X ~ ( Mean, 2.15) ( Suppose we do not have information of Mean of the population) Question 1----------------------------------------------------------------------------- Use calculator go to math---> PRB-------> random sample 1. * Pick one (only-one) sample size of 5 (As we did in lab 2) ** calculate Mean of this sample and use the S.D. = 2.15/ sqrt(5) *** Calculate EBM **** Find C.I. for mean at 95% C.L. 2. Repeat the same process for n=10 and n = 20 3. Study what happened to Confidence Interval when we increased the sample size. Question 2------------------------------------------------------------------------------------------------- 4. * Find C.I. For n=20 for C.L. 90% ** Study what happened to Confidence Interval when we increased and C.L. from 90% to 95% Question 3---------------------------------------------------------------------------------------- From the previous lab we know mean for recipe data was 3.57. Write a comment about what you notice in Question 1 and Question 2

Are low-fat diets or low-carb diets more effective for weight loss? A sample of 70 subjects went on a low-carb At the end of that time, the sample mean weight loss was 10.5 pounds with a sample standard deviation of 7.09 pounds. A second sample of 76 went on a low-fat diet. Their sample mean weight loss was 18.0 with a standard deviation of 7.26. Can you conclude that their was no difference in the mean weight loss between the two diets? Use a = .05 level.

Be sure to show all 5 steps of the Hypothesis test.

Evaluate using a P-Value hypothesis test, Explain why you chose your decision.

(1) Many people consider their smart phone to be essential! Communication, news, Internet, entertainment, photos, and just keeping current are all conveniently possible with a smart phone. However, the battery better be charged or the phone is useless. Battery life of course depends on the frequency, duration, and type of use. One study involving heavy use of the phones showed the mean of the battery life to be 11.75 hours with a standard deviation of 3.4 hours. Then the battery needs to be recharged. Assume the battery life between charges is normally distributed. (a) Find the probability that with heavy use, the battery life exceeds 12 hours. (Round your answer to four decimal places.) (b) You are planning your recharging schedule so that the probability your phone will die is no more than 5%. After how many hours should you plan to recharge your phone? (Round your answer to the nearest tenth of an hour.) -------hours (2) The University of Montana ski team has six entrants in a men's downhill ski event. The coach would like the first, second, and third places to go to the team members. In how many ways can the six team entrants achieve first, second, and third places? (3) Find z such that 15% of the area under the standard normal curve lies to the right of z. (Round your answer to two decimal places.) Incorrect: Your answer is incorrect.

- Assume that the variable Z is distributed as a standardized normal with a mean of 0 and a standard deviation of 1. Calculate the following:

a. Pr (Z > 1.06)

b. Pr (Z < - 0.29)

c. Pr (-1.96 < Z < -0.29)

d. The value of Z for which only 8.08% of all possible values of Z are larger than.

- Assume that the variable X is distributed as normal with μ= 51 and σ = 4. Calculate the following:

a. Pr (X = 51)

b. Pr (X > 46)

c. Pr (X < 43)

d. Pr ( 43 < X < 46)

e. Two values of X, symmetric around the mean, between which 70% of all the possible values of X lie.

7. The following data represent the number of workdays absent during the past year, y, and the number of years employed by the company x, for seven employees randomly selected from a large company. Assume data is normally distributed.

Y 2 0 5 6 4 9 2

X 7 8 2 3 5 3 7

The slope estimate (b1) was found to be = -1.09 That is:

b 1= nExy - (Ex) (EY)

---------------------------- = - 1.09

  nEx^{2 - (EX)2}

a) Using appropriate statistical techniques and the above-added information, find the least squares estimate of the regression equation. i. e. What is the linear regression equation?

b) Using the proper (hypothesis) statistical test, support or refute the assumption above, i.e. there is a linear relationship between years employed and absenteeism. [

Note: S(b) = 0.28425

The owner of Showtime Movie Theaters, Inc., would like to predict weekly gross revenue as a function of advertising expenditures. Historical data for a sample of eight weeks follow.

(a)

Develop an estimated regression equation with the amount of television advertising as the independent variable. (Round your numerical values to two decimal places. Let x₁ represent the amount of television advertising in $1,000s and y represent the weekly gross revenue in $1,000s.)

ŷ =

(b)

Develop an estimated regression equation with both television advertising and newspaper advertising as the independent variables. (Round your numerical values to two decimal places. Let x₁ represent the amount of television advertising in $1,000s, x₂ represent the amount of newspaper advertising in $1,000s, and y represent the weekly gross revenue in $1,000s.)

ŷ =

(c)

Is the estimated regression equation coefficient for television advertising expenditures the same in part (a) and in part (b)?

---Select--- Yes No , it is  in part (a) and  in part (b).

Interpret the coefficient in each case.

