A cell phone company offers two plans to its subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 40 subscribers to Plan A is $50,500 with a standard deviation of $6,700. For a sample of 25 subscribers to Plan B, the mean income is $53,300 with a standard deviation of $6,300.

At the 0.02 significance level, is it reasonable to conclude the mean income of those selecting Plan B is larger? Hint: For the calculations, assume the Plan A as the first sample.

The test statistic is ____. (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)

HW8#11

A genetic experiment involving peas yielded one sample of offspring consisting of 430 green peas and 168 yellow peas. Use a 0.01 significance level to test the claim that under the same​ circumstances, 27​% of offspring peas will be yellow. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method and the normal distribution as an approximation to the binomial distribution.

A. Identify the null and alternative hypotheses.

B. What is the test statistic?

   z=

   Round two decimal places as needed.

C. What is the P-value?

D. Fail to reject OR Reject the null hypothesis?

E. There is or is not sufficient evidence?

The file medinc.mtw contains data on the median incomes (medinc) of census dissemination areas in Toronto. (a) Treating this set of data as the population, use Minitab to calculate the population mean and the population standard deviation for the medinc variable. Set aside all population information until parts (d) and (e). (b) Use Minitab (Calc Menu – Random Data – Sample from Columns) to draw twenty samples of size n = 40 from the Toronto medinc population. This procedure must be replicated twenty times (note that if you open up the same sampling dialog box each time from the menu, then you only have to replace the last destination column with the next one. In Minitab Express, you must specify the sample size each time but you do not need to specify the destination column). For each sample, use Minitab to calculate the sample mean. Note that when you can use the Descriptive Statistics function, you can specify the calculation of only the sample means of all your samples at once. (c) Copy the sample means into a new column and calculate the mean and standard deviation of your sample means. (d) The sampling distribution of the sample mean has an average equal to the population mean. Explain why the mean in part (c) may be different from the population mean in part (a). (e) How would you use the sample standard deviation in part (c) to estimate the population standard deviation in part (a)? How close is your estimate to the true value? I am only looking for how we calculate e) I have all the others done and my sample standard deviation is 1593.5 with a mean of 25251.7. I'm using n=40 and perhaps that is where I am going wrong

3.2 Our overworked student trudges to his 3rd exam of the day, a true/false exam with 100 questions. Again,he just guesses the answers. Notice that this time he has a 50% chance of getting each particular question correct. (10 Marks: 2 Marks for each of parts a), b), c), d) and e).

a. What is the probability that the student gets at least one question correct?

b. What is the probability that the student gets between 15 and 35 questions (inclusive) correct?

c. What is the probability that the student gets between 40 and 60 questions (inclusive) correct?

d. What is the probability that the student passes the exam (gets 60 or more questions correct)?

e. Why does the student have a better chance of passing the exam in question 3.2 than the one in question 3.1?

Read the “Evaluating the Quality of Open Source Software” article (Spinellis, et al., 2009). What are the strengths and weaknesses of this particular study? (For example, does the study do an adequate job of collecting and analyzing representative samples?) What would you suggest that the authors do to remedy those weaknesses and why? Use both your text and the article An Introduction to Research Design (Ganster, 2003). You may also draw on any book, article, website, or other reliable resource in answering this question.

Choose one of the tools listed here that was not utilized in the course (SAS, R, or JMP,). Provide a brief description of this tool including its strengths and weaknesses. Describe a scenario where you would make use of this tool and explain why this tool would be preferred in that case.

HW8#10

A certain drug is used to treat asthma. In a clinical trial of the​ drug,17 of 288 treated subjects experienced headaches​ (based on data from the​ manufacturer). The accompanying calculator display shows results from a test of the claim that less than 10​% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts​ (a) through​ (e) below.

A. Is the test two-tailed, Left-tailed, or right tailed?

B. What is the test statistic?

   z=

   Round two decimal places as needed.

C. What is the P-value?

Round four decimal places as needed.

D. What is the null hypothesis, and what do you conclude about it?

E. What is the final conclusion?

   There is or is not sufficient evidence?

