Questions
(20.37) Researchers claim that women speak significantly more words per day than men. One estimate is...

(20.37) Researchers claim that women speak significantly more words per day than men. One estimate is that a woman uses about 20,000 words per day while a man uses about 7,000. To investigate such claims, one study used a special device to record the conversations of male and female university students over a four- day period. From these recordings, the daily word count of the 20 men in the study was determined. Here are their daily word counts:

28403 10087 15926 21694 37784
10581 12888 11080 17809 13175
8927 6490 8144 7010 4434
10056 4004 12632 10981 5260

What value we should remove from observation for applying t procedures?

A 90% confidence interval (±±10) for the mean number of words per day of men at this university is from  to  words.

Is there evidence at the 10% level that the mean number of words per day of men at this university differs from 7000?

Yes

No

In: Math

You started taking the bus to work. The local transit authority says that a bus should...

You started taking the bus to work. The local transit authority says that a bus should arrive at your bus stop every five minutes. After a while, you notice you spend a lot more than five minutes waiting for the bus, so you start to keep a record.

You spend the next two months recording how long it takes for the bus to arrive to the bus stop. This give a total of sixty observations that denote the number of minutes it took for the bus to arrive (rounded to the nearest minute). These observations are hosted at

   https://mattbutner.github.io/data/bus_stop_time.csv

Load these data into R as a data frame titled bus_stop_time

Create a histogram of the time_until_bus varaible. Would you say that five minutes is a reasonable guess for the average arrival time based on this picture alone?

Create 95% confidence interval for the bus arrival times using the Z distribution. Does 5 minutes fall within the 95% confidence interval?

How would you communicate your finding to the local transit authority?

In: Math

Shipments of TV sets that arrive at a factory have a varying levels of quality. In...

Shipments of TV sets that arrive at a factory have a varying levels of quality. In order to decide whether to accept a particular shipment, inspectors randomly select a sample of 15 TVs and test them; if no more than one TV in the sample is defective, the shipment is accepted. Let X be a random variable representing the number of defective staples in the random sample 15.

a. Explain why X may be treated as a binomial random variable:

•Identify n (the number of trails):
•Specify in words which event would be defined as a “success”
•Explain why the trails may be considered independent:
•Give the value of p (probability of a success)
b. What is the probability that shipment is accepted? ( Use a table or the formula)
c. What is the expected value of the number of defective TV set in the sample?
d. Fill this sentence: According to the Law of Large Numbers, if we have obtained many different simple random samples of size___ from this shipment, the average number of defective TV sets per sample would be approximately ___.

In: Math

For this problem, carry at least four digits after the decimal in your calculations. Answers may...

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

In a random sample of 63 professional actors, it was found that 39 were extroverts.

(a) Let p represent the proportion of all actors who are extroverts. Find a point estimate for p. (Round your answer to four decimal places.)


(b) Find a 95% confidence interval for p. (Round your answers to two decimal places.)

lower limit    
upper limit    


Give a brief interpretation of the meaning of the confidence interval you have found.

We are 5% confident that the true proportion of actors who are extroverts falls within this interval.We are 95% confident that the true proportion of actors who are extroverts falls outside this interval.    We are 5% confident that the true proportion of actors who are extroverts falls above this interval.We are 95% confident that the true proportion of actors who are extroverts falls within this interval.


(c) Do you think the conditions np > 5 and nq > 5 are satisfied in this problem? Explain why this would be an important consideration.

Yes, the conditions are satisfied. This is important because it allows us to say that is approximately binomial.No, the conditions are not satisfied. This is important because it allows us to say that is approximately normal.    Yes, the conditions are satisfied. This is important because it allows us to say that is approximately normal.No, the conditions are not satisfied. This is important because it allows us to say that is approximately binomial.

In: Math

A study  examined whether or not noses continue to grow throughout a person’s lifetime.  The study included many...

A study  examined whether or not noses continue to grow throughout a person’s lifetime.  The study included many measurements (including size of the nose as measured by total volume) and included multiple tests.  For each of the tests described below:

            a)  State the null and alternative hypotheses.

            b) Give a formal decision using a 5% significance level, and interpret the conclusion in context.