In part (a) it represents the change in revenue due to a one-unit increase in television advertising expenditure. In part (b) it represents the change in revenue due to a one-unit increase in television advertising with newspaper advertising held constant.In part (a) it represents the change in revenue due to a one-unit increase in television advertising expenditure with newspaper advertising held constant. In part (b) it represents the change in revenue due to a one-unit increase in newspaper advertising with television advertising held constant.    In part (a) it represents the change in revenue due to a one-unit increase in newspaper advertising expenditure with television advertising held constant. In part (b) it represents the change in revenue due to a one-unit increase in television advertising with newspaper advertising held constant.In part (a) it represents the change in revenue due to a one-unit increase in television advertising expenditure. In part (b) it represents the change in revenue due to a one-unit increase in newspaper advertising with television advertising held constant.In part (a) it represents the change in revenue due to a one-unit increase in television advertising with newspaper advertising held constant. In part (b) it represents the change in revenue due to a one-unit increase in television advertising expenditure.

(d)

Predict weekly gross revenue (in dollars) for a week when $3,500 is spent on television advertising and $1,800 is spent on newspaper advertising. (Round your answer to the nearest cent.)

$

Directions Use the Chi-Square option in the Nonparametric Tests menu to answer the questions based on the following scenario. (Assume a level of significance of .05 and use information from the scenario to determine the expected frequencies for each category).

Scenario

During the analysis of the district data, it was determined that one high school had substantially higher Graduate Exit Exam scores than the state average and the averages of high schools in the surrounding districts. To better understand possible reasons for this difference, the superintendent conducted several analyses. One analysis examined the population of students who completed the exam. Specifically, the superintendent wanted to know if the distribution of special education, regular education, and gifted/talented test takers from the local high school differed from the statewide distribution. The obtained data are provided below.

Description

Number of students from local high school who took the graduate exam.

Special Education (10) Regular Education (104) Gifted/Talented (26)

Percent of test-taking students state wide who took the graduate exam

Special Education (7) Regular Education (77) Gifted/Talented (16)

1. If the student distribution for the local high school did not differ from the state, what would be the expected percentage of students in each category?

2. What were the actual percentages of local high school students in each category? (Report final answer to two decimal places)

3. State an appropriate null hypothesis for this analysis.

4. What is the value of the chi-square statistic?

5. What are the reported degrees of freedom?

6. What is the reported level of significance?

7. Based on the results of the one-sample chi-square test, was the population of test taking students at the local high school statistically significantly different from the statewide population?

8. Present the results as they might appear in an article. This must include a table and narrative statement that reports and interprets the results of the analysis.

A genetic experiment with peas resulted in one sample of offspring that consisted of 441green peas and 166 yellow peas.

a. Construct a 95​% confidence interval to estimate of the percentage of yellow peas.

__ <P< __

b. It was expected that​ 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations?

Construct a scattergram for each data set. Then calculate r and r2 for each data set. Interpret their values. Complete parts a through d.

Calculate r.

r=. 9853.​(Round to four decimal places as​ needed.)

Calculate r2.

r2=0.9709​(Round to four decimal places as​ needed.)

Interpret r. Choose the correct answer below.

A.There is not enough information to answer this question.

B.There is a very strong negative linear relationship between x and y.

C.x and y are not related.

D.There is a very strong positive linear relationship between x and y.  answer is correct.

Interpret r2.

97.09​% of the total sample variability around overbar y is explained by the linear relationship between x and y.

​(Round to two decimal places as​ needed.)

Calculate r.

r=. 9853 ​(Round to four decimal places as​ needed.)

Calculate r2.

r2=0.9709 ​(Round to four decimal places as​ needed.)

Interpret r. Choose the correct answer below.

A.There is not enough information to answer this question.

B.There is a very strong negative linear relationship between x and y.

C.x and y are not related.

D.There is a very strong positive linear relationship between x and y. Your answer is correct.

Interpret r2.

97.09% of the total sample variability around ove rbar y is explained by the linear relationship between x and y. ​(Round to two decimal places as​ needed.)

Calculate r.

requals=negative 0.9934−0.9934

​(Round to four decimal places as​ needed.)

Calculate

r squaredr2.

r squaredr2equals=. 9868.9868

​(Round to four decimal places as​ needed.)

Interpret r. Choose the correct answer below.

A.

There is a very strong negative linear relationship between x and y.

Your answer is correct.

B.

There is not enough information to answer this question.

C.

There is a very strong positive linear relationship between x and y.

D.

x and y are not related.

Interpret

r squaredr2.

98.6898.68​%

of the total sample variability around

y overbary

is explained

by the linear relationship between x and y.

​(Round to two decimal places as​ needed.)

Calculate r.

requals=negative 0.9934−0.9934

​(Round to four decimal places as​ needed.)

Calculate

r squaredr2.

r squaredr2equals=. 9868.9868

​(Round to four decimal places as​ needed.)

Interpret r. Choose the correct answer below.

A.There is a very strong negative linear relationship between x and y.  answer is correct.

B. There is not enough information to answer this question.

C There is a very strong positive linear relationship between x and y.

D.x and y are not related.

Interpret r2.

98.68% of the total sample variability around over bar y is explained by the linear relationship between x and y.​(Round to two decimal places as​ needed.)