Question 1:In the layout of a printed circuit board for an electronic product, there are 12 different locations that can accomodate chips. (a) If Five different types of chips are to be placed on the board, how many different layouts are possible? (b) What is the probability that five chips that are placed on the board are of the same type? Question 2: A Web ad can be designed from 4 different colors, 3 font types, 5 font sizes, 3 images, and 5 text phrases. A specific design is randomly generated by the Web server when you visit the site. Let A denote the event that the design color is red and let B denote the event that the font size is not the smallest one. Find P(A ∪ B0 ) and P(B|A). Question 3: A batch of 500 containers for frozen orange juice contains 10 defective containers. Three are selected randomly without replacement from the batch. (a) What is the probability that the second one is defective given that the third one is defective? (b) What is the probability that the third one is defective? (c) What is the probability that the first one is defective given that the other ones are not defective? Question 4: A player of a video game is confronted with a series of 4 opponents and an 80% probability of defeating each opponent. Assume that the results are independent (and that when the player is defeated by an opponent the game ends). (a) What is the probability that a player defeats all 4 opponents in a game? (b) If the game is played 4 times, what is the probability that the player defeats all 4 opponents at most twice. Question 5: A credit card contains 16 digits. It also contains a month and year of expiration. Suppose there are one million credit card holders with unique numbers. A hacker randomly selects a 16-digit credit card number. (a) What is the probability that it doesn’t belong to a real user? (b) Suppose a hacker has a 10% chance of correctly guessing the month 2 of the expiry and randomly selects a year from 2018 to 2025. What is the probability that the hacker correctly selects the month and year of expiration (all the years are equally likely)? Question 6: An optical inspection system is to distinguish among different parts. The probability of a correct classification of any part is 98%. Suppose that 3 parts are inspected and that the classifications are independent. Let the random variable X denote the number of parts that are correctly classified. (a) Determine the probability mass function X. (b) Find P(X ≤ 2)

A loan officer compares the interest rates for 48-month fixed-rate auto loans and 48-month variable-rate auto loans. Two independent, random samples of auto loan rates are selected. A sample of eight 48-month fixed-rate auto loans and a sample of five variable-rate auto loans had the following loan rates:

Fixed(%)	Variable(%)
4.29	3.59
3.75	2.75
3.5	2.99
3.99	2.5
3.75	3
3.99
5.4
4

Answer the following questions:

Let's define μFμF as the population mean loan rate for fixed-rate auto loans and

μVμV as the population mean loan rate for variable-rate auto loans.

a. The loan officer thinks the mean loan rates for fixed-rate auto loans is at least 1% greater than the mean loan rates for the variable-rate auto loans. Set up the null and alternative hypotheses needed to determine whether this claim is correct.

H0: μFμF-μVμV (Click to select)≠0<1≠1≤ 1=0≥1>1>0≥0≤ 0=1<0

Ha: μFμF-μVμV (Click to select)≤1>1≠0=0≤0>0 ≠1<0<1≥1=1

b. What is the critical value rejection rule. (Assume unequal variances and a significance level of 0.05.Answers should be in 4 decimals.)

Critical value rule: Reject H0 if (Click to select)t<-t alphat > t alphat > t alpha/2 or t <-t alpha/2 where the critical value is .

c. What is the t test statistic and p-value? (4 decimals)

t=

p-value=

What is the value in the numerator of the test statistic?

What is the value in the denominator of the test statistic?

Which one of the following is the correct p-value calculation?(Click to select)2 x (1-t.inv(0.95,10))1-t.dist(0.430,10,true)t.dist(0.430,10,true)2 x (1-t.dist(0.430,10,true))t.inv(0.95,10)1-t.inv(0.95,10)

d. With 95% confidence we (Click to select)cancannot  conclude that the mean loan rate for fixed-rate auto-loans is at least 1% greater than the loan rate for variable-rate auto-loans.

e. We have (Click to select)extremely strongsomenovery strongstrong evidence that the mean loan rates for fixed-rate auto loans is at least 1% greater than the mean loan rates for the variable rate auto loans.

f. Calculate a 95 percent confidence interval for the difference between the mean rates for fixed-rate and variable-rate 48-month auto loans.

_____ ≤  μFμF-μVμV  ≤ _____ (4 decimals).

In the book Essentials of Marketing Research, William R. Dillon, Thomas J. Madden, and Neil H. Firtle discuss a research proposal in which a telephone company wants to determine whether the appeal of a new security system varies between homeowners and renters. Independent samples of 140 homeowners and 60 renters are randomly selected. Each respondent views a TV pilot in which a test ad for the new security system is embedded twice. Afterward, each respondent is interviewed to find out whether he or she would purchase the security system.

Results show that 25 out of the 140 homeowners definitely would buy the security system, while 9 out of the 60 renters definitely would buy the system.

a. Letting p₁ be the population proportion of homeowners who would buy the security system, and letting p₂ be the population proportion of renters who would buy the security system, set up the null and alternative hypotheses needed to determine whether the proportion of homeowners who would buy the security system differs from the proportion of renters who would buy the security system.

