Questions
Year Quarter SPeClty 2006 3 4480.77 4 5827.48 2007 1 5602.84 2 5269.74 3 4945.67 4...

Year Quarter SPeClty
2006 3 4480.77
4 5827.48
2007 1 5602.84
2 5269.74
3 4945.67
4 5124.22
2008 1 4996.31
2 5147.61
3 4903.32
4 4960.84
2009 1 4864.16
2 4785.21
3 4684.43
4 4732.58
2010 1 4947.63
2 4826.21
3 4801.46
4 4789.05
2011 1 4812.69
2 4787.65
3 4871.31
4 4756.61
2012 1 4583.15
2 4659.09
3 4587.55
4 4746.01
2013 1 4556.63
2 4632.57
3 4561.03
4 4719.49
2014 1 4516.85
2 4592.78
3 4521.24
4 4844.6
2015 1 4588.54
2 4571.84
3 4671.24
4 5011.6
2016 1 4821.54
2 4916.84
3 4791.31
4 5183.49
2017 1 5045.38
2 5138.97
3 4915.37
4 5369.34
2018 1 5379.83
2 5481
3 5265.31
4 5579.14
2019 1 5743.98
2 5750.01
3 5542.86


a) in a paragraph write about Trend and Seasonality of this store.

b) write about Regression model(s) and examine hypotheses

c) what is the least square regression line?

d) write about standard error and interpret it in a short paragraph.

* please just do it with excel and explain how you are solving it (preferred to use excel screenshots)

In: Statistics and Probability

Find the critical value, tc, for c=0.95 and n=41.

Find the critical value, tc, for c=0.95 and n=41.

In: Statistics and Probability

General Social Survey asked participants if they thought it was “OK” for a woman to get...

General Social Survey asked participants if they thought it was “OK” for a woman to get an abortion for any reason and also asked them for their political party affiliation. The table below summarizes the data.

Yes, “Ok”

No, Opposed

Strong Democrat

145

123

Independent

128

282

Strong Republican

46

162

Consider the following two events:

A = the event a person is a Strong Republican,

B = the event a person is ok to a woman getting an abortion for any reason.

Based on these data, select the correct answer about these two events.

Question 1 options:

Two events are disjoint because there are no people belong to both of these events

Two events are disjoint because there are 46 people belong to both of these events

Two events are not disjoint because there are no people belong to both of these events

Two events are not disjoint because there are 46 people belong to both of these events

Question 2 (0.5 points)

A General Social Survey asked participants if they thought it was “OK” for a woman to get an abortion for any reason and also asked them for their political party affiliation. The table below summarizes the data.

Yes, “Ok”

No, Opposed

Strong Democrat

145

123

Independent

128

282

Strong Republican

46

162

Select the correct answer for the probability that a randomly selected person is a Strong Republican and opposed to a woman getting an abortion for any reason.

Question 2 options:

208/567

162/886

208/886

162/567

Question 3 (0.5 points)

Saved

A General Social Survey asked participants if they thought it was “OK” for a woman to get an abortion for any reason and also asked them for their political party affiliation. The table below summarizes the data.

Yes, “Ok”

No, Opposed

Strong Democrat

145

123

Independent

128

282

Strong Republican

46

162

Select the correct answer for the probability that a randomly selected person is a Strong Democrat who is also opposed to a woman getting an abortion for any reason.

Question 3 options:

123/268

123/886

268/567

Question 4 (0.5 points)

A General Social Survey asked participants if they thought it was “OK” for a woman to get an abortion for any reason and also asked them for their political party affiliation. The table below summarizes the data.

Yes, “Ok”

No, Opposed

Strong Democrat

145

123

Independent

128

282

Strong Republican

46

162

Select the correct answer for the probability that a randomly selected person is a Strong Democrat or opposed to a woman getting an abortion for any reason.

Question 4 options:

(268/886)+(567/886)+(123/886) = 958/886

(268/886)+(567/886) = 835/886

(268/886)+(567/886)-(123/886) = 712/886

Question 5 (0.5 points)

A General Social Survey asked participants if they thought it was “OK” for a woman to get an abortion for any reason and also asked them for their political party affiliation. The table below summarizes the data.

Yes, “Ok”

No, Opposed

Strong Democrat

145

123

Independent

128

282

Strong Republican

46

162

If a randomly selected person supports a woman getting an abortion for any reason, find the probability that this person is a Strong Republican?

Select the correct answer.

