On the basis of her newly developed technique a student believes she can reduce the amount of time schizophrenics spend in an institution. As director of training at a nearby institution, you agree to let her try her method on 20 schizophrenics, randomly sampled from your institution. The mean duration that schizophrenics stay at your institution is 85 weeks and the scores are normally distributed. The results of the experiment show that the patients treated by the student's new method stay a mean duration of 78 weeks with a standard deviation of 15 weeks.
MULTIPLE CHOICE!!!

The H₁ states that                             [ Select ]                       ["the population mean is less then 78", "the population is greater than or equal to 78", "the population mean is greater than or equal to 85", "the population mean is less than 85", "directional hypothesis", "non-directional hypothesis", "z-test", "t-test", "one tailed", "two tailed", "-1.645", "-1.729", "3.354", "-2.09", "reject the null hypothesis", "fail to reject the null hypothesis", "her new method may work", "her new method probably does not work"]         

The H₀ states that                             [ Select ]                       ["the population mean is less then 78", "the population is greater than or equal to 78", "the population mean is greater than or equal to 85", "the population mean is less than 85", "directional hypothesis", "non-directional hypothesis", "z-test", "t-test", "one tailed", "two tailed", "-1.645", "-1.729", "3.354", "-2.09", "reject the null hypothesis", "fail to reject the null hypothesis", "her new method may work", "her new method probably does not work"]      

This would be an example of a                             [ Select ]                       ["the population mean is less then 78", "the population is greater than or equal to 78", "the population mean is greater than or equal to 85", "the population mean is less than 85", "directional hypothesis", "non-directional hypothesis", "z-test", "t-test", "one tailed", "two tailed", "-1.645", "-1.729", "3.354", "-2.09", "reject the null hypothesis", "fail to reject the null hypothesis", "her new method may work", "her new method probably does not work"]         

The best statistical test to use is                            [ Select ]                       ["the population mean is less then 78", "the population is greater than or equal to 78", "the population mean is greater than or equal to 85", "the population mean is less than 85", "directional hypothesis", "non-directional hypothesis", "z-test", "t-test", "one tailed", "two tailed", "-1.645", "-1.729", "3.354", "-2.09", "reject the null hypothesis", "fail to reject the null hypothesis", "her new method may work", "her new method probably does not work"]         

It would best to use a                            [ Select ]                       ["the population mean is less then 78", "the population is greater than or equal to 78", "the population mean is greater than or equal to 85", "the population mean is less than 85", "directional hypothesis", "non-directional hypothesis", "z-test", "t-test", "one tailed", "two tailed", "-1.645", "-1.729", "3.354", "-2.09", "reject the null hypothesis", "fail to reject the null hypothesis", "her new method may work", "her new method probably does not work"]         probability

The critical value at an alpha of .05 would be                            [ Select ]                       ["the population mean is less then 78", "the population is greater than or equal to 78", "the population mean is greater than or equal to 85", "the population mean is less than 85", "directional hypothesis", "non-directional hypothesis", "z-test", "t-test", "one tailed", "two tailed", "-1.645", "-1.729", "3.354", "-2.09", "reject the null hypothesis", "fail to reject the null hypothesis", "her new method may work", "her new method probably does not work"]         

Value of the obtained statistic is                            [ Select ]                       ["the population mean is less then 78", "the population is greater than or equal to 78", "the population mean is greater than or equal to 85", "the population mean is less than 85", "directional hypothesis", "non-directional hypothesis", "z-test", "t-test", "one tailed", "two tailed", "-1.645", "-1.729", "3.354", "-2.09", "reject the null hypothesis", "fail to reject the null hypothesis", "her new method may work", "her new method probably does not work"]         

My decision based on the statistical results                            [ Select ]                       ["the population mean is less then 78", "the population is greater than or equal to 78", "the population mean is greater than or equal to 85", "the population mean is less than 85", "directional hypothesis", "non-directional hypothesis", "z-test", "t-test", "one tailed", "two tailed", "-1.645", "-1.729", "3.354", "-2.09", "reject the null hypothesis", "fail to reject the null hypothesis", "her new method may work", "her new method probably does not work"]         

Based on that decision, I conclude that                            [ Select ]                       ["the population mean is less then 78", "the population is greater than or equal to 78", "the population mean is greater than or equal to 85", "the population mean is less than 85", "directional hypothesis", "non-directional hypothesis", "z-test", "t-test", "one tailed", "two tailed", "-1.645", "-1.729", "3.354", "-2.09", "reject the null hypothesis", "fail to reject the null hypothesis", "her new method may work", "her new method probably does not work"]         

Kind of Cookie

Observed Frequency

Chocolate chip

187

Peanut butter

168

Cheese cracker

155

Lemon flavored

161

Chocolate mint

211

Vanilla filled

165

Observed x=

Cookie Sales is    by kind of cookie.

