## Cost Curve Anaylsis

You have been presented with the following cost data table and asked to fit a statistical cost function.

Quantity | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100

Total Cost | 104 | 107 | 109 | 111.5 | 114.5 | 118 | 123 | 128.5 | 137 | 150

Question 1: fit (estimate) three possible statistical cost functions to the data. Use straight-line, quadratic, and cubic formulas.

2: discuss the statistical results you obtained in question 1. Include in your discussion R^2 , the coefficients, and the statistical significance of the coefficients. For the statistical significance test, use the p-test rather than the t-test.

3: If the data represent 10 months of production for one plant of a specific company, would you consider this to be a short-run analysis?

4: how would you answer to question 3 change if you were told that the data represent 10 different plants during a particular month of the year?

How would I answer this in a word doc and an Excel? Please help.

## Hypothesis Testing Sample Questions

1. Measurements of the left handed and right handed gripping strengths of twelve left handed persons are recorded as follows.

PERSON 1 2 3 4 5 6 7 8 9 10 11 12

LEFT HAND 140 90 125 130 95 121 85 97 131 110 98 109

RIGHT HAND 138 87 110 132 96 120 86 90 129 100 101 117

We wish to test whether left handed persons have greater gripping strength in their dominant (left) hand. Assume normality of the appropriate population(s).

1a) What is the observed value of the test statistic?

Answer

a. .39

b. 1.17

c. 2.08

d. 3.16

1b) How many degrees of freedom does the test statistic have, under the null hypothesis?

Answer

a. 9

b. 10

c. 11

d. 20

1c) What is the p-value for these data?

Answer

a. <.01

b. .01-.05

c. .05-.10

d. >.10

1d) Find a 95% confidence interval for the difference in left and right hand gripping strength.

Answer

a. (-.31, 7.51)

b. (.03,7.17)

c. (-8.75, 15.95)

d. (-14.9,22.1)

e. (-1.84, 6.00)

2. A test of null hypothesis Ho vs the alternative Ha has p-value .047. Will the null hypothesis be rejected when the test is at level alpha=.05?

Answer

a. yes

b. no

3. Two spreadsheet programs (A and B) are compared to see which is easier to use. 10 students in the Faculty of Management were given a standard set of finance problems to solve using Program A and 16 students from the faculty were told solve the same problems using Program B. For each student, the total time taken to complete all the problems is recorded. The times for the students using Program A had mean 51.5 minutes and standard deviation 10.46 minutes; for Program B the mean was 38.0 minutes and the standard deviation 8.67 minutes.

3a) What is the pooled variance Sp**2 for these data? (Note: the notation Sp**2 means the square of Sp)

Answer

a. 8.62

b. 9.36

c. 9.57

d. 88.01

e. 81.24

f. 88.34

g. 92.29

4. The study in problem 4 was repeated in the Faculty of Science. Samples of size 10 and 15 were used for program A and program B. In this case the mean time for the students using program A was 52 minutes, and the mean time for students using program B was 42 minutes. The pooled standard deviation s_p was equal to 9 minutes. Using this faculty of Science data, test the alternative hypothesis that the mean time for program B is less than the mean time for program A.

4a) What is the value of the test statistic?

Answer

a. 0.12

b. 1.11

c. 2.72

d. 6.84

e. 10.0

4b) How many degrees of freedom does the test statistic have under the null hypothesis?

Answer

a. 10

b. 13

c. 16

d. 23

e. 25

f. 24

4c) What is the p-value for the test?

Answer

a. < .01

b. .01-.05

c. .05-.10

d. > .10

4d) Find a 95% confidence interval for the difference in population mean completion times between Program A and Program B.

Answer

a. (2.42, 17.58)

b. (5.5, 14.5)

c. (8.15, 11.85)

d. (8.18, 11.82)

e. (-0.31, 20.31)

5. If a test rejects Ho: mu = mu_0 in favour of Ha: mu not = mu_0 at level alpha, then the test will also reject Ho in favour of Ha: mu > mu_0 at level alpha.

Answer

a. true

b. false

c. insufficient information to decide

6. In testing for a difference between a new and old production process, the null hypothesis states that the new process is at least as good as the old one. A type I error is committed if:

Answer

a. It is concluded that the new process is at least as good when in fact, it is not.

b. It is concluded that the old process is better when in fact, it is.

c. It is concluded that the old process is better when in fact, it is not.

d. It is concluded that the new process is at least as good when in fact, it is.