  • In a test to see whether males, on average, have bigger noses than females, the study indicates that “p < 0.01.”
  • In a test to see whether there is a positive linear relationship between age and nose size, the study indicates that “p < 0.001.”

In: Math

The failure rate in a statistics class is 20%. In a class of 30 students, find...

The failure rate in a statistics class is 20%. In a class of 30 students, find the probability that exactly five students will fail. Use the normal distribution to approximate the binomial distribution.

In: Math

The mean number of pets per household is 2.96 with standard deviation 1.4. A sample of...

The mean number of pets per household is 2.96 with standard deviation 1.4. A sample of 52 households is drawn. Find the probability that the sample mean is less than 3.11.

a.

0.2245

b.

0.5676

c. 0.3254

d.

0.7726

In: Math

The pH of 20 randomly selected lakes is measured. Their average pH is 5.7. Part A....

The pH of 20 randomly selected lakes is measured. Their average pH is 5.7.

Part A. Historically the standard deviation in the pH values is 0.9. Use this standard deviation for the following questions.
Part Ai. Build a 95% confidence interval for the population mean lake pH.
Part Aii. Build a 90% confidence interval for the population mean lake pH.
Part Aiii. Build an 80% confidence interval for the population mean lake pH.
Part Aiv. Compare the intervals you created in Ai, Aii and Aiii. What effect does changing the level of confidence have on the interval?

Please solve only part B

Part B. For the 20 measured lakes the standard deviation in the pH values is 0.9. Use this standard deviation for the following questions.
Part Bi. Build a 95% confidence interval for the population mean lake pH.
Part Bii. Compare the confidence intervals in Bi and Ai. What effect does not knowing the value of ? have on the interval?
Part Biii. Test whether the population mean pH differs from 6.

In: Math

For 300 trading​ days, the daily closing price of a stock​ (in $) is well modeled...

For 300 trading​ days, the daily closing price of a stock​ (in $) is well modeled by a Normal model with a mean of

​$197.12197.12 and a standard deviation of

​$7.187.18. According to this​ model, what is the probability that on a randomly selected day in this​ period, the stock price closed as follows.

​a) above ​$204.30204.30​?

​b) below ​$211.48211.48​?

​c) between ​$182.76182.76 and ​$211.48211.48​?

In: Math

Hale's TV Productions is considering producing a pilot for a comedy series in the hope of...

Hale's TV Productions is considering producing a pilot for a comedy series in the hope of selling it to a major television network. The network may decide to reject the series, but it may also decide to purchase the rights to the series for either one or two years. At this point in time, Hale may either produce the pilot and wait for the network's decision or transfer the rights for the pilot and series to a competitor for $250,000. Hale's decision alternatives and profits (in thousands of dollars) are as follows:

State of Nature
Decision Alternative Reject, S1 1 Year, S2 2 Years, S3
Produce pilot, d1 -150 50 450
Sell to competitor, d2 250 250 250

The probabilities for the states of nature are P(S1) = 0.20, P(S2) = 0.30, and P(S3) = 0.50. For a consulting fee of $45,000, an agency will review the plans for the comedy series and indicate the overall chances of a favorable network reaction to the series. Assume that the agency review will result in a favorable (F) or an unfavorable (U) review and that the following probabilities are relevant:

P(F) = 0.63 P(S1|F) = 0.07 P(S1|U) = 0.41
P(U) = 0.37 P(S2|F) = 0.27 P(S2|U) = 0.38
P(S3|F) = 0.66 P(S3|U) = 0.21

a. What is the expected value?

b. What is the expected value of perfect information?

c. What is the expected value of the agency's information? Round your answer to two decimal places.

d. What is the maximum that Hale should be willing to pay for the information? Round your answer to two decimal places.

In: Math

In each of parts​ (a)-(c), we have given a likely range for the observed value of...