Question 5 options:

208/886

46/162

46/319

46/886

Question 6 (0.5 points)

A General Social Survey asked participants if they thought it was “OK” for a woman to get an abortion for any reason and also asked them for their political party affiliation. The table below summarizes the data.

Yes, “Ok”

No, Opposed

Strong Democrat

145

123

Independent

128

282

Strong Republican

46

162

Consider the following two events:

A = the event a person is a Strong Democrat,

B = the event a person is ok to a woman getting an abortion for any reason.

Based on these data, decide whether these two events are independent according to probability.

Question 6 options:

Two events are independent because P(A)xP(B) = P(A and B) is true

Two events are not independent because P(A)xP(B) = P(A and B) is true

Two events are independent because P(A)xP(B) = P(A and B) is not true

Two events are not independent because P(A)xP(B) = P(A and B) is not true

In: Statistics and Probability

A customer service operator receives, on average, 6 calls per week asking for assistance. a. What...

A customer service operator receives, on average, 6 calls per week asking for assistance.
a. What is the type of distribution? Why?
b. For any randomly selected week, find the probability that the operator will receive
i) at least 6 calls
ii) 4 or more calls
iii) At most 5 calls.

In: Statistics and Probability

Given two independent random samples with the following results: n1=13x‾1=180s1=21   n2=9x‾2=163s2=30 Use this data to find...

Given two independent random samples with the following results: n1=13x‾1=180s1=21   n2=9x‾2=163s2=30 Use this data to find the 99% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 3: Find the point estimate that should be used in constructing the confidence interval.

Step 2 of 3: Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Step 3 of 3: Construct the 99%99% confidence interval. Round your answers to the nearest whole number.

In: Statistics and Probability

Answer the following questions: 1. In the following scenarios, identify whether or not it is uniform,...

Answer the following questions:

1. In the following scenarios, identify whether or not it is uniform, binomial, or hypogeometric probability distribution. Briefly explain your choice.
[Do not solve the problem]

a) A spinner is split into 7 equal sections. Each section is coloured differently. One of the sections is coloured purple. What is the probability that the spinner lands on purple

b) If you roll a die 14 times. What is the probability that you will roll an even number 4 times?

2. a) What is the key characteristic that differentiates a binomial distribution from a hypergeometric distribution?

b) Using cards as the scenario, create a question (that doesn’t need to be answered) that illustrate a binomial distribution

c) Using cards as the scenario, create a question (that doesn’t need to be answered) that illustrate a hypergeometric distribution    

In: Statistics and Probability

A psychology quiz consists of ten multiple choice questions (each question is of worth 1 mark)...

A psychology quiz consists of ten multiple choice questions (each question is of worth 1 mark) with 5 options per question. A student knows the correct answer to 5 of the questions, can rule out 1 option in 3 questions, can rule out 2 options in 1 question, can not rule out any options in 1 question. What are the expected marks that the student would get?

In: Statistics and Probability

Suppose the mean and the standard deviation of the waiting times of passengers at the bus...

Suppose the mean and the standard deviation of the waiting times of passengers at the bus station near the Cross Harbour Tunnel are 11.5 minutes and 2.2 minutes, respectively.

(a) Assume the waiting times are normally distributed. How likely a randomly selected passenger waits less than 15 minutes at the bus station?

(b) Assume the waiting times are normally distributed. 82% of the passengers at the bus station are expected to wait more than k minutes. Find k.

(c) For a random sample of 36 passengers, find the probability that their mean waiting time will be less than 11 minutes. Does your calculation require assuming the waiting times to be normally distributed? Explain.

In: Statistics and Probability

Given that Z is a standard normal random variable, find z for each situation using the...

Given that Z is a standard normal random variable, find z for each situation using the Norm.S.Inv function in excel.

a. The area to the right of z is .01

b. The area between -z and z is .9030

c. The area between -z and z is .9948

e. The area to the right of z is .025

f. The area to the right of z is .3300

In: Statistics and Probability

A dentist wants to find the average time taken by one of her hygienists to take...

A dentist wants to find the average time taken by one of her hygienists to take X-ray and clean the tooth for her patients. She recorded the time to serve 24 randomly selected patients by the hygienist. The data of this experiment are save in the column “Time” of the SPSS file Ass_2Q1.sav.

Time

38

33

38

40

37

37

37

39

36

33

36

38

34

38

36

36

34

33

32

35

33

35

35

32

  1. Compute the appropriate point estimate for population (true) mean time taken by this hygienist.
  2. Construct a 98% confidence interval for the true average time taken by this hygienist.
  3. The standard average for all the hygienists for this type of job is 34 minutes. Do the sample data provides evidence that the average time taken by the hygienist is longer than standard average for this job type? Use 1% significance level.                 