## Hypothesis Testing – Heating and Oil Company

The director of Heating and Oil at Superior Oil Company is concerned about the high cost of home heating oil being offered to their customers for the upcoming Fall season in Philadelphia, PA. The company has no possibility of modifying the oil price under the current economic conditions. He believes that the company should offer a low cost maintenance contract to foster good will. The director decides to survey a sample of the surrounding towns to see what people are paying for similar home maintenance contracts. The telephone sample of 25 homes indicated that the customers are paying X (bar) = $225.4 and s = $25.20 for a yearly contract.

a. Using the 0.01 level of significance, is there evidence that the population mean is above $225?

b. What is your answer in (a) if s = $50 and the 0.10 level of significance is used?

c. What is your answer in (a) if X (bar) = $279.00 and s = $20.20?

## Probability, hypothesis testing and confidence intervals

What is an important similarity between the uniform and normal probability distributions?

A. The mean, median and mode are all equal.

B. The mean and median are equal.

C. They are negatively skewed.

D. About 68% of all observations are within one standard deviation of the mean.

The distribution of the annual incomes of a group of middle management employees approximated a normal distribution with a mean of $37,200 and a standard deviation of $800. About 68 percent of the incomes lie between what two incomes?

A. $30,000 and $40,000

B. $36,400 and $38,000

C. $34,800 and $39,600

D. $35,600 and $38,800

A marketing firm is studying consumer preferences for winter fashions in four different months. From a population of women, 18-21 years of age, a random sample of 100 women was selected in January. Another random sample of 100 women was selected in March. Another random sample of 100 women was selected in June. Another random sample of 100 women was selected in September.

A. The number of samples was 4.

B. The number of samples was 100.

C. The number of samples was 400.

D. The number of samples was 1.

Truck tire life is normally distributed with a mean of 60,000 miles and a standard deviation of 4,000 miles. You bought four tires. What is the probability that the average mileage of the four tires exceeds 66,000 miles?

A. 0.0013

B. 0.9987

C. 0.4987

D. 0.9544

A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week. Assuming a population standard deviation of 6 hours, what is the required sample size if the error should be less than ½ hour with a 95% level of confidence?

A. 554

B. 130

C. 35

D. 393

A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week. Assuming a population standard deviation of 3 hours, what is the required sample size if the error should be less than ½ hour with a 99% level of confidence?

A. 196

B. 240

C. 15

D. 16

Assuming a population standard deviation of 3 hours, what is the required sample size if the error should be less than ½ hour with a 99% level of confidence?

A. 196

B. 240

C. 15

D. 16

A survey of 50 retail stores revealed that the average price of a microwave was $375 with a standard error of $20. Assuming the population is normally distributed, what is the 95% confidence interval to estimate the true cost of the microwave?

A. $323.40 to $426.60

B. $328.40 to $421.60

C. $350.80 to $395.80

D. $369.31 to $380.69

To conduct a test of hypothesis with a small sample, we need to be able to make an assumption that:

A. a larger computed value of t will be needed to reject the null hypothesis.

B. the region of acceptance will be wider than for large samples.

C. the confidence interval will be wider than for large samples.

D. the population is normally distributed.

If the critical z-value for a test statistic equals 2.45, what value of the test statistic would provide the least chance of making a Type I error?

A. 3.74

B. 10,000

C. 2.46

D. 4.56

A recent study focused on the number of times men and women send a Twitter message in a day. The information is summarized below.

Men Sample size-25 Sample mean-20 Population standard deviation-5

Women Sample size-30 Sample mean-30 Population standard deviation-10

At the .01 significance level, is there a difference in the mean number of times men and women in a day? What is the test statistic for this hypothesis?

A. z-statistic

B. t-statistic

C. p-statistic

D. df-statistic

A recent study focused on the number of times men and women send a Twitter message in a day. The information is summarized below.

Men Sample size-25 Sample mean-20 Population standard deviation-5

Women Sample size-30 Sample mean-30 Population standard deviation-10

At the .01 significance level, is there a difference in the mean number of times men and women send a Twitter message in a day? What is the p-value hypothesis test?

A. Reject the null hypothesis and conclude the means are different.

B. Reject the null hypothesis and conclude the means are the same.

C. Fail to reject the null hypothesis and conclude the means are the same.

D. Fail to reject the null hypothesis and conclude the means are different.

When testing the difference between two population means, the variances are pooled when

A. the population standard deviations are known and equal.

B. the population means are known.

C. the population standard deviations are assumed unequal and unknown.

D. the population standard deviations are assumed equal but unknown.

When is it appropriate to use the paired difference t-test?

