ONE SAMPLE T-TEST
Assignment 1: Change in Mood Following a Romantic Movie Clip
A researcher asks participants to watch a short romantic movie clip. The movie clip depicts a romantic scene ending with two long-lost lovers embracing in a kiss. After the short romantic movie clip, participants are asked to indicate how the romantic movie clip has affected their mood on a bipolar scale ranging from -3 (much worse mood) to +3 (much better mood), with 0 indicating no change in mood. The results are given below. It was assumed that the average participant would give a rating of 0 if there were no change in mood. Test whether or not participants reported a significant change in mood at a .05 level of significance using a twotailed test.
SCORES: -3, -3, -3, -3, -3, -3, -3, -3, -3,
-2, -2, -2, -2, -2, -2, -2, -2, -2,
2,
-1, -1, -1, -1, -1, -1, -1
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
With regard to the SPSS exercise, answer the following questions:
Based on the SPSS output, state the following values:
Sample size, Sample mean, Sample standard deviation, Estimated standard error
Based on the table shown in SPSS, state the following values associated with the test statistic:
Mean difference, t obtained (t), Degrees of freedom (df), Significance (2-tailed)
In: Math
2,006 persons are cross classified by disease and exposure.
Our outcome of interest is having the disease.
Use the following 2 x 2 table to answer questions below.
Disease |
No Disease |
|
Exposed |
9 |
992 |
Not Exposed |
16 |
989 |
a. What percent of people have disease?
b. What percent of the exposed persons have disease?
c. What is the odds of disease among the exposed?
d. What is the odds of disease among the unexposed?
e. What is the odds ratio of disease among the exposed and unexposed. Interpret the odds ratio and comment whether the exposure is deleterious or protective.
In: Math
Traditional definition of independence says that events A and B
are
independent if and only if P(A n B)=P(A)×P(B). Show that P(A n
B)=P(A)×P(B) if
and only if
a. P(A n B’) = P(A) × P(B’)b. P(A’ n B’) = P(A’) × P(B’)
You may use these without proof in your solutions:
o P(A n B’) = P(A) – P(A n B). [This can be proven by first showing
that
(A ∩ B ′ ) ∪̇ (A ∩ B) = A and using the Addition Rule.]
o P(A n B) = 1 – P(A’ n B’). [This can be proven by noting that
under de
Morgan’s Law, (A n B)’ = A’ n B’.]
In: Math
A friend of mine is giving a dinner party. His current wine supply includes 12 bottles of zinfandel, 8 of merlot, and 10 of cabernet (he only drinks red wine), all from different wineries.
(a) If he wants to serve 3 bottles of zinfandel and serving
order is important, how many ways are there to do this?
ways
(b) If 6 bottles of wine are to be randomly selected from the 30
for serving, how many ways are there to do this?
ways
(c) If 6 bottles are randomly selected, how many ways are there to
obtain two bottles of each variety?
ways
(d) If 6 bottles are randomly selected, what is the probability
that this results in two bottles of each variety being chosen?
(Round your answer to three decimal places.)
(e) If 6 bottles are randomly selected, what is the probability
that all of them are the same variety? (Round your answer to three
decimal places.)
In: Math
Assume the binomial distribution is appropriate. If N = 10 and P = 0.30, the probability of getting at least 8 P events is _________.
In: Math
Two players (player A and player B) are playing a game against each other repeatedly until one is bankrupt. When a player wins a game they take $1 from the other player. All plays of the game are independent and identical. Suppose player A starts with $6 and player B starts with $6. If player A wins a game with probability 0.5, what is the probability the game ends (someone loses all their money) on exactly the 10th play of the game?
In: Math
Determine the area under the standard normal curve that lies between:
a. z= -.31 and z=1.61
b. z=-2.01 and z=.26
c. z=-1.20 and z=2.11
d. z=1.92 and z=2.43
Find the value of za
a. z.56=
b. Z.23=
c. Z.81=
d. Z.06=
In: Math
5.) For the data set
2 |
4 |
4 |
5 |
7 |
8 |
9 |
10 |
12 |
12 |
13 |
13 |
16 |
16 |
16 |
16 |
17 |
19 |
19 |
20 |
23 |
24 |
24 |
24 |
25 |
26 |
26 |
27 |
28 |
28 |
29 |
31 |
32 |
34 |
34 |
36 |
37 |
38 |
42 |
44 |
45 |
46 |
47 |
47 |
48 |
50 |
52 |
53 |
53 |
54 |
55 |
56 |
56 |
57 |
58 |
(a) Find the 80th percentile.
