1. The following data were obtained from a two-factor independent-measures experiment with n

= 5 participants in each treatment condition.

B1                       B2                         B3

1. State the hypotheses for each of the three separate tests included in the two-factor ANOVA.
2. Calculate degrees of freedom and locate the critical region for each of the three tests.
3. Calculate the three F-ratios.
4. State a conclusion for each test (Alpha level = .05).

Let P_Sand P_Drepresent the prices charged for each standard golf bag and deluxe golf bag respectively. Assume that “S” and “D” are demands for standard and deluxe bags respectively.

S = 2250 – 15P_S                                                                                                                                                                    (8.1)

D = 1500 – 5P_D                                                                                                                                                                      (8.2)

Revenue generated from the sale of S number of standard bags is P_S*S. Cost per unit production is $70 and the cost for producing S number of standard bags is 70*S.

So the profit for producing and selling S number of standard bags = revenue – cost = P_SS – 70S                            (8.3)

By rearranging 8.1 we get                                  

15P_S= 2250 – S or

                                    P_S= 2250/15 – S/15 or

                                    P_S= 150 – S/15                                                                                                                                                                     (8.3a)

Substituting the value of P_Sfrom 8.3a in 8.3 we get the profit contribution of the standard bag:

                                    (150 –S/15)S – 70S = 150S – S²/15 – 70S = 80S – S²/15                                                                                 (8.4)

Revenue generated from the sale of D number of deluxe bags is P_D*D. Cost per unit production is $150 and the cost for producing D number of deluxe bags is 150*D.

So the profit for producing and selling D number of deluxe bags = revenue – cost = P_DD – 150D                           (8.4a)

By rearranging 8.2 we get                                  

5P_D= 1500 – D or

                                    P_D= 1500/5 – D/5 or

                                    P_D= 300 – D/5                                                                                                                                                                       (8.4b)

Substituting the value of P_Dfrom 8.4b in 8.4a we get the profit contribution of the deluxe bags:

                                    (300 -D/5)D – 150D = 300D – D²/5 – 150D = 150D – D²/5                                                                          (8.4c)

By adding 8.4 and 8.4c we get the total profit contribution for selling S standard bags and D deluxe bags.

                                    Total profit contribution = 80S –S²/15 + 150D – D²/5                                                                                 (8.5)

Homework assignment:

Reconstruct new objective function for 8.5 by changing “15P_S” to “8P_S” in 8.1, “5P_D” to “10P_D” in 8.2, cost per unit standard bagfrom 70 to 91 and cost per unit deluxe bag from 150 to 125. Keep other parameter values unchanged. Use up to 2 decimal points accuracy. Substitute the new expression for 8.5 in the excel solver workbook as explained in the class and solve for the optimal combination values for S and D..Instructor will not accept any homework late or submitted outside the class. Make sure you submit the results (just one page excel printout). Write/type your full name (first name first) in upper case, last 4 of your student ID, and, your new objective function expression (like equation 8.5 above) on the printout. Use S and D instead of b15 or c15 in the formulation. If you fail to follow the instructions, you will lose points.

*PLEASE Show also Excel Solution*

Determine the following z-scores: a. z0.1 = __________ b. z0.025 = __________

I don't understand the answer which has been posted. Thanks.

PART 2:

The researchers hypothesized that mice consuming more saturated fats will have a higher probability of becoming obese and developing symptoms of diabetes. To test for dietary effects on the development of metabolic disease, the researchers collected additional data from the two cohorts of mice. They measured the fasting blood glucose levels and they performed an insulin tolerance test (effect of a dose of insulin on blood glucose level). Data are reported as means (±SEM).

• Using a t-test, determine whether the difference in the fasting blood glucose levels for the lard-fed and the fish oil-fed mice is significant.

• Plot the glucose over time data.

• Did the two groups of mice differ in their response to the same dose of insulin? At which time points?

• What conclusions can you make about the effect of the type of dietary fat on carbohydrate metabolism?

PART 3:

To determine whether the difference in the microbiota is responsible for the difference in the weight gain or just a consequence of the different diet, the researchers raised mice for 11 weeks on either a lard or fish-oil diet. Then they took two new groups of mice and treated them with antibiotic to kill their gut microbiota. These mice then received transplants of the microbiota from the mice that had been fed either lard or fish oil for 11 weeks. All of the “transplanted” mice were maintained on a lard diet for 3 weeks to see if there was a difference in their weight gain.

• Was there a significant difference in the weight gain in these two groups of mice?

• What conclusion can you make about the effect of the microbiota composition versus the diet on weight gain?

• What additional information would you want to have to support your conclusion?

A survey found that 62​% of callers complain about the service they receive from a call center. State the assumptions and determine the probability of each event described below.

​(a) The next three consecutive callers complain about the service.

​(b) The next two callers​ complain, but not the third.

​(c) Two out of the next three calls produce a complaint.

​(d) None of the next 10 calls produces a complaint.

The probability that the next three callers complain is about?