A. Four samples are compared at once

B. Any two samples are compared

C. Two independent samples are compared

D. Two dependent samples are compared

20 randomly selected statistics students were given 15 multiple-choice questions and 15 open-ended questions – all on the same material. The professor was interested in determining which type of questions the students scored higher. This experiment is an example of

A. a one sample test of means.

B. a two sample test of means

C. a paired t-test

D. a test of proportions

What is the variable used to predict the value of another called?

A. Independent variable

B. Dependent variable

C. Correlation variable

D. Variable of determination

What does a coefficient of correlation of 0.70 infer?

A. Almost no correlation because 0.70 is close to 1.0

B. 70% of the variation in one variable is explained by the other

C. Coefficient of determination is 0.49

D. Coefficient of nondetermination is 0.30

A hypothesis test is conducted at the .05 level of significance to test whether or not the population correlation is zero. If the sample consists of 25 observations and the correlation coefficient is 0.60, what is the computed value of the test statistic?

A. 1.96

B. 2.07

C. 2.94

## Review of a Statistics Article on Death Certificates

Using the attached article, please help me respond to the following questions:

1. What is the main purpose and hypothesis of the study?

2. Develop a pictorial representation of the relationship between the variables that the authors studied. Include the level of measurement (nominal, ordinal, interval/ratio) for each identified variable.

3. What statistical test(s) or epidemiological concepts do the author(s) use?

4. Discuss whether the statistical and epidemiological tests were applied correctly?

5. Would you do anything differently from a presentation of results, testing or measurement of variables point of view?

6. What real-life conclusions can be drawn from the study? What does the study mean for patients or practice?

## Multiple Choice Statistics Problems

3. A chemical engineer is investigating the effect of process operating temperature on product yield. The study results in the following data; use your knowledge of least squares regression to construct a linear model for predicting yield from temperature. These data apply throughout question 3.

Temp(Celsius) Yield (grams)

100 43.6215

110 46.6231

120 58.2752

130 58.8906

140 65.4354

150 74.5960

160 72.5202

170 79.0639

180 83.6280

190 84.4351

Calculate the slope and intercept of the least squares regression model by hand, which requires only the means and standard deviations of X and Y, and the correlation coefficient (here r = 0.9805).

3 a) If the yield were measured in ounces instead of grams (note that 1 gram is 0.35274 ounces), the slope would increase by a factor of:

Answer

a. 0.35274

b. 1/0.35274

c. would not change

3 b) If the yield were measured in ounces instead of grams (note that 1 gram is 0.35274 ounces), the correlation coefficient would increase by a factor of:

Answer

a. 0.35274

b. 1/0.35274

c. would not change

4. (Look at the file attachment for the table)

Results of a two year study of the effects of calcium supplementation on bone loss are summarized below. The rate of bone loss, computed for each subject, was expressed as a percentage of their initial bone mass. Subjects were randomly allocated to three treatment groups. Group 1 received estrogen creme and a calcium placebo – Estrogen Group. Group 2 received placebo estrogen creme and 200 mg/day calcium – Calcium Group. Group 3 received placebo estrogen creme and a calcium placebo – Placebo Group.

Use one-way ANOVA to compare mean bone mass change per year for the three treatment groups and perform an F test to see if the treatment group means differ. Proceed in a step-by-step fashion doing the computations by hand (with a calculator), and answer throughout question 4.

4 a) What is the value of the grand mean computed from the above data?

Answer

a. 1.50 – 2.00

b. 2.01 – 2.25

c. 2.26 – 2.50

d. 2.51 – 2.75

e. 2.76 – 3.00

4 b) What is the value of the error sum of squares (SSE) for the above data?

(Pick the interval containing the best answer.)

Answer

a. 150 – 160

b. 161 – 170

c. 171 – 180

d. 181 – 190

e. 191 – 200

4 c) Assuming a significance level of 0.05, what is the critical value of F (F crit) for this test?

Answer

a. 3.0 – 3.5

b. 4.0 – 4.5

c. 5.0 – 5.5

d. 6.0 – 6.5

e. 7.0 – 7.5

4 d) What is the value of the F statistic for this sample of data (Fdata)?

Answer

a. 3.50 – 4.00

b. 4.01 – 4.50

c. 4.51 – 5.00

d. 5.01 – 5.50

e. 5.51 – 6.00

4 e) Is the strength of evidence against the H0 provided by the data strong enough to reject it in favor of the alternative hypothesis?

Answer

a. yes, reject H0

b. no, do not reject H0

c. can’t tell.