H0: p1−p2p1−p2(Click to select)><=≠≤≥ 0

Ha: p1−p2p1−p2(Click to select)=≠≤<>≥ 0

b. Find the test statistic z and the p-value for testing the hypotheses of part a. (α=0.05)

z= , p-value=

c. Determine the rejection rule:

    Critical Value Rule:

     Reject H₀ if (Click to select)t> t alpha/2 or t < -t alpha/2t > t alphat < t alpha/2z> z alpha/2 or z < -z alpha/2z > z alphaz < z alpha

d. What is the meaning of p-value for the hypothesis test in part a if pˆ1−pˆ2p^1−p^2 = 0.1786-0.15= 0.0286? (Hint: p-value is the probability that we will get the sample values (or more extreme) from a population where the null hypothesis holds. THIS IS A TWO-SIDED TEST.)

(Click to select)The probability that the difference in sample proportions is greater than 0.0286.The probability that the difference in sample proportions is less than 0.0286The probability that the difference in sample proportions is more extreme than 0.0286 if the difference in population proportions is equal to 0.

e. We have (Click to select)someextremely strongvery strongnostrong evidence that the proportions of homeowners and renters differ.

f. Calculate a 90 percent confidence interval for the difference between the proportions of homeowners and renters who would buy the security system.

≤ p1−p2p1−p2 ≤

g. Can we conclude at the 90% confidence that the difference between the proportions of homeowners and renters who would buy the security system is greater than 0?

(Click to select)Yes, because the whole interval is above 0. No, because 0 is within the confidence interval. Yes, because the whole interval is below 0.

h. Can we conclude at the 90% confidence that the difference between the proportions of homeowners and renters who would buy the security system is less than 0.1?

(Click to select)Yes, because the whole interval is above 0.1. Yes, because the whole interval is below 0.1.No, because 0.1 is within the confidence interval.

1. From a box containing 10 red, 30 white, 20 blue and 15 orange balls, 5 are drawn at random. Find the probability that:
a) All are white - 0.008256723
b) Be red, white or blue - 0.009169617
c) They are neither red nor blue - 0.070788

2. A company will launch three new shampoos. It is believed that the probability that the first one is successful is 0.45, that the second one has it is 0.55 and that the third one has it is 0.75. What is the probability that
a) The three succeed? - 0.185625
b) At least two succeed? -
c) None succeed?

3. You have three boxes. In the first there are 6 rabbits and 2 pigeons; in the second, 8 rabbits and 5 pigeons, and in the third 4 rabbits and 6 pigeons. What is the probability of
a) take out a rabbit?
b) that the rabbit comes from the third box?
4. An exam has 10 multiple-choice questions. There are four possible answers for each question. Luis has not studied for the exam and has decided to do it at random. Assuming that you need 70% of the correct questions to pass, calculate the probability that Luis will pass the exam.

5. On average, four imperfections are observed in each 50m stretch of network cable. Calculate the probability that, in a 25 m section of this type of network cable:
a) More than three imperfections are observed b) At least one imperfection

6. According to one study, 30% of adults suffer from insomnia. If you find a group of randomly selected adults, what is the probability that:
a) the sixth revised is the first to suffer insomnia?
b) the fifth review is the third to suffer insomnia?

7. The weight of a newborn baby in a certain country is a random variable that follows a normal distribution with an average of 3.2 kg. And standard deviation of 0.4 kg.
a) Determine the percentage of newborn babies weighing 3.5 kg or more - 0.2266
b) Calculate the conditional probability that a newborn baby weighs more than 3.5 kg, if it is known to weigh at least 3 kg - 0.3277
c) From what weight is 10% of babies born weighing more? - 3.7126 Kg
   
8. In a hospital, 22% of transplants are cornea. Use the approximation to the binomial distribution by normal to calculate the probability that, in the following 120 transplants:
a) 30 are cornea - 0.0641
b) at least 30 are from the cornea - 0.1831

9. In a company of 5000 workers, it is desirable to know if the positive assessment of management management has varied greatly, which last year was conclusively concluded that it was 80% of the workers. For this, a sample of size 200 is carried out, resulting in the positive assessment being considered by 55% of the workers surveyed. Can we affirm that the valuation has varied with the probability of being 1% wrong? - Yes it has varied

10. A company producing pasteurized milk has as a rule not to accept raw milk with a fat content exceeding 34 g / 100g. A sample of 36 liters of milk obtained from as many cows belonging to the same farm gave an average value of fat content in milk of 35.2 g / 100g with a deviation of 4.1 g / 100g. Can milk be accepted by the pasteurizer? The company admits an error level of 1%. - Yes milk is accepted.
11. Calculate Zα / 2 in the following sections

a) With a confidence level of 96% - 2.05
b) With a confidence level of 99.5% and 92% - 2.81 and 1.75
c) with a level of significance of 2% and 7% - 2.32 and 1.81
d) With α = 0.03 and α = 0.11 - 2.17 and 1.6