Most vertebrates have testosterone, and have behaviors that are mediated by this hormone. Testosterone can be helpful to animals, by enhancing (e.g.) territory acquisition, or harmful, by (e.g.) causing physiological stress. When male blackbirds are exposed to other male blackbirds, their testosterone levels change. In order to understand the impacts of testosterone on male blackbirds, researchers followed a few individual males, to monitor testosterone changes after encountering another male.   A pre-exposure measurement was made (in nanograms/deciliter) and a post-exposure measurement was taken, data below. Researchers will test the hypothesis that pre-exposure testosterone levels are the same as post-testosterone levels.

(2 points) 1. Does this hypothesis test depend on a t distribution, a Z distribution, or a χ² distribution?

(2 points) 2. What are the df for the test you will do?

(2 points) 3a. Is this a one-tailed, or a two-tailed test?

(2 points) 3b. How would you rephrase the hypothesis test to make it the other way (for instance, if you chose a one-tailed test in 3a, how would you re-phrase my original question to make it a two-tailed test?)

(2 points) 4. What is the hypothesized difference?

(2 points) 5. Calculate your test statistic here:

(2 points) 6. Do you reject or fail to reject the null hypothesis?

(2 points) 7. How would you articulate your conclusion to my grandmother, who would not like to hear about rejecting (or failing to reject) null hypotheses, but would be interested to know about blackbirds?

!!!!!!ANSWER ALL!!!!

A paper company requires that the true median height of pine trees exceed 40 feet before they are harvested. A penalty is assigned if the median height is less than 40 feet. The management wants to avoid this penalty. A sample of 24 trees in one large plot is selected, and 7 of them are over 40 feet. Conduct the appropriate hypothesis test to determine if the median height of pine trees in the plot is less than 40 feet. The estimated median is 35 feet. Do not assume normality. Let a = 0.05.

Which of the following statements is correct?

a. sigma subscript x with bar on top end subscript equals fraction numerator sigma over denominator square root of n end fraction, where sigma = (sampled) population standard deviation and n = sample size--assuming n/N less or equal than .05, where N = (sampled) population size.

b. The sampling distribution of x with bar on top is either normally distributed, when the sampled population distribution is normal, or can be approximated by a normal distribution as the sample size becomes large, when the sampled population distribution is not normal.

c. By increasing the sample size, we increase the probability of obtaining a sample mean that is closer to the (sampled) population mean.

d. All of the above.

When you looking for the anxiety levels in males and females and are running a t test for two categorical variables what type of T test would you run?

How do you write the Ho and Ha when the question asks if there is a significant difference in the means?

Exhibit: Vaccination.

A public health official is planning for the supply of influenza vaccine needed for the upcoming flu season. She took a poll of 280 local citizens and found that only 113 said they would be vaccinated.

(For this exhibit, avoid rounding intermediate steps and round your final solutions to 4 digits)

1. Assuming that the population of the city is very large, construct the 91% confidence interval for the true proportion of people who plan to get the vaccine and provide the lower and upper bound of the confidence interval below. (Give your solution as a decimal fraction)

2. Now, assume that the town’s population is 1,400. Construct the 91% confidence interval and provide the lower and upper bound of the confidence interval below. (Give your solution as a decimal fraction)

3. How would you be wrong if the town’s population was in fact 1,400 but you computed the confidence interval not taking into account the finite population correction factor? Explain.

For the following data values below, construct a 90% confidence interval if the sample mean is known to be 0.719 and the population standard deviation is 0.366. (Round to the nearest thousandth) (Type your answer in using parentheses!Use a comma when inputing your answers! Do not type any unnecessary spaces! List your answers in ascending order!) for example: (0.45,0.78)

0.56, 0.75, 0.10, 0.95, 1.25, 0.54, 0.88

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.83 ​hours, with a standard deviation of 2.49 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.47 ​hours, with a standard deviation of 1.89 hours. Construct and interpret a 95​% confidence interval for the mean difference in leisure time between adults with no children and adults with children (mu 1 - mu 2)

Let mu 1μ1 represent the mean leisure hours of adults with no children under the age of 18 and mu 2μ2

represent the mean leisure hours of adults with children under the age of 18.

The 95​% confidence interval for left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2 is the range from   

hours to hours.

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

What is the interpretation of this confidence​ interval?

A.There is 95​% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of ainsufficient evidence of asignificant difference in the number of leisure hours.

B.There is a 95​%probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

C.There is 95​% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

D. There is an 95​%probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours.

Gautam is interested in learning what types of ethical messages are most effective.

He has three conditions, integrity, confidentiality, professional behavio, and

he randomly assigns 20 people to each of these conditions. After each person reads the message, they rate its effectiveness

on a 1-10 scale.He runs a one-way ANOVA and finds the following results:

Condition Mean

Integrity               6

Confidentiality           5

Professional Behavior   9

Source SS df MS F

Between   3000 2

Within

Total 11,500 59

a)Complete the one-way ANOVA table.

b) Determine the critical F at a=.01

What conclusion should Gautam make?

c) Is it appropriate to calculate a Tukey HSD test for this problem? Why? If it is appropriate, calculate a Tukey HSD test for each of the three possible comparisons between two means. What groups are significantly different from each other at a=.01

(integrity, confidentiality, professional behavior) according to this test?

In a random sample of 19 residents of the state of Montana, the mean waste recycled per person per day was 1.4 pounds with a standard deviation of 0.7 pounds. Determine the 90% confidence interval for the mean waste recycled per person per day for the population of Montana. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.