer to four decimal places.)

You are evaluating the capability of a process. Standard deviation (s) is 5. Customers have said that they will accept units ranging between 35 and 60. The location of the mean is unknown, but your boss wants to know if the process is capable of performing to customer specifications if the mean of the data was centered between specification limits. What would you say?

A: The process isn't capable no matter how you try to center it

B: The process is capable, but not acceptable

C: The process is capable and acceptable

D: I need to know more information before I can give you an answer

Use the data from question 1. Conduct a hypothesis test at α = .01 to determine if the population variance is greater than 904.75.

Question 1: 1. Consider the following sampled data: s 2 = 906.304, n = 31. Calculate the following confidence intervals for the population variance: (a) 90% (b) 95% (c) 99%

In 1997 a woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard (Genessey v. Digital Equipment Corporation). The jury awarded about $3.5 million for pain and suffering, but the court then set aside that award as being unreasonable compensation. In making this determination, the court identified a "normative" group of 27 similar cases and specified a reasonable award as one within 2 standard deviations of the mean of the awards in the 27 cases. The 27 award amounts (in thousands of dollars) are in the table below.

What is the maximum possible amount that could be awarded under the "2-standard deviations rule"? (Round all intermediate calculations and the answer to three decimal places.)
___________ (in thousands of $)

Discuss the following three questions:

• What Big Data trends in the field of personal health are more exciting to you? why?
• What new Big Data trends in the field of sports do you think will impact the sport and audiences the most? why?
• How, do you think, the music industry can benefit from the application of Big Data technology?

3. Use the data from question 1. Conduct a hypothesis test at α = .05 to determine if the population variance is less than 909.00.

Question 1- 1. Consider the following sampled data: s 2 = 906.304, n = 31. Calculate the following confidence intervals for the population variance: (a) 90% (b) 95% (c) 99%

1.A soccer ball manufacturer wants to estimate the mean circumference of soccer balls within 0.05 in. ​(a) Determine the minimum sample size required to construct a 95​% confidence interval for the population mean. Assume the population standard deviation is 0.20 in. ​(b) Repeat part​ (a) using a population standard deviation of 0.30 in. Which standard deviation requires a larger sample​ size? Explain.

2.A soccer ball manufacturer wants to estimate the mean circumference of​ mini-soccer balls within 0.05 inch. Assume the population of circumferences is normally distributed. ​(a) Determine the minimum sample size required to construct a 95​% confidence interval for the population mean. Assume the population standard deviation is 0.30 inch. ​(b) Repeat part​ (a) using a population standard deviation of 0.40 inch. ​(c) Which standard deviation requires a larger sample​ size? Explain.

2. Using your answers from question 1, determine the following confidence intervals for the population standard deviation:

(a) 90%

(b) 95%

(c) 99%

QUESTION 1:

Question 1: 1. Consider the following sampled data: s 2 = 906.304, n = 31. Calculate the following confidence intervals for the population variance: (a) 90% (b) 95% (c) 99%

The mean of a normal probability distribution is 380; the standard deviation is 16. About 68% of the observations lie between what two values