1.You conducted a mail survey in the City of Pasadena regarding a proposal to start a ferry service to a nearby tourist destination. Your survey results revealed that only 25% of the population supports the project. If you had a confidence interval of 95% with a +/- 5% margin of error, this means you are 95% confident that if you were to survey the entire population in the City of Pasadena, those who would support the ferry service would be between 5% and 25%.

2. You conducted a mail survey in the City of Pasadena regarding a proposal to start a ferry service to a nearby tourist destination. Your survey results revealed that only 25% of the population supports the project. If you had a confidence interval of 95% with a +/- 5% margin of error, this means you are 95% confident that if you were to survey the entire population in the City of Pasadena, those who would support the ferry service would be between 5% and 25%.

This data set was obtained by collecting information on a randomly selected sample of 93 employees working at a bank.

SALARY-  starting annual salary at the time of hire

EDUC  -  number of years of schooling at the time of the hire

EXPER -  number of months of previous work experience at the time of hire

TIME   -  number of months that the employee has been working at the bank until now

2. Use the least squares method to fit a simple linear model that relates the salary (dependent variable) toeducation (independent variable).

a)  What is your model? State the hypothesis that is to be tested, the decision rule, the test statistic, and your decision, usinga level of significance of 5%.

b)  What percentage of the variation in salary has been explained by the regression?

c) Provide a 95% confidence interval estimate for the true slope value.

d) Based on your model, what is the expected salary of a new hire with 12 years of education

e ) What is the 95% prediction interval for the salary of a new hire with 12 years of education? Use the fact that the distance value = 0.011286

The LDL cholesterol level of all men aged 20 to 34 follows the Normal distribution with mean μ=120μ=120 milligrams per deciliter and the standard deviation is σ=30σ=30 milligrams per deciliter. If you choose an SRS of 25 men from this population, what is the sampling distribution of mean cholesterol level of the 25 selected men.

Select one:

a. x¯∼N(120,1.2)x¯∼N(120,1.2)

b. Couldn't decide

c. x¯∼N(120,30)x¯∼N(120,30)

d. x¯∼N(120,6)

The average number of hours slept per week for college seniors is believed to be about 60 hours. A researcher asks 10 random seniors at a University how many hours they sleep per week. Based on the following results, should the researcher conclude that college seniors do or do not sleep an average of 60 hours per week?

Data (average number of hours slept per week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55

A: What is the null hypothesis for this study?

B: What is the alternative hypothesis for this study?

C: What is the standard deviation of this sample?

D: How many degrees of freedom are there for this data?

E: For an alpha level of 0.05 (two-tailed), what is the critical t-value?

F: Should the researcher accept or reject the null hypothesis?

Problem 11-11 (Algorithmic)

Agan Interior Design provides home and office decorating assistance to its customers. In normal operation, an average of 2.9 customers arrive each hour. One design consultant is available to answer customer questions and make product recommendations. The consultant averages 10 minutes with each customer.

1. Compute the operating characteristics of the customer waiting line, assuming Poisson arrivals and exponential service times. Round your answers to four decimal places. Do not round intermediate calculations.
   
   L_q =  
   
   L =  
   
   W_q =  minutes
   
   W =  minutes
   
   P_w =  
   
2. Service goals dictate that an arriving customer should not wait for service more than an average of 6 minutes. Is this goal being met? If not, what action do you recommend?
   
   No. Firm should increase the mean service rate u for the consultant or hire a second consultant.  
   
3. If the consultant can reduce the average time spent per customer to 8 minutes, what is the mean service rate? Round your answer to four decimal places. Do not round intermediate calculations.
   
   µ =  customers per hour
   
   W_q =  minutes
   
   Will the service goal be met?
   
   Yes

The following table shows ceremonial ranking and type of pottery sherd for a random sample of 434 sherds at an archaeological location.