The 80t percentile is = |
|
(a) Find the 42nd percentile.
The 42nd percentile is = |
|
(a) Find the 17th percentile.
The 17th percentile is = |
|
(a) Find the 65th percentile.
The 65th percentile is = |
|
6.) Internet providers: In a survey of 672 homeowners with high-speed Internet, the average monthly cost of a high-speed Internet plan was $52.8 with standard deviation $12.42 Assume the plan costs to be approximately bell-shaped. Estimate the number of plans that cost between $40.38 and $65.22
The number of plans that cost between $40.38 and $65.22 is =. |
|
In: Math
The table below lists a random sample of 50 speeding tickets on
I-25 in Colorado.
0-10 mph over limit | 10-20 mph over limit | More than 20 mph over limit | Total | |
Male | 9 | 9 | 17 | 35 |
Female | 2 | 9 | 4 | 15 |
Total | 11 | 18 | 21 | 50 |
Round to the fourth.
a) If a random ticket was selected, what would be the probability
that the driver was female?
b) Given that a particular ticket had a male offender, what is the
probability that they were more than 20 mph over the limit?
c) Given that a particular ticket was 10-20 mph over the limit,
what is the probability that the driver was female?
d) If a random ticket was selected, what would be the probability
that the driver was a male?
e) If a random ticket was selected, what would be the probability
that the driver is a female and driving 10-20 mph over the
limit?
f) If a random ticket was selected, what would be the probability
that the driver is a female or driving 0-10 mph over the limit?
In: Math
Chapter 3 – Numerically Summarizing Data
pH in Water:
The acidity of alkalinity of a solution is measured using pH. A pH less than 7 is acidic; a pH greater than 7 is alkaline. The following data represent the pH in samples of bottled water and tap water.
Tap |
7.64 |
7.45 |
7.47 |
7.50 |
7.68 |
7.69 |
7.45 |
7.10 |
7.56 |
7.47 |
7.52 |
7.47 |
|
Bottled |
5.15 |
5.09 |
5.26 |
5.20 |
5.02 |
5.23 |
5.28 |
5.26 |
5.13 |
5.26 |
5.21 |
5.24 |
pH in Water:
Use the same pH data table in the above question to answer the following.
The Empirical Rule:
SAT Math scores have a bell-shaped distribution with a mean of 515 and a standard deviation of 114.
You Explain it – Percentiles & Quartiles:
One variable that is measured by online homework systems is the amount of time a student spends on homework for each section of the text. The following is a summary of the number of minutes a student spends for each section of the text for the fall 2014 semester in a college statistics class at UHWO.
Q1 = 42 Q2 = 51.5 Q3 = 72.5
Got a Headache?
The following data represent the weight in grams of a random sample of 25 Tylenol tablets.
0.608 |
0.601 |
0.606 |
0.602 |
0.611 |
0.608 |
0.610 |
0.610 |
0.607 |
0.600 |
0.608 |
0.608 |
0.605 |
0.609 |
0.605 |
0.610 |
0.607 |
0.611 |
0.608 |
0.610 |
0.612 |
0.598 |
0.600 |
0.605 |
0.603 |
In: Math
(A review on SAS data management) The following is a data set of 12 individuals. And we want to relate the heart rate at rest (Y) to kilograms body weight (X). X Y 90 62 87 41 87 63 73 46 73 53 86 55 100 70 75 47 76 49 87 69 79 41 78 48 Write a program in SAS which read the given data set into SAS library. In your program, create a new data set which is a copy of the first, with the addition of a new variable which represents the weight in pounds (1 kilogram≈2.2 pounds). Use “proc print” to view the data for all three variables. (c) Create a new variable called weight_group which divides the data set into four groups based on the estimated quartiles for the variable weight. The estimated quartiles may be requested using the keywords q1, median, and q3, in the “proc means” statement. (d) Use “proc freq” to summarize the frequency of subjects in each of the four groups. (e) Obtain the mean and standard deviation for the heart rate at rest for each of the four groups. Comment on the results. (f) Produce a scatterplot of the heart rate at rest versus body weight in kilograms. Describe what you see. (g)Fit a straight line model for heart rate at rest versus body weight in kilograms. Comment on the fit of the line.
