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Lecture for the Final ExamStatistics For Decision

Making

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Week 3 Quiz

Some Things to Remember

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This data shows the Lab Report scores of 8 selected students and the number of hours they spent preparing their Statistics Lab Report. 40 was the highest score the student could make.

(hours, scores), (3,34), (2,30), (4,38), (4,40), (2,32), (3,33),

(4,37), (5,39)

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1) Find the equation of the regression line for the given data.

2) What is the r2 for the data?3) What is the r for the data?4) What does correlation say about causation?5) Predict a Lab Report Score for someone

who spent one hour on it.

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Predicted score for someone who spent one hour would be:y = 3.158(1) +24.71y = 27.9 or I would say 28 since all scores are in whole numbersAlso, correlation says nothing about causation!

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0 +1

-1

Stronger Positive Correlation

Stronger Negative Correlation

“r”Not to be used, posted, etc. without my expressed permission. B Heard

Know the difference between Binomial and Poisson

Remember we talked about this in a previous lecture!

A State Trooper notes that at a certain intersection, an average of three cars run the red-light per hour.  What is the probability that the next time he is there exactly two cars run the red-light?

 Poisson with average of 3. want P(2)

P(2) = .2240 (Use Excel or Excel Template)

Final Exam Review

The probability that a house in a neighborhood has a dog is 40%.  If 50 houses in the neighborhood are randomly selected what is the probability that one (or a certain number) of the houses will have a dog?

 a. Is this a binomial experiment?  b. Use the correct formula to find the probability

that, out of 50 houses, exactly 22 of the houses will have dogs. Show your calculations or explain how you found the probability.

Answer Follows

Final Exam Review

a) Fixed number of independent trials, only two possible outcomes in each trial {S,F} (dog or not), probability of success is the same for each trial, and random variable x counts the number of successful trials. So YES it is.

 b) n = 50; p = .40 = P(success) = house has a

dog    We want P(22) --> P(22) = .0959 (I used a Binomial Template) (Also be able to use the “at least” column etc.)

Final Exam Review

Know basic terms like mean, median, mode, standard deviation, variance, etc.

Mean is the “average”Median is the “center”Mode is the “most frequently occurring”Know the variance is the standard deviation

squaredKnow the standard deviation is the square

root of the variance

Be able to understand the normal distribution and how it relates to the mean, standard deviation, variance, etc.

For example, I did an analysis and found the mean number of failures was 7 and the standard deviation was 1.5. Answer the two questions below.

How many standard deviations is 10 from the mean? 10 – 7 = 3, 3/1.5 = 2 (your answer)

How many standard deviations is 6.25 from the mean? 6.25 – 7 = - .75, - .75/1.5 = -0.5 (your answer)

Final Exam Review

Be able to use the Standard Normal Distribution Tables or

Excel to find probability values and z scores.Examples:Find the following probability involving the

Standard Normal Distribution.  What is P(z<1.55)?

.9394 (from the table or use Excel “=NORMDIST(1.55,0,1,TRUE)”)

Find the following probability involving the Standard Normal Distribution.  What is P(z > -.60)?

1 – 0.2743 = 0.7257 (using table or Excel 1 – “=NORMDIST(-0.6,0,1,TRUE)”)

Final Exam Review

ExplanationGo to Template Sitehttp://highered.mcgraw-hill.com/sites/007062

0164/student_view0/excel_templates.htmlSave “Normal Distribution File” to your

computerOpen the file and make sure you “Unprotect”

as I described in previous presentationsTake out all the values in the green cells

Explanation

I put 0 in for the mean and 1 for the standard deviation.

I worked the first one on the top left and the second on the top middle. (Only input into green cells)

The mean number of teachers in a Virginia public school is said to be 42.7. A hypothesis test is performed at a level of significance of 0.05 and a P-value of .06. How would you interpret this?

Final Exam Review

The mean number of teachers in a Virginia public school is said to be 42.7. A hypothesis test is performed at a level of significance of 0.05 and a P-value of .06. How would you interpret this?

Fail to reject the null hypothesis, because there is not enough evidence to reject the claim that there are 42.7 teachers per school.

Final Exam Review

I am buying parts for a new project. I have two vendors to choose from. Vendor X has a customer satisfaction rating of 8.7 with a standard deviation of 1.9. Vendor Y has a customer satisfaction rating of 8.6 with a standard deviation of 0.2 Which should I choose?

Final Exam Review

I am buying parts for a new project. I have two vendors to choose from. Vendor X has a customer satisfaction rating of 8.7 with a standard deviation of 1.9. Vendor Y has a customer satisfaction rating of 8.6 with a standard deviation of 0.2 Which should I choose?

I think I would go with Vendor Y who seems to be more consistent (standard deviation)

Final Exam Review

I am playing a game that has four different outcomes in terms of how much money I could win. Determine my expected gain if I played this game 5 times.

Outcomes/Probability $10 (10%), $6 (20%), $2 (30%), $1 (40%)

Final Exam Review

I am playing a game that has four different outcomes in terms of how much money I could win. Determine my expected gain if I played this game 5 times.

Outcomes/Probability $10 (10%), $6 (20%), $2 (30%), $1 (40%)

10(0.10)+6(0.20) + 2(0.30) + 1 (0.40) = $3.20

For five times it would be 5($3.20) = $16.00

Final Exam Review

How would you describe the following stem and leaf plot?

2| 53| 24| 18995| 136686| 0227Not to be used, posted, etc. without my expressed permission. B Heard

How would you describe the following stem and leaf plot?

