|
Descriptives |
||||||||
|
Current stress level |
||||||||
|
N |
Mean |
Std. Deviation |
Std. Error |
95% Confidence Interval for Mean |
Minimum |
Maximum |
||
|
Lower Bound |
Upper Bound |
|||||||
|
Psychologist |
5 |
1.0000 |
1.73205 |
.77460 |
-1.1506 |
3.1506 |
.00 |
4.00 |
|
Doctor |
5 |
5.0000 |
2.23607 |
1.00000 |
2.2236 |
7.7764 |
2.00 |
8.00 |
|
Lawyer |
5 |
6.0000 |
1.87083 |
.83666 |
3.6771 |
8.3229 |
4.00 |
9.00 |
|
Total |
15 |
4.0000 |
2.87849 |
.74322 |
2.4059 |
5.5941 |
.00 |
9.00 |
|
Test of Homogeneity of Variances |
|||
|
Current stress level |
|||
|
Levene Statistic |
df1 |
df2 |
Sig. |
|
.170 |
2 |
12 |
.845 |
|
ANOVA |
|||||
|
Current stress level |
|||||
|
Sum of Squares |
df |
Mean Square |
F |
Sig. |
|
|
Between Groups |
70.000 |
2 |
35.000 |
9.130 |
.004 |
|
Within Groups |
46.000 |
12 |
3.833 |
||
|
Total |
116.000 |
14 |
|||
|
*. The mean difference is significant at the 0.05 level. |
1. Write up the results from this analysis using APA format. Make sure you include each group mean (and SD) exam score, the correct F stat with the correct degrees of freedom and p-value, decision regarding the null, and the follow-up comparisons (if necessary). Use the example in the SPSS ANOVA lecture to help guide you. [3 points]
In: Math
1) Suppose that E(Y∣X)=X^2. Then E(Y/X) is equal to which of the following?
a) 1 b) E(X) c) E(X^2) d) E(Y)
2)Var(Y∣X=x) is less than or equal to Var(Y) unless Var(Y)=0. True or False?
In: Math
8-5 (30) X and Y are independent random variables, and both are normally distributed with mean zero and variance one. Two new random variables. Z and W are defined by Z = X2 + Y2 , W = X/Y
(14) a). Find fz.w(z.w)indicating the domains over which it is defined.
(6) b). Are Z and W independent? Explain your answer.
(10) c) Find the marginal densities fW (w) and fZ (z) and the domain of each.
In: Math
Neutropenia is an abnormally low number of neutrophils in the blood. Chemotherapy often reduces the number of neutrphils to a level that makes patients susceptible to fever and infections. G. Bucaneve et al. published a study of such cancer patients in the paper "Levofloxacin to Prevent Bacterial Infection in Patients With Cancer and Neutropenia" (New England Journal of Medicine, Vol. 353, No. 10, pp. 977-987). For the study, 375 patients were randomly assigned to receive a daily dose of levofloxacin, and 363 were given a placebo. In the group receiving levofloxacin, fever was present in 243 patients for the duration of neutropenia, whereas fever was experienced by 308 patients in the placebo group. (Source: Elementary Statistics, Weiss, 8th Edition)
Calculate the margin of error for a 95% confidence interval for the difference in the proportion of patients with fever between patients taking levofloxacin and those not taking anything while suffering from neutropenia. (round proportions to hundredths place, round the critical value to the hundredths place, the standard error to the thousandths place , and the margin of error to the thousandths place)
In: Math
In clinical tests of adverse reactions to the drug
Viagra, 51 of the 734 subjects in the treat
ment group experienced dyspepsia (indigestion) and
22 of the 725
subjects in the placebo group experienced dyspepsia (based on data from Pfizer Pharmaceuticals). Using a
0.05 significance level, test the claim that the proportion of dyspepsia cases among V
iagra users (group 1)
is greater than the proportion of dyspepsia cases among non
-Viagra users (group 2).
a)
What proportion of Viagra users experienced dyspepsia?
