In 2008 the Better Business Bureau settled 75% of complaints they received (USA Today, March 2, 2009). Suppose you have been hired by the Better Business Bureau to investigate the complaints they received this year involving new car dealers. You plan to select a sample of new car dealer complaints to estimate the proportion of complaints the Better Business Bureau is able to settle. Assume the population proportion of complaints settled for new car dealers is .75, the same as the overall proportion of complaints settled in 2008.

Based upon a sample of 450 complaints, what is the probability that the sample proportion will be within .04 of the population proportion? (Round to four decimal places) Answer

Based upon the smaller sample of only 200 complaints, what is the probability that the sample proportion will be within .04 of the population proportion? (Round to four decimal places)

For this, problem, please include the excel procedure you used to arrive at an answer (screenshots, explanation, actual file, etc.).

The American Association of Individual Investors (AAII) On-Line Discount Broker Survey polls members on their experiences with electronic trades handled by discount brokers. As part of the survey, members were asked to rate their satisfaction with the trade price and the speed of execution, as well as provide an overall satisfaction rating. Possible responses (scores) were no opinion (0), unsatisfied (1), somewhat satisfied (2), satisfied (3), and very satisfied (4). For each broker, summary scores were computed by computing a weighted average of the scores provided by each respondent. A portion the survey results follow (AAII web site, February 7, 2012).

A. Develop an estimated regression equation using trade price and speed of execution to predict overall satisfaction with the broker. Interpret the coefficient of determination.

B. Use the t test to determine the significance of each independent variable. What are your conclusions at the 0.05 level of significance?

C. Interpret the estimated regression parameters. Are the relationships indicated by these estimates what you would expect?

D. Finger Lakes Investments has developed a new electronic trading system and would like to predict overall customer satisfaction assuming they can provide satisfactory service levels (3) for both trade price and speed of execution. Use the estimated regression equation developed in part (a) to predict overall satisfaction level for Finger Lakes Investments if they can achieve these performance levels.

E. What concerns (if any) do you have with regard to the possible responses the respondents could select on the survey

Clinical trials involved treating flu patients with Tamiflu, which is a medicine intended to attack the influenza virus and stop it from causing flu symptoms. Among 724 patients treated with Tamiflu, 72 experienced nausea as an adverse reaction. (a) Construct a 95% confidence interval for the (population) proportion of patients treated with Tamiflu that experienced nausea as an adverse reaction.  

(b) It is reported that the rate of nausea among patients treated with Tamiflu is greater than 7%. Using the current data, test the claim that the rate of nausea among patients treated with Tamiflu is greater than 7%. Use all steps for hypothesis testing to make a conclusion at a significance level of 5%.

(c) What assumptions or conditions have you made in statistical inference in parts (a) and (b) respectively? Are they being satisfied?

Compute in excel

Text Question 8.3.8 (Confidence Interval for a Mean, not a proportion) (adapted) The Table below contains the percentage of women receiving prenatal care in 2009 for a sample of countries. (i) Find the 90% confidence interval for the average percentage of woman receiving prenatal care in 2009 using the normal distribution. (Express as percentages with two digits, as shown in the data.) (ii) Find the 90% confidence interval for the average percentage of woman receiving prenatal care in 2009 using the t-distribution. (Express as percentages with two digits, as shown in the data.) (iii) Are you answers similar? Why or why not? Note: You’ll find the data below reproduced in the Excel answer template. 70.08 72.73 74.52 75.79 76.28 76.28 76.65 80.34 80.60 81.90 86.30 87.70 87.76 88.40 90.70 91.50 91.80 92.10 92.20 92.41 92.47 93.00 93.20 93.40 93.63 93.68 93.80 94.30 94.51 95.00 95.80 95.80 96.23 96.24 97.30 97.90 97.95 98.20 99.00 99.00 99.10 99.10 100.0 100.0 100.00 100.00 100.00 (b) Text Question 8.2.5 (Confidence Interval for a Proportion) In 2013, the Gallup poll asked 1039 U.S. adults if they agreed that a conspiracy was behind the assassination of President Kennedy. And found that 634 did so. Using a 98% confidence interval, estimate the proportion of Americans who believe in this conspiracy. Express both lower and upper bounds as a percentage, rounding to 2 digits, e.g., 57.29, 65.34.

Based on data from a​ college, scores on a certain test are normally distributed with a mean of 1532 1532 and a standard deviation of 316 316. LOADING... Click the icon to view the table with standard scores and percentiles for a normal distribution. a. Find the percentage of scores greater than 1848 1848. nothing ​% ​ (Round to two decimal places as​ needed.) b. Find the percentage of scores less than 900 900. nothing ​% ​ (Round to two decimal places as​ needed.) c. Find the percentage of scores between 1374 1374 and 2006 2006. nothing ​% ​ (Round to two decimal places as​ needed.)

Standard score   Percent
-3.0   0.13
-2.5   0.62
-2   2.28
-1.5   6.68
-1   15.87
-0.9   18.41
-0.5   30.85
-0.1   46.02
0   50.00
0.10   53.98
0.5   69.15
0.9   81.59
1   84.13
1.5   93.32
2   97.72
2.5   99.38
3   99.87
3.5   99.98

Part 2: Analyzing Study Time per Week

You spoke with your instructor and she claimed that the average number of hours that you should study has to be more than 5 hours per week which will help you achieve an above average grade on any subject. She also suggested as a practice that you can test her claim and let her know what your conclusion is. So you decided to contact your peers and gather information to conduct a hypothesis to test your instructors claim.