In: Math
What does the sigma level capability mean for a process? Explain at least two reasons for the importance of achieving six-sigma capability. explain in detail
In: Math
From your analysis of Desertia, you conclude that citizens there average $1,000 per month in household disposable income. The Minister of Development for Mountania says that disposable income is the same in his country. To see if the data supports this, your company has sampled 100 households from Mountania and obtained data on household disposable income. Use the Data provided to:
a. Derive a frequency distribution of income. Divide by 8
b.Create a histogram of income
c.Obtain and fully interpret/explain the sample mean and standard
deviation
d.Test the hypothesis, at the 5% significance level, that
Mountania also has a mean of $1,000
Disposable Income |
602 |
822.65 |
842.79 |
840.95 |
820.94 |
958.15 |
910.1 |
957.48 |
995.66 |
867.17 |
802.62 |
1008.77 |
745.97 |
905.88 |
940.84 |
754.77 |
960.76 |
897.26 |
835.38 |
925.79 |
948.25 |
872.22 |
719.48 |
883.08 |
1028.29 |
738.2 |
996.32 |
919.36 |
819.94 |
1004.23 |
861.31 |
1067.62 |
847.87 |
1041.8 |
791.61 |
906.73 |
1033.35 |
935.21 |
983.94 |
1004.91 |
993.6 |
970.68 |
794.23 |
948.29 |
949.45 |
804.84 |
936.08 |
986.73 |
784.51 |
922.03 |
1014.22 |
963.19 |
829.64 |
806.42 |
860.37 |
891.54 |
731.69 |
888.42 |
868.95 |
1025.6 |
976.48 |
878.54 |
871.55 |
772.61 |
953.84 |
830.44 |
859.37 |
781.73 |
880.33 |
915.49 |
939.17 |
867.59 |
804.17 |
924.46 |
847.51 |
882.95 |
818.04 |
807.13 |
892.76 |
824.78 |
977.04 |
1079.09 |
895.89 |
872.66 |
953.17 |
994.84 |
842.77 |
1033.59 |
766.47 |
976.48 |
846.98 |
942.39 |
840.77 |
856.07 |
1041.24 |
782.38 |
884.3 |
970.45 |
878.92 961.69 |
In: Math
In: Math
You may need to use the appropriate technology to answer this question.
Test the following hypotheses by using the
χ2
goodness of fit test.
H0: | pA = 0.40, pB = 0.40, and pC = 0.20 |
Ha: | The population proportions are not pA = 0.40, pB = 0.40, and pC = 0.20. |
A sample of size 200 yielded 80 in category A, 20 in category B, and 100 in category C. Use α = 0.01 and test to see whether the proportions are as stated in
H0.
(a)
Use the p-value approach.
Find the value of the test statistic.
Find the p-value. (Round your answer to four decimal places.)
p-value =
State your conclusion.
Reject H0. We conclude that the proportions are equal to 0.40, 0.40, and 0.20.Reject H0. We conclude that the proportions differ from 0.40, 0.40, and 0.20. Do not reject H0. We cannot conclude that the proportions differ from 0.40, 0.40, and 0.20.Do not reject H0. We cannot conclude that the proportions are equal to 0.40, 0.40, and 0.20.
(b)
Repeat the test using the critical value approach.
Find the value of the test statistic.
State the critical values for the rejection rule. (If the test is one-tailed, enter NONE for the unused tail. Round your answers to three decimal places.)
test statistic ≤test statistic ≥
State your conclusion.
Do not reject H0. We cannot conclude that the proportions are equal to 0.40, 0.40, and 0.20.Do not reject H0. We cannot conclude that the proportions differ from 0.40, 0.40, and 0.20. Reject H0. We conclude that the proportions are equal to 0.40, 0.40, and 0.20.Reject H0. We conclude that the proportions differ from 0.40, 0.40, and 0.20.
In: Math