A. 0.030
B. 0.069
C. 0.073
D. 0.050
b) Based on the description above, w
hat
are the
null
and alternative hypothes
es, respectively
?
A.
p
1
=
p
2
,
p
1
≠
p
2
B.
p
1
>
p
2
, p1 < p2
C.
p
1
≤
p
2
,
p
1
>
p
2
D.
p
1
<
p
2
,
p
1
≥
p
2
c)
What is the value of the calculated z test statistic use
d to test the given hypothesis?
A. 0.048
B. 0.59
C. 3.43
D. 11.75
5
d)
What is the
p
-
value (corresponding to your test statistic)?
A. 0.0003
B. 0.05
C. 0.95
D. 0.9997
E. Cannot be determined from the given information
e)
What conclusion should you make based on the given data?
A. Reject H
0
and conclude there is not sufficient evidence to support the claim that dyspepsia occurs at a higher
rate among Viagra users than those who do not use Viagra.
B. Reject H
0
and conclude there is sufficient evidence to support the
claim that dyspepsia occurs at a
higher rate among Viagra users than those who do not use Viagra.
C. Accept H
0
and conclude there is sufficient evidence to support the claim that dyspepsia occurs at a
higher rate among Viagra users than those who do not
use Viagra.
D. Fail to reject H
0
and conclude there is not sufficient evidence to support the claim that dyspepsia
occurs at a higher rate among Viagra users than those who do not use Viagra.
In: Math
The time taken by an automobile mechanic to complete an oil change is random with mean 29.5 minutes and standard deviation 3 minutes.
a. What is the probability that 50 oil changes take more than 1500 minutes?
b. What is the probability that a mechanic can complete 80 or more oil changes in 40 hours?
c. The mechanic wants to reduce the mean time per oil change so that the probability is 0.99 that 80 or more oil changes can be completed in 40 hours. What does the mean time need to be? Assume the standard deviation remains 3 minutes.
In: Math
Distinguish between a survey and experimental design in quantitative research.
In: Math
Does Red Increase Men’s Attraction to Women?
A study1 examines the impact of the color red on how
attractive men perceive women to be. In the study, men were
randomly divided into two groups and were asked to rate the
attractiveness of women on a scale of 1 (not at all attractive) to
9 (extremely attractive). Men in one group were shown pictures of
women on a white background while the men in the other group were
shown the same pictures of women on a red background. The results
are shown in Table 1 and the data for both groups are reasonably
symmetric with no outliers.
| Color | n | x¯ | s |
|---|---|---|---|
| Red | 15 | 7.2 | 0.6 |
| White | 12 | 6.1 | 0.4 |
Table 1 Does red increase men’s attraction to women?
To determine the possible effect size of the red background over
the white, find a 99% confidence interval for the difference in
mean attractiveness rating μR-μW, where μR represents the mean
rating with the red background and μW represents the mean rating
with the white background.
Round your answers to two decimal places.
The 99% confidence interval is
In: Math
calculate the probability of a person liking a dark-colored imported car over a light-colored imported car. Your answers are probabilities. Show your work.
The Dependent Variable (DV) is "Prefers Dark colored
imported car." This measure is labeled"PrefDark" in the
data
= 0 if preference is for a light colored car,
= 1 if preference is for a dark-colored car.
Here are the Independent Variables (IVs):
Age in years (no intervals – labeled "Age" in the data)
Gender (measure is labeled "Gender" in the data)
= 0 if male,
= 1 if female.
Education level (measure is labeled EducLevel in the data)
= 0 if completed high school only
= 1 if completed Associate's degree (Community College)
= 2 if completed Undergraduate degree (BA or BS)
= 3 if completed a Graduate degree
Income per year (in Euros, measure is labeled Income))
Consider, also, these coefficients for each measure (data point), calculated by running a Logit analysis on the data sample for the DV, PrefDark:
Coefficients and Constant
Age
0.101
Gender 0.34
EducLevel –5.1
Income 0.000142
Constant 3.22
Assume all coefficients and the constant are statistically significant (you can't ignore them).