Calculate r.

r=____________​(Round to four decimal places as​ needed.)

the accompanying data represent the miles per gallon of a random sample of cars with a​ three-cylinder, 1.0 liter engine. ​(a) compute the​ z-score corresponding to the individual who obtained 38.7 miles per gallon. interpret this result. ​(b) determine the quartiles. ​(c) compute and interpret the interquartile​ range, iqr. ​(d) determine the lower and upper fences. are there any​ outliers?39.939.9 42.442.4 34.634.6 36.336.3 38.138.1 38.938.9 40.540.5 42.842.8 34.734.7 37.537.5 38.338.3 39.439.4 41.441.4 43.643.6 35.235.2 37.637.6 38.538.5 39.739.7 41.641.6 49.049.0

9.16 A bottled water distributor wants to determine whether the mean amount of water contained in 1-gallon bottles purchased from a nationally known water bottling company is actually 1

gallon. You know from the water bottling company specifications that the standard deviation of the amount of water per bottle is 0.02 gallon. You select a random sample of 50 bottles, and

the mean amount of water per 1-gallon bottle is 0.995 gallon.

A) Is there evidence that the mean amount is different from 1.0 gallon? (Use α=0.01.)

B) Compute the p-value and interpret its meaning.

C) Construct a 99% confidence interval estimate of the population mean amount of water per bottle.

D) Compare the results of (a) and (c). What conclusions do you reach?

To study how social media may influence the products consumers​ buy, researchers collected the opening weekend box office revenue​ (in millions of​ dollars) for 23 recent movies and the social media message rate​(average number of messages referring to the movie per​ hour). The data are available below. Conduct a complete simple linear regression analysis of the relationship between revenue​ (y) and message rate​ (x).

The least squares regression equation is y=−0.031+l0.086x. ​(Round to three decimal places as​ needed.)

Check the usefulness of the hypothesized model. What are the hypotheses to​ test?

A.H0​: β1≠0 against Ha​:β1=0

B.H0​β0:=0 againstHa​:β0≠0

C.H0​:β0≠0 against Ha​:β0=0

D.H0​:β1=0 againstHa​:β1≠0 Your answer is correct.

Determine the estimate of the standard deviation.

s=9.59 ​(Round to two decimal places as​ needed.)

What is the test statistic for the​ hypotheses?

t=15.14 ​(Round to two decimal places as​ needed.)

What is the​ p-value for the test​ statistic?

​p-value=0​(Round to three decimal places as​ needed.)State the conclusion at α=0.05.

Since the​ p-value is less than α​, there is sufficient evidence to reject H0.

Conclude there is

a linear relationship between revenue and message rate.What is the value for the coefficient of determination

r2​?

r2=0.92 ​(Round to two decimal places as​ needed.)Interpret the value of r2 in the context of this problem.

A.r2 is the proportion of the total sample variability around message rate that is explained by the linear relationship between the revenue and the message rate.

B.r2 is the proportion of the sample points that do not fit within the​ 95% confidence interval.

C. r2 is the proportion of the sample points that fit on the estimated linear regression line.

D.r2 is the proportion of the total sample variability around mean revenue that is explained by the linear relationship between the revenue and the message rate.

1. A. An union representing the Bottle Blowers of America (BBA) is considering a proposal to merge with the Teamsters Union. According to BBA bylaws, at least three-fourths of the union membership must approve any merger. A random sample of 2,000 current BBA members reveals 1,600 plan to vote for the merger proposal. Develop a 95% confidence interval for the population proportion. Show point estimate, standard error and margin of error and finally your confidence interval.

Basing your decision on this sample proportion, can you conclude that the necessary         proportion of BBA members favor the merger? Why?                                                       

      B. The estimate of the population proportion is to be within plus or minus 0.10, with a 99% confidence coefficient level. The best estimate of the population proportion is given as 45%. How large a sample is required under these specifications?                                          

. At a large factory, the employees were surveyed and classified according to their level of education and whether they smoked. The data are shown in the table below:

If an employee is selected at random, find these probabilities.

a. The employee smokes, given that he or she graduated from high school.

b. Given that the employee did not graduate from high school, he or she is not a smoker.

c. The employee does not smoke and has a graduate level degree (MS or PhD) or is a college graduate.

The accompanying data represent the total compensation for 12 randomly selected chief executive officers​ (CEOs) and the​ company's stock performance.

Company   Compensation   Return
A   14.98   74.48
B   4.61   63.62
C   6.15   148.21
D   1.11   30.35
E   1.54   11.94
F   3.28   29.09
G   11.06   0.64
H   7.77   64.16
I   8.23   50.41
J   4.47   53.19
K   21.39   21.94
L   5.23   33.68

​(a) Treating compensation as the explanatory​ variable, x, use technology to determine the estimates of β0 and β1.

The estimate of β1 is −0.217.

​(Round to three decimal places as​ needed.)

The estimate of beta 0β0 is 50.1

​(Round to one decimal place as​ needed.)

​(b) Assuming that the residuals are normally distributed​, test whether a linear relation exists between compensation and stock return at the α=0.01level of significance.

What are the null and alternative​ hypotheses?

B.

H0​: β1=0

H1​: β1≠0

Your answer is correct.

Compute the test statistic using technology.

__?__

​(Round to two decimal places as​ needed.)