## Statistics Problem: Are Visa’s new ads successful?

In 2006, Visa wanted to move away from its long running television advertising them of “Visa, it’s everywhere you want to be.” During the Winter Olympics, Visa featured Olympians in commercials with a broader message, including security; check cards, and payment technologies such as contactless processing. One of the first commercials featured snowboarder Lindsey Jacobellis being coached to calm down before a big race by imagining that her Visa Check Card got stolen. A key metric for the success of television advertisements is the proportion of viewers who “like the ads a lot.” Harris Ad Research Service conducted a study of 903 adults who viewed the new Visa advertisement and reported that 54 indicated that they “like the ad a lot.” According to Harris, the proportion of a typical television advertisement receiving the “like the ad a lot” score is 0.21.

A) Use the six-step critical value approach to hypothesis testing and a 0.05 level of significance to try to prove that the new Visa ad is less successful than a typical television advertisement.

B) Use the five-step p-value approach to hypothesis testing and a 0.05 level of significance to try to prove that the new Visa ad is less successful than a typical television advertisement.

C) Compare the results of (a) and (b).

## Statistics and Hypothesis Testing

I need help understanding the results. I can’t interpret the results and their differences. I attached the whole project so you can see what I am talking about.

Results Section :

a. The data should be explained in narrative form including what (of importance) was or was not statistically significant.

c. Address the hypotheses and whether or not they are accepted or rejected. Use .05 as the alpha (level of significance).

Discussion

a. You need to interpret the statistical findings, not just what they are, but what they mean. You must synthesize and explain.

b. What is the clinical significance of the findings and are they important.

c. What are the strengths and limitations of the study?

d. What suggestions do you have for further research?

Conclusion

a. Briefly summarize the purpose and findings

b. Briefly restate the meaning of the results for practice.

## Analyzing research data

Attached is a simple spreadsheet I’ve compiled about the 21 commercial aircraft accidents in the US since 1972 that have at least one survivor and one death. I need help coming up with some good stats. Meaning, please look at the data and crunch the numbers to determine where/if it is safer to sit in the front or the back during a flight. It’s apparent looking at it that the back is safer but I need some interesting statistical compilations. Please don’t just say do this or that, I’m looking for data using this small sample size. Thanks.

## Assembly Line Question

Scenario: The manager of a package courier service believes that packages shipped at the end of the month are heavier than those shipped early in the month. As an experiment, he weighed a random sample of 20 packages at the beginning of the month. He found that the mean weight was 20.25 pounds and the standard deviation was 5.84 pounds. Ten packages randomly selected at the end of the month had a mean weight of 24.80 pounds and a standard deviation of 5.67 pounds. At the .05 significance level, we will test to see if we can we conclude that the packages shipped at the end of the month weigh more.

Questions:

– What is the alternate hypothesis?

– What is the critical value? (a.) 1.701, b.) 0.50, c.) 1.645, d.) 1.96

– What is the value of the test statistic? (a.) -2.03, b.) 11.75, c.) 2.03, d.) -11.05 What is the decision? (a.) Reject Ho, b.) Do not reject Ho, c.) Indeterminate, d.) Reject H1)

– What is the conclusion? (a.) evidence packages shipped at end of month weigh more, b.) no evidence packages shipped at end of month weigh more, c.) evidence packages weigh less, d.) no evidence of a difference in the mean weight).

## Evaluating Hypothesis Testing

Scenario: The manager of a package courier service believes that packages shipped at the end of the month are heavier than those shipped early in the month. As an experiment, he weighed a random sample of 20 packages at the beginning of the month. He found that the mean weight was 20.25 pounds and the standard deviation was 5.84 pounds. Ten packages randomly selected at the end of the month had a mean weight of 24.80 pounds and a standard deviation of 5.67 pounds. At the .05 significance level, we will test to see if we can we conclude that the packages shipped at the end of the month weigh more.

Questions:

– What is the alternate hypothesis?

– What is the critical value? (a.) 1.701, b.) 0.50, c.) 1.645, d.) 1.96

– What is the value of the test statistic? (a.) -2.03, b.) 11.75, c.) 2.03, d.) -11.05 What is the decision? (a.) Reject Ho, b.) Do not reject Ho, c.) Indeterminate, d.) Reject H1)

– What is the conclusion? (a.) evidence packages shipped at end of month weigh more, b.) no evidence packages shipped at end of month weigh more, c.) evidence packages weigh less, d.) no evidence of a difference in the mean weight).

## Hypothesis Testing on Motor Vehicle Accidents

A random sample of 400 college students was selected and 120 of them had at least one motor vehicle accident in the previous two years. A random sample of 600 young adults not enrolled in college was selected and 150 of them had at least one motor vehicle accident in the previous two years. At the .05 level, you are testing whether there is a significant difference in the proportions of college students and similar aged non college students having accidents during the two year period.