In: Statistics and Probability

Health researchers believe that the neonatal mortality rate is higher at home (MRH) than that in...

Health researchers believe that the neonatal mortality rate is higher at home (MRH) than that in the health centers (MRC). The neonatal mortality rates (both MRH and MRC) for 20 countries are recorded in SSPS file Ass_2Q2.sav.

  1. Do the data support researchers’ beliefs? Use α=0.05.
  2. Find a 95% confidence interval to investigate the researchers’ belief and explain its meaning.
  3. Check the assumptions of the above procedure.

Neonatal -mortality

group

9

3

4

7

6

6

3

7

10

6

7

8

9

6

9

3

9

4

7

6

2

0

8

5

4

3

0

2

1

4

4

6

3

1

2

2

3

3

1

4

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

           

In: Statistics and Probability

Your friend owns a retail store where she plans to give arriving customers a short verbal...

Your friend owns a retail store where she plans to give arriving customers a short verbal ”commercial” message regarding products for sale. Your friend asks you to determine if this will boost the average dollar value of sales to each customer, µ, which is currently $5.00. You may assume that the standard deviation per transaction is $1.00 and that the population is of unlimited size. You monitor a random sample of n = 200 customers given the short verbal ”commercial” message upon arrival at the store and determine the amount purchased by each. (a) Your friend will only adopt the policy of giving short verbal ”commercial” messages to each newly arriving customer if the sample indicates that the values of sales per customer will improve. Formulate the null and alternative hypotheses. [2 marks] (b) Suppose you wish to test the null hypothesis while ensuring that your friend only has a 5% chance of getting a recommendation to adopt the policy of giving short verbal ”commercial” messages when that policy actually does not improve the value of sales per customer. Describe the test statistic you would use to conduct the test, compute the critical region for the test, and describe the decision rule you would use. [4 marks] (c) You collect the sample data. What recommendation would you make to your friend if (i) X¯=$5.37, (ii) X¯=$5.05, (iii) X¯=$4.97, and (iv) X¯=$5.20 ? Express your recommendations as formal hypothesis testing conclusions. [4 marks]

In: Statistics and Probability

Some students do homework with the tv on. some researchers want to see if people can...

Some students do homework with the tv on. some researchers want to see if people can work as effectively with as without distraction. the researchers will time some volunteers to see how long it takes them to complete some relatively easy crossword puzzles with and without tv on.

a) design an experiment that will require a two sample independent test

b) design experiment that will require dependent test

In: Statistics and Probability

Births are approximately Uniformly distributed between the 52 weeks of the year. They can be said...

Births are approximately Uniformly distributed between the 52 weeks of the year. They can be said to follow a Uniform distribution from 1 to 53 (a spread of 52 weeks). Round answers to 4 decimal places when possible.

  1. The mean of this distribution is
  2. The standard deviation is
  3. The probability that a person will be born at the exact moment that week 52 begins is P(x = 52) =
  4. The probability that a person will be born between weeks 20 and 34 is P(20<x<34)P(20<x<34) =
  5. The probability that a person will be born after week 14 is P(x > 14) =
  6. P(x > 6 | x < 19) =
  7. Find the 23rd percentile.
  8. Find the minimum for the upper quartile.

In: Statistics and Probability

For an adult male the average weight is 179 pounds and the standard deviation is 29.4...

For an adult male the average weight is 179 pounds and the standard deviation is 29.4 pounds. We can assume that the distribution is Normal (Gaussian). Answer the following questions either via simulations (use 10000 points) or via “rule of thumbs”. I). What is the approximate probability that a randomly picked adult male will weigh more than 180 pounds? Pick the closest answer. (6.66 points) a. About 15% b. About 30% c. About 50% d. About 65% II) What would be the range [A to B], which would contain about 68% of the adult males? Pick the closest answer. (6.66 points) a. Between 149.6 and 208.4 pounds b. Between 155.5 and 211.4 pounds c. Between 120.2 and 237.8 pounds d. Between 130.1 and 233.2 pounds III) A random adult male is chosen, which of these outcomes is the least likely? (6.66 points) a. The person will weigh more than 210 pounds b. The person will weigh more than 225 pounds c. The person will weigh less than 150 d. The person will weigh between 170 and 200 pounds

In: Statistics and Probability