12. A factory produces steel cables, whose resilience follows a normal distribution of unknown mean and standard deviation σ = 10 KJ / m3. A sample of 100 pieces was taken and through a statistical study a confidence interval (898.04 - 901.96) was obtained for the average resilience of the steel cables produced in the factory.

a) Calculate the value of the average resilience of the 100 pieces of the sample. - 900
b) Calculate the level of confidence with which this interval has been obtained. - 95% confidence
c) If you would like to have an error of 3 KJ / m3 in the confidence interval, what size should the sample be? A sample of

13. It is known that the average electric power consumption in a certain province is 721 Kwh.

A technology company in the region believes that its employees consume more than the provincial average. Collect information on the consumption of 15 randomly chosen employees, and obtain the following data:

710 774 814 768 823
732 675 755 770 660
654 757 736 677 797

If the distribution of monthly electricity consumption is normal:

a) Is there evidence to state that the average household electric power consumption of the company's employees is higher than the average consumption at the provincial level? Use a significance level of 10%. - Yes there is evidence.
b) What is the p-value of the decision? - 0.0878

14. An electric company manufactures cell phone batteries that have a duration that is distributed approximately normally with an average of 800 hours and a standard deviation of 40 hours. A random sample of 30 batteries has an average duration of 785 hours.

a) Do the data show sufficient evidence to say that the average duration is less than 800? Use a significance level of 5%. - Yes there is evidence to say that it is minor.
b) What is the probability of deciding that the average is 800 hours when in reality it is 780 hours? - 0.1357 (they don't come like this on the exam, it's difficult and they have to deduce several things)

15. A cake factory manufactures, in its usual production, 3% of defective cakes. A customer receives an order of 500 cakes from the factory. Calculate the probability that ...
a) Find more than 5% of defective cakes. - 0.004375
b) Find between 1% and 3% of defective cakes. - 0.4956
c) Calculate a 95% confidence interval for the percentage of defective cakes. (0.015-0.4495)

16. It is assumed that the distribution of the temperature of the human body in the population has an average of 37ºC and a standard deviation of 0.85ºC. A sample of 105 people is chosen, calculate the following probabilities:
a) That the average is less than or equal to 36.9 ºC - 0.114
b) That the average is greater than 38.5 ºC - 0
c) Find from what temperature 10% of the hottest bodies are found. - 38.09
d) Find from what temperature 5% of the coldest bodies are found. - 35.6

Consider the following hypothesis tests. (Give your answer bounds exactly.)

(a) Calculate the p-value for H_a: σ² ≠ 22, n = 16, χ² = 28.2.
_______< p <_______

(b) Calculate the p-value for H_a: σ² > 28, n = 26, χ² = 32.6.
_______< p <________

(c) Calculate the p-value for H_a: σ² ≠ 44, df = 30, χ² = 39.
_______< p < _______

(d) Calculate the p-value for H_a: σ² < 13, df = 40, χ² = 26.5.
_______< p < ________

A headline read, “More than half of Americans Say Federal Taxes Too High.” The headline was based on a random sample of 1026 adult Americans in which 534 stated the amount of federal tax they have to pay is too high.

a. State the null and alternative hypotheses.

b. Check whether np0(1-p0) is greater or equal to 10.

c. Manually calculate the test statistic.

d. Find p-value.

(Question: Is the combined scores data approximately normal? Yes or no?)

Data:

Race	Position	Oral exam results	Written exam results	Combined results
B	Lieutenant	45.83	46	45.932
B	Lieutenant	52.92	49	50.568
B	Captain	54.76	49	51.304
H	Lieutenant	48.33	58	54.132
B	Lieutenant	52.08	56	54.432
H	Lieutenant	40.83	64	54.732
B	Captain	60	53	55.8
W	Captain	53.81	58	56.324
B	Lieutenant	58.75	55	56.5
W	Lieutenant	54.58	58	56.632
B	Lieutenant	55.83	58	57.132
W	Lieutenant	45.42	65	57.168
H	Lieutenant	44.17	66	57.268
H	Captain	42.86	67	57.344
W	Lieutenant	44.58	66	57.432
W	Lieutenant	51.25	62	57.7
W	Lieutenant	57.92	58	57.968
B	Captain	70.48	50	58.192
B	Lieutenant	60.83	58	59.132
H	Lieutenant	46.25	68	59.3
H	Captain	57.14	61	59.456
W	Lieutenant	60	60	60
W	Lieutenant	49.58	67	60.032
B	Lieutenant	51.25	66	60.1
B	Lieutenant	55	64	60.4
B	Captain	67.62	56	60.648
H	Lieutenant	51.25	67	60.7
H	Lieutenant	42.5	73	60.8
W	Captain	48.57	69	60.828
B	Lieutenant	56.25	64	60.9
B	Lieutenant	56.67	64	61.068
W	Lieutenant	56.25	66	62.1
H	Captain	58.57	65	62.428
H	Lieutenant	50.42	71	62.768
W	Lieutenant	71.67	57	62.868
W	Captain	55.24	68	62.896
H	Lieutenant	51.67	71	63.268
B	Lieutenant	69.17	60	63.668
H	Lieutenant	57.5	70	65
W	Lieutenant	50.42	75	65.168
B	Lieutenant	66.25	65	65.5
W	Lieutenant	55	73	65.8
H	Captain	67.14	65	65.856
H	Lieutenant	56.25	73	66.3
W	Lieutenant	79.17	59	67.068
B	Captain	52.38	77	67.152
W	Lieutenant	64.58	69	67.232
B	Lieutenant	70.83	65	67.332
W	Captain	57.14	75	67.856
W	Lieutenant	51.67	79	68.068
H	Lieutenant	62.5	72	68.2
W	Lieutenant	51.25	80	68.5
W	Lieutenant	73.75	66	69.1
W	Lieutenant	58.75	76	69.1
H	Lieutenant	70.83	68	69.132
H	Captain	60.48	75	69.192
W	Captain	59.05	76	69.22
W	Captain	71.43	68	69.372
W	Captain	71.43	68	69.372
W	Lieutenant	64.58	73	69.632
W	Captain	78.57	64	69.828
W	Captain	62.38	75	69.952
W	Lieutenant	71.67	69	70.068
W	Lieutenant	73.75	68	70.3
B	Captain	70.95	70	70.38
W	Lieutenant	77.5	66	70.6
H	Lieutenant	53.75	82	70.7
W	Lieutenant	65.83	74	70.732
W	Lieutenant	58.33	79	70.732
H	Lieutenant	69.17	72	70.868
H	Lieutenant	55	82	71.2
W	Captain	56.67	81	71.268
W	Captain	82.38	64	71.352
B	Captain	68.57	74	71.828
W	Lieutenant	73.33	71	71.932
W	Lieutenant	72.5	72	72.2
B	Lieutenant	92.08	59	72.232
W	Lieutenant	66.67	76	72.268
B	Lieutenant	70.83	74	72.732
W	Lieutenant	78.33	70	73.332
W	Captain	70	76	73.6
W	Captain	73.33	74	73.732
B	Lieutenant	65.83	80	74.332
W	Lieutenant	87.5	66	74.6
B	Captain	82.38	70	74.952
W	Captain	76.67	74	75.068
W	Lieutenant	63.33	83	75.132
W	Captain	73.81	77	75.724
W	Lieutenant	74.17	77	75.868
B	Lieutenant	80.42	73	75.968
H	Captain	79.05	74	76.02
B	Lieutenant	61.25	86	76.1
W	Captain	80	74	76.4
W	Captain	87.62	69	76.448
B	Lieutenant	77.5	76	76.6
W	Lieutenant	58.75	89	76.9
W	Captain	84.29	72	76.916
W	Lieutenant	73.75	81	78.1
W	Captain	73.81	81	78.124
H	Captain	70	84	78.4
W	Lieutenant	69.58	86	79.432
W	Captain	76.19	82	79.676
H	Captain	76.19	82	79.676
W	Captain	76.19	84	80.876
W	Captain	88.57	76	81.028
W	Lieutenant	68.33	91	81.932
W	Lieutenant	63.75	95	82.5
W	Lieutenant	80.83	84	82.732
W	Lieutenant	73.33	89	82.732
W	Lieutenant	73.75	91	84.1
W	Lieutenant	80	87	84.2
W	Lieutenant	85	84	84.4
W	Captain	82.38	87	85.152
W	Lieutenant	77.5	91	85.6
W	Lieutenant	87.5	87	87.2
W	Captain	80	95	89
W	Lieutenant	88.75	91	90.1
W	Captain	89.52	95	92.808

A statistician would like to construct a hypothesis test of the mean using the P-value approach. Rank the steps that the statistician can follow to carry out this hypothesis test at a level of significance of α. Note that there may be more than one correct answer. Step Rank Collect sample data Define the null and alternative hypotheses Draw a conclusion Calculate the P-value Define the distribution of the test statistic Calculate the test statistic