In each of parts​ (a)-(c), we have given a likely range for the observed value of a sample proportion p. Based on the given​ range, identify the educated guess that should be used for the observed value of p to calculate the required sample size for a prescribed confidence level and margin of error.

a. 0.2 to 0.3

b. 0.1 or less

c. 0.3 or greater

In: Math

Physical activity of obese young adults. In a study on the physical activity of young adults,...

Physical activity of obese young adults. In a study on the physical activity of young adults, pediatric researchers measured overall physical activity as the total number of registered movements (counts) over a period of time and then computed the number of counts per minute (cpm) for each subject (International Journal of Obesity, Jan. 2007). The study revealed that the overall physical activity of obese young adults has a mean of μ = 320 cpm μ = 320   cpm and a standard deviation of σ = 100 c p m . σ = 100 c p m . (In comparison, the mean for young adults of normal weight is 540 cpm.) In a random sample of n = 100 n = 100 obese young adults, consider the sample mean counts per minute, ¯ x x ‾ . Describe the sampling distribution of ¯ x x ‾ . What is the probability that the mean overall physical activity level of the sample is between 300 and 310 cpm? What is the probability that the mean overall physical activity level of the sample is greater than 360 cpm?

In: Math

Please Double Check answers I've recived 3 wrong answers on three diffrent questions today thank you...

Please Double Check answers I've recived 3 wrong answers on three diffrent questions today thank you

CNNBC recently reported that the mean annual cost of auto insurance is 1006 dollars. Assume the standard deviation is 245 dollars. You take a simple random sample of 73 auto insurance policies.

Find the probability that a single randomly selected value is less than 973 dollars. P(X < 973) =

Find the probability that a sample of size n = 73 is randomly selected with a mean less than 973 dollars. P(M < 973) =

In: Math

A sample containing years to maturity and yield for 40 corporate bonds are contained in the...

A sample containing years to maturity and yield for 40 corporate bonds are contained in the data given below.

Years to Maturity Yield Years to Maturity Yield
23.50 4.757 3.75 2.769
21.75 2.473 12.00 6.293
21.50 4.464 17.50 7.411
23.50 4.684 18.00 3.558
27.00 4.799 8.25 0.945
18.25 3.755 23.25 2.966
15.75 7.068 14.75 1.476
2.00 7.043 10.00 1.382
8.75 6.540 23.00 6.334
5.25 7.000 15.25 0.887
11.25 4.823 4.75 4.810
25.75 1.874 18.00 1.238
14.25 5.654 3.00 6.767
19.25 1.745 9.50 3.745
25.00 8.153 17.50 4.186
6.75 6.571 17.00 5.991
23.00 7.506 9.50 7.322
19.00 2.857 5.50 4.871
10.75 8.010 27.50 2.403
21.25 4.214 26.00 4.500

a. What is the sample mean years to maturity for corporate bonds and what is the sample standard deviation?

Mean ? (to 4 decimals)
Standard deviation ? (to 4 decimals)

b. Develop a 95% confidence interval for the population mean years to maturity. Round the answer to four decimal places.

( , ) years

c. What is the sample mean yield on corporate bonds and what is the sample standard deviation?

Mean ?(to 4 decimals)
Standard deviation ?(to 4 decimals)

d. Develop a 95% confidence interval for the population mean yield on corporate bonds. Round the answer to four decimal places.

( , )percent

In: Math

As a data scientist of a company, you want to analyze the following data collected by...

As a data scientist of a company, you want to analyze the following data collected by your company which relates the advertising expenditure A in thousands of dollars to total sales S in thousands of dollars. The following table shows this relationship

Advertising Expenditure (A) Total Sales (S)
18.6 312
18.8 322
18.8 333
18.8 317
19 301
19 320
19.2 305

Using Advertising expenditure (A) as the domain and Total Sales (S) as the range, the data is not a function because the value 18.8 and 19 appear in the domain more than once with a different corresponding value of the range each time.

--Interpret the slope and y-intercept of this equation.

--Express this equation as a function S of A and find its domain.

--Predict the sales if the advertising expenditure is $25000.

In: Math