The document contains data on lab report scores of 8 students and the number of hours they spent preparing. It shows the highest possible score was 40. It then provides the regression equation to predict scores based on hours and defines the correlation coefficient. It predicts a score of 28 for someone who spent 1 hour preparing. It also defines the correlation coefficient and what it indicates about the relationship between variables.

1. B Heard
Not to be used, posted, etc. without my expressed permission. B Heard

2. Some Things to Remember
Not to be used, posted, etc. without my expressed permission. B Heard

3. 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)
Not to be used, posted, etc. without my expressed permission. B Heard

4. 1) Understand the equation of the regression line for
the given data.
2) What does the correlation coefficient “r” for the data
mean?
3) Predict a Lab Report Score for someone who spent
one hour on it.
Not to be used, posted, etc. without my expressed permission. B Heard

5. Predicted score for someone who spent one hour would be:
y = 3.158(1) +24.71
y = 27.9 or I would say 28 since all scores are in whole numbers
“r” of 0.9248 means strong positive correlation.
Not to be used, posted, etc. without my expressed permission. B Heard

6. Stronger Negative Correlation Stronger Positive Correlation
-1 0 +1
“r”
Not to be used, posted, etc. without my expressed permission. B Heard

7.  Know the difference between Binomial and Poisson
 Remember we talked about these in a previous lecture!

8.  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 Minitab) and also be able to find
probability values for less than, less than or equal
to, etc.

9.  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

10. 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 (Use Minitab)
(Also be able to find “at least” , “at most”, “or” etc.

11.  Know basic terms like mean, median, mode, standard
deviation, variance, range etc.
 Mean is the “average”
 Median is the “center”
 Mode is the “most frequently occurring”
 Know the variance is the standard deviation squared
 Know the standard deviation is the square root of the
variance

12.  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)

13. Be able to use the Standard Normal Distribution Tables or
Minitab to find probability values and z scores.
Examples:
 Find the following probability involving the Standard
Normal Distribution. What is P(z<1.55)?
.9394 (Use Minitab)
 Find the following probability involving the Standard
Normal Distribution. What is P(z > -.60)?
1 – 0.2743 = 0.7257 (Use Minitab)

14.  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?

15.  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.

16.  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?

17.  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 (smaller standard deviation)

18.  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%)

19.  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 (this would
be the expected gain for playing the game once)
 For five times it would be 5($3.20) = $16.00

20. How would you describe
the following stem and
leaf plot?
2| 5
3| 2
4| 1899
5| 13668
6| 0227
Not to be used, posted, etc. without my expressed permission. B Heard

21. How would you describe
the following stem and
leaf plot?
2| 5
3| 2
4| 1899
5| 13668
6| 0227
Not to be used, posted, etc. without my expressed permission. B Heard

22.  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.

23.  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)

24.  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 = 12
we want top 15% or an area greater than 1 - .15 or .85
z = 1.04 ---> x = (1.04)(12) + 60 = 72.48 or
need a score of 73 (Round it up)

25.  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?

26.  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

27.  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

28. Using Tables or Minitab (I used Minitab to show
you some previously):
a) P(550 < x < 650) = 0.5953
b) P(x < 700) =.9522
c) P(x > 630) = .3085

29. 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?
Not to be used, posted, etc. without my expressed permission. B Heard

30. y = -250+ 16a + 30b
Substitute 7 for “a” and 10 for “b”
y = -250+ 16*7 + 30*10
y = -250 + 112 + 300
y = -250 + 412
y = 162
Not to be used, posted, etc. without my expressed permission. B Heard

31. Be able to write the null and alternative hypothesis and
know which is the claim.
Not to be used, posted, etc. without my expressed permission. B Heard

32.  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

33. 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)

34.  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

35. (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 = 51.67
xbar - 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

36.  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 = 169.3
median =190
mode = 200
variance = 4553.6
range = 185
Doubtful it came from a normal, compare mean, median,
mode, etc.

37. Drop by and see me on
Facebook
http://www.facebook.com/statcave
You can find a link to these charts
there!