Questions:

1. Contact at least 15 peers (they don’t have to be from the same class) and obtain how much time they study each week.
   • Jorielyn - 8hours per week
   • Honey Bell - 4 hours per week
   • Shanaia - 4
   • Adona- 10
   • Christian- 6
   • Vincent- 3
   • Jam- 6
   • Joie- 3
   • Christine-4
   • Joseph- 3
   • Paolo- 5
   • Joriz-9
   • Leomer- 6
   • Angel- 5
   • Danika- 3

2. If you are to conduct a hypothesis test using the data you gathered,
   • Which test (t test or z test) do you think be appropriate and why?
     (Hint: think about whether you have the population or sample standard deviation available)

• Is this a test on population proportion or mean?

• Calculate the sample statistics (sample mean or sample proportion based on the information given) (1 mark)

• Use the data set you gathered to conduct a hypothesis test to determine your instructor’s claim. Use 0.05 level of significance. Please show your work, the 5 steps as described in the textbook.

3. Construct a 95% confidence interval for the population mean number of hours. Show your work.

Compute the confidence interval for the difference of two population means. Show your work.

Sample Mean 1= 17

Population standard deviation 1= 15

n1= 144

Sample Mean 2= 26

Population Standard Deviation 2= 13

n2 = 121

Confidence Level= 99

The variance in drug weights is critical in the pharmaceutical industry. For a specific drug, with weights measured in grams, a sample of 20 units provided a sample variance of s^2 = 0.49.

a)  construct a 90% confidence interval estimate for the population variance

b) construct a 90% confidence interval estimate of the population standard deviation

Find the​ t-value such that the area in the right tail is 0.025 with 9 degrees of freedom. nothing ​(Round to three decimal places as​ needed.) ​(b) Find the​ t-value such that the area in the right tail is 0.25 with 29 degrees of freedom. nothing ​(Round to three decimal places as​ needed.) ​(c) Find the​ t-value such that the area left of the​ t-value is 0.15 with 7 degrees of freedom.​ [Hint: Use​ symmetry.] nothing ​(Round to three decimal places as​ needed.) ​(d) Find the critical​ t-value that corresponds to 60​% confidence. Assume 13 degrees of freedom. nothing ​(Round to three decimal places as​ needed.)

1. Utilizing the sample size chart, what would be the minimum sample size for the following situations?

a. one sample test              ES = .8,*a = .05/2, 1- B = .90

b. two sample test (independent)   ES = .8, *a = .01, 1- B = .80

c. two sample test (independent)    ES = .2, *a = .05/2, 1- B = .95

d. one sample test      ES = .5, *a = .01, 1- B = .80

e. two sample test       ES = .8, *a = .05/2, 1- B = .95

f. one sample test        ES = .5, *a = .01, 1- B = .90

The eigenvalue is a measure of how much of the variance of the observed variables a factor explains. Any factor with an eigenvalue ≥1 explains more variance than a single observed variable, so if the factor for socioeconomic status had an eigenvalue of 2.3 it would explain as much variance as 2.3 of the three variables. This factor, which captures most of the variance in those three variables, could then be used in another analysis. The factors that explain the least amount of variance are generally discarded. How do we determine how many factors are useful to retain?

Consider the following results for two independent random samples taken from two populations.

Sample 1:

n1 = 40

x̅1 = 13.9

σ1 = 2.3

Sample 2:

n2 = 30

x̅2 = 11.1

σ2 = 3.4

What is the point estimate of the difference between the two population means? (to 1 decimal)

Provide a 90% confidence interval for the difference between the two population means (to 2 decimals).

Provide a 95% confidence interval for the difference between the two population means (to 2 decimals).

Last rating period, the percentages of viewers watching several channels between 11 P.M. and 11:30 P.M. in a major TV market were as follows: WDUX (News) WWTY (News) WACO (Cheers Reruns) WTJW (News) Others 15% 21% 25% 17% 22% Suppose that in the current rating period, a survey of 2,000 viewers gives the following frequencies: WDUX (News) WWTY (News) WACO (Cheers Reruns) WTJW (News) Others 280 401 504 354 461

(a) Show that it is appropriate to carry out a chi-square test using these data. Each expected value is ≥

(b) Test to determine whether the viewing shares in the current rating period differ from those in the last rating period at the .10 level of significance. (Round your answer to 3 decimal places.) χ2 χ 2 H0. Conclude viewing shares of the current rating period from those of the last.

The weight of trout in a fish farm follows the distribution N(200,502). A trout is randomly selected. (a) What is the probability that its weight does not exceed 175g? (b) What is the probability that its weight is greater than 230g? (c) What is the probability that its weight is between 225g and 275g? (d) What is the probability that out of eight trout selected randomly from the fish farm, less than three of them will not weigh more than 175g?

1. At a particular amusement park, most of the live characters have height requirements of a minimum of 57 in. and a maximum of 63 in. A survey found that​ women's heights are normally distributed with a mean of 62.4 in. and a standard deviation of 3.6 in. The survey also found that​ men's heights are normally distributed with a mean of 68.3 in. and a standard deviation of 3.6 in.
   
   Part 1:
   Find the percentage of men meeting the height requirement.
   
   The percentage of men who meet the height requirement is ____?____.
   ​(Round answer to nearest hundredth of a percent - i.e. 23.34%)
   
   What does the result suggest about the genders of the people who are employed as characters at the amusement​ park?
   Since most men___?___ the height​ requirement, it likely that most of the characters are ___?___ .
   (Use "meet" or "do not meet" for the first blank and "men" or "women" for the second blank.)
   
   Part 2: I was able to solve part 2 on my own.
   If the height requirements are changed to exclude only the tallest​ 50% of men and the shortest​ 5% of​ men, what are the new height​ requirements?
   The new height requirements are a minimum of 62.4 in. and a maximum of 68.3 in.
   ​(Round to one decimal place as​ needed.)