1. Is this a 1 or 2 tail test?

2. Is this a one sample or two sample test?

3. Is this a test of sample means or sample proportions?

a.) sample proportions, b.) sample means, c.) both, d.) neither

4. What are your critical values? a.) +-1.65, b.) +-1.96, c.) 1.74, d.) 1.79

5. What is the value of your pooled proportion? a.) .27, b.) .73, c.) .55 d.) .05

6. What is the value of your test statistic? a.) 1.96, b.) 1.65, c.) 1.74, d.) .05

7. What is your decision? a.) Reject Ho, b.) Reject H1, c.) So not reject Ho, d.) none of the above

## One Sample Hypothesis Testing on Average Weekly Tips

The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.20 liters. A sample of 10 adults after the campaign shows the following consumption in liters:

– 1.34 1.32 1.38 1.40 1.70 1.40 1.32 1.90 1.38 1.36

At the 0.10 significance level, can we conclude that water consumption has increased? Calculate and interpret the p-value.

(a) State the null hypothesis and the alternate hypothesis. (Round your answers to 2 decimal places.)

(b) State the decision rule for .10 significance level. (Round your answer to 3 decimal places.)

(c) Compute the value of the test statistic. (Round your intermediate and final answer to 3 decimal places.)

(d) At the 0.10 level, can we conclude that water consumption has increased?

(e) Estimate the p-value.

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $72 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $2.45. Over the first 34 days she was employed at the restaurant, the mean daily amount of her tips was $73.07. At the .10 significance level, can Ms. Brigden conclude that her daily tips average more than $72?

(a) State the null hypothesis and the alternate hypothesis.

(b) State the decision rule.

(c) Compute the value of the test statistic. (Round your answer to 2 decimal places.)

(d) What is your decision regarding H0?

(e) What is the p-value? (Round your answer to 4 decimal places.)

## Hypothesis Testing: One Sample Tests

The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.20 liters. A sample of 10 adults after the campaign shows the following consumption in liters:

– 1.34 1.32 1.38 1.40 1.70 1.40 1.32 1.90 1.38 1.36

At the 0.10 significance level, can we conclude that water consumption has increased? Calculate and interpret the p-value.

(a) State the null hypothesis and the alternate hypothesis. (Round your answers to 2 decimal places.)

(b) State the decision rule for .10 significance level. (Round your answer to 3 decimal places.)

(c) Compute the value of the test statistic. (Round your intermediate and final answer to 3 decimal places.)

(d) At the 0.10 level, can we conclude that water consumption has increased?

(e) Estimate the p-value.

## Evaluating Types of Therapy & Results Reported in a Quantitative Study

I’m having problems with these questions. I would like to have the explanations along with solutions.

1. Foa, Rothbaum, Riggs, and Murdock (1991) conducted a study evaluating four different types of therapy for rape victims. The Stress Inoculation Therapy (SIT) group received instructions on coping with stress. The Prolonged Exposure (PE) group went over the events in their minds repeatedly. The Supportive Counseling (SC) group were taught a general problem-solving technique. Finally, the Waiting List (WL) control group received no therapy. Data with the same characteristics as theirs follow, where the dependent variable was the severity rating of a series of symptoms.

Group n Mean S.D.

SIT 14 11.07 3.95

PE 10 15.40 11.12

SC 11 18.09 7.13

WL 10 19.50 7.11

a) The analysis of variance (ANOVA) was run, and the results are in the following table. Draw whatever conclusions are warranted. Interpret what the conclusions mean.

Source df SS MS F

Treatment 3 507.840 169.280 3.04*

Error 41 2279.067 55.587

Total 44 2786.907

* p < .05

b) The Bonferroni test was run to compare the WL group with each of the other three groups. The results are in the following table. What would you conclude? How does this compare to the answer to part a?

WL versus SIT WL versus PE WL versus SC

t = 2.73 t = 1.23 t = 0.433

The critical value of the Bonferroni test is 2.50.

2. Examine the following results reported in a quantitative study:

“The scores varied for band members (M=3.5), choir members (M=3.9), and for student athletes (M=5.4) for attitudes toward engaging in school activities during the 3-5 p.m. period of time. A comparison of the groups, at an alpha of .05, showed a statistically significant difference among the three groups, F (3, 8) = 9.87, p = .031, effect size = .91 SD.”

As you examine this statement, you conclude: (state whether the statements below are NO or YES).

The null hypothesis was rejected.

The level of significance showed a probability of rejecting set at 5 out of 100 times.

The statistical test used was a t-test.

The magnitude of differences among the groups was over one standard deviation.

Band members differed significantly from student athletes in their attitudes.