2| 53| 24| 18995| 136686| 0227Not to be used, posted, etc. without my expressed permission. B Heard

Determine the minimum required sample size if you want to be 90% confident that the sample mean is within 5 units of the population mean given sigma = 8.4.  Assume the population is normally distributed.

Final Exam Review

Determine the minimum required sample size if you want to be 90% confident that the sample mean is within 5 units of the population mean given sigma = 8.4.  Assume the population is normally distributed.

n = (Zc*sigma/E)^2 = [(1.645 * 8.4)/ 5]^2 = (2.7636)^2= 7.64= 8 (always round up sample sizes)

Final Exam Review

Scores on an exam for entering a military training program are normally distributed, with a mean of 60 and a standard deviation of 12. To be eligible to enter, a person must score in the top 15%. What is the lowest score you can earn and still be eligible to enter?

mu = 60; sigma = 12we want top 15% or an area greater than 1 - .15 or .85z = 1.04 ---> x = (1.04)(12) + 60 = 72.48 orneed a score of 73 (Round it up)

Final Exam Review

Explanation

I put 60 in for the mean and 12 for the standard deviation.

Put in 0.15 for the bottom middle to get 72.44 (remember to round up to 73).

An airplane has 50 passengers. There are 4 celebrities on the plane. How many ways can a reporter choose 3 of these passengers at random and not pick a celebrity?

Final Exam Review

An airplane has 50 passengers. There are 4 celebrities on the plane. How many ways can a reporter choose 3 of these passengers at random and not pick a celebrity?

This is a Combination 46C3 which is 15180

Final Exam Review

The average (mean) monthly grocery cost for a family of 4 is $600.  The distribution is known to be “normal” with a standard deviation = 60.  A family is chosen at random. 

a) Find the probability that the family’s monthly grocery cost purchases will be between $550 and $650.

b)Find the probability that the family’s monthly grocery cost purchases will be less than $700.

c) What is the probability that the family’s monthly grocery cost purchases will be more than $630?

Answers follow

Final Exam Review

Using Tables or EXCEL (I used an Excel Template like one I showed you):a) P(550 < x < 650) = 0.5953 b) P(x < 700) =.9522  c) P(x > 630) = .3085 

Final Exam Review

Explanation

I put 600 in for the mean and 60 for the standard deviation.

I worked the part a on the top right, part b on the top left and part c on the top middle. (Only input into green cells AND don’t forget to hit the “Enter” key)

As an instructor, I have been collecting data to see if I can model a student’s performance on a standardized entrance exam. I determined that the multiple regression equation y = -250+ 16a + 30b, where a is a student’s grade on a quiz, b is the student’s rank on a class list, gives y, the score on a standardized entrance exam. Based on this equation, what would the standardized entrance exam score for a student who makes a 7 on the quiz and had a ranking of 10 be?

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y = -250+ 16a + 30bSubstitute 7 for “a” and 10 for “b”

y = -250+ 16*7 + 30*10y = -250 + 112 + 300y = -250 + 412y = 162

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Be able to write the null and alternative hypothesis and know which is the claim.

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A Pizza Delivery Service claims that it will get its pizzas delivered in less than 30 minutes. A random selection of 49 service times was collected, and their mean was calculated to be 28.6 minutes. The standard deviation was 4.7 minutes. Is there enough evidence to support the claim at alpha = .10. Perform an appropriate hypothesis test, showing each important step.  (Note:  1st Step:  Write Ho and Ha; 2nd Step:  Determine Rejection Region; etc.)

Answer following chart

Final Exam Review

Ho: mu >= 30 min.

Ha: mu < 30 min. (claim). Therefore, it is a left-tailed test.

n=49; x-bar=28.6; s=4.7; alpha=0.10

Since alpha = 0.10, then the critical z value will be zc = -1.28

since n>30 then s can be used in place of sigma.

Standardized test statistic z = (x-bar - mu)/(s/sqrt(n))                                       z = (28.6-30)/(4.7/sqrt(49))                                      z = -2.085

since -2.085 < -1.28, we REJECT Ho. 

That is, at alpha = 0.10, There is enough evidence to support the

Pizza Delivery Service’s claim.

(p-value method could have also been used)

Final Exam Review

A polling company wants to estimate the average amount of contributions to their candidate. For a sample of 100 randomly selected contributors, the mean contribution was $50 and the standard deviation was $8.50.

(a)  Find a 95% confidence interval for the mean amount given to the candidate

(b)  Interpret this confidence interval and write a sentence that explains it.

Answer Follows

Final Exam Review

(a).  Since sample size = n = 100> 30, we can use a z-value.  For a 95% confidence level, z-value = 1.96.  Also, sample mean = xbar = 50; population standard deviation is estimated by sample standard deviation (since n > 30) = s = 8.50

E = z * s / sqrt(n) = 1.96 * 8.50/sqrt(100) = 1.666

xbar + E = 50.00 + 1.67 = 51.67xbar - E = 50.00 - 1.67 = 48.33

Thus, 95% confidence interval = ($48.33,$51.67)

(b)  We are 95% confident that the population mean amount contributed is between $48.33 and $51.67

Final Exam Review

Explanation From part 3 of the Week 6 lab (Excel provided there – tab labeled “Conf Intervals”) Put in 95% (or your value may already be there) and see the z-score. Then use formula to complete.

The failure times of a component are listed in hours.  {100, 95, 120, 190, 200, 200,280}.Find the mean, median, mode, variance, and range.Do you think this sample might have come from a normal population?  Why or why not?

mean = 169.3median =190mode = 200variance = 4553.6range = 185Doubtful it came from a normal, compare mean,

median,mode, etc.

Final Exam Review

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