Use a chi-square test to determine if ceremonial ranking and pottery type are independent at the 0.05 level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: Ceremonial ranking and pottery type are not independent.
H₁: Ceremonial ranking and pottery type are independent.H₀: Ceremonial ranking and pottery type are independent.
H₁: Ceremonial ranking and pottery type are not independent.     H₀: Ceremonial ranking and pottery type are not independent.
H₁: Ceremonial ranking and pottery type are not independent.H₀: Ceremonial ranking and pottery type are independent.
H₁: Ceremonial ranking and pottery type are independent.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo     

What sampling distribution will you use?

Student's tuniform     chi-squarebinomialnormal

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)

p-value > 0.1000.050 < p-value < 0.100     0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.     Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is sufficient evidence to conclude that ceremonial ranking and pottery type are not independent.At the 5% level of significance, there is insufficient evidence to conclude that ceremonial ranking and pottery type are not independent.    

Identify and describe two types of non-probability sampling methods and two types of probability sampling methods

Approximately 93% of all corn is genetically modified. Assume that you have randomly acquired 5 different corn seed packets for your garden. (c) Give an interpretation of the value you found for LaTeX: P\left(X\ge1\right)P ( X ≥ 1 ).

4. The height of maple trees are distributed normally with a mean of 31 meters and a standard deviation of 4 meters.

a) What is the probability of a tree being taller than 33 meters? Represent this graphically as well as numerically.

b) A group of 20 trees is selected at random. What is the probability that the average height of these 20 trees is more than 33 meters? Represent this graphically as well as numerically.

c) A group of 20 trees is selected at random. What is the probability that the average height of these trees is between 30 and 32 meters? Represent this graphically as well as numerically.

The Conference Board produces a Consumer Confidence Index (CCI) that reflects people’s feelings about general business conditions, employment opportunities, and their own income prospects. Some researchers may feel that consumer confidence is a function of the median household income. Shown here are the CCIs for nine years and the median household incomes for the same nine years published by the U.S. Census Bureau. Determine the equation of the regression line to predict the CCI from the median household income. Compute the standard error of the estimate for this model. Compute the value of r². Does median household income appear to be a good predictor of the CCI? Why or why not? Conduct the five step hypothesis test for both the model and regressor, using .05 level of significance.

Please provide the 5 steps for both the model and the regressor test, the Minitab output for each hypothesis test, and state the business implication based upon your analysis. You must use Minitab and the 5 step hypothesis testing process.

CCI            Median Household Income ($1,000)

116.8        37.415

91.5          36.770

68.5          35.501

61.6          35.047

65.9          34.700

90.6          34.942

100.0        35.887

104.6        36.306

125.4        37.005

Please provide the 5 steps and Minitab output, and make a decision about the data. You must use Minitab and the 5 step hypothesis testing process.

1. A researcher is interested in whether music tastes differ by personality type. The table below presents the observed and expected frequencies. Does a statistically significant relationship exist between personality type and music preference? Conduct a hypothesis test to see. Follow the five-step model with an alpha level of 0.05. Make sure to go through each step and interpret the results in terms of the research question above.

Round to twoplaces past the decimal point for your calculations and final answers.

True or False, explain your answer:

a) An observation with a studentized residual of more than 10 is probably an outlier.

b) If assumptions are met, least squares residuals are not correlated with the fitted values.

c) It is possible to reject a null hypothesis when the null hypothesis is true.

You wish to test the claim that the first population mean is less than the second population mean at a significance level of α=0.02α=0.02.     

Ho:μ1=μ2Ho:μ1=μ2
Ha:μ1<μ2Ha:μ1<μ2

You obtain the following two samples of data.

1. What is the test statistic for this sample?
   
   test statistic =  Round to 3 decimal places.
   
2. What is the p-value for this sample?
   
   p-value =  Use Technology Round to 4 decimal places.
   
3. The p-value is...
   • less than (or equal to) αα
   • greater than αα
   
   
4. This test statistic leads to a decision to...
   • reject the null
   • accept the null
   • fail to reject the null
   
   
5. As such, the final conclusion is that...
   • There is sufficient evidence to warrant rejection of the claim that the first population mean is less than the second population mean.
   • There is not sufficient evidence to warrant rejection of the claim that the first population mean is less than the second population mean.
   • The sample data support the claim that the first population mean is less than the second population mean.
   • There is not sufficient sample evidence to support the claim that the first population mean is less than the second population mean.