Part 1:
Now consider this person, Respondent 1:
Age = 24
Gender = 1 (female)
EducLevel = 2 (Undergraduate degree)
Income/year = Euros 38000
What is the probability this person prefers a dark-colored
imported car?
Part 2:
Additionally, consider this other person, Respondent 2:
54 year old male, with a graduate degree, earning Euros 58000 per
year.
What is the probability this person prefers a dark-colored
imported car?
Hint: Use the formula given in the video for calculating P(Yi=yi).
Show your work, please.
Part 3:
Which Respondent has a higher probability of preferring a
dark-colored car?
This is quite straightforward if you have Parts 1 and 2
correct.
In: Math
The process has an average of 40grams and a standard deviation of
2grams. With a confidence level of 90%, determine the average was
reduced. Samples: 38, 39, 42, 40, 39, 41, 42, 40, 41, 42. Determine
the 5 steps and the p-value of the hypothesis test.
In: Math
solve the problem make sure to explain in words what you did with the problem and state your conclusions in terms of the problem.
Part I: Choose to do one of the following: 1) Test the claim that the mean Unit 3 Test scores of data set 7176 is greater than the mean Unit 3 Test scores of data set 7178 at the .05 significance level
| Class 7176 | Class 7178 | |||||
| Unit Test 3 | Course Grade | Attendance | Unit Test 3 | Course Grade | Attendance | |
| 238 | 63 | 96 | 291 | 95 | 100 | |
| 208 | 55 | 48 | 301 | 91 | 83 | |
| 258 | 89 | 96 | 261 | 68 | 87 | |
| 264 | 84 | 96 | 0 | 53 | 91 | |
| 324 | 98 | 100 | 0 | 23 | 44 | |
| 0 | 62 | 44 | 284 | 93 | 96 | |
| 0 | 56 | 66 | 307 | 77 | 78 | |
| 274 | 87 | 96 | 0 | 44 | 70 | |
| 274 | 83 | 96 | 208 | 72 | 57 | |
| 0 | 0 | 18 | 0 | 56 | 78 | |
| 179 | 71 | 100 | 0 | 73 | 91 | |
| 268 | 86 | 100 | 0 | 28 | 39 | |
| 241 | 60 | 87 | 231 | 64 | 91 | |
| 0 | 8 | 26 | 307 | 87 | 100 | |
| 278 | 84 | 96 | 301 | 87 | 96 | |
| 307 | 89 | 87 | 228 | 52 | 74 | |
| 294 | 87 | 100 | 255 | 73 | 70 | |
| 175 | 76 | 74 | 304 | 85 | 83 | |
| 129 | 66 | 87 | 0 | 37 | 44 | |
| 284 | 82 | 100 | 0 | 60 | 48 | |
| 297 | 90 | 79 | 0 | 25 | 35 | |
| 255 | 74 | 74 | 0 | 37 | 48 | |
| 268 | 88 | 100 | 0 | 67 | 100 | |
| 215 | 77 | 39 | 301 | 94 | 87 | |
| 146 | 71 | 87 | 284 | 71 | 57 | |
| 304 | 88 | 100 | 321 | 87 | 100 | |
| 311 | 91 | 96 | ||||
| 274 | 83 | 96 | ||||
| 307 | 97 | 100 | ||||
| 278 | 92 | 91 | ||||
| 0 | 52 | 91 | ||||
In: Math
Comparing Population Measures of Center and Dispersion
The HDL cholesterol (in mg/dL) of 10 males and 10 females were recorded for random samples of Americans as part of a National Center for Health Statistics survey.
|
Female |
Male |
|
74 |
44 |
|
56 |
41 |
|
70 |
71 |
|
40 |
41 |
|
67 |
57 |
|
96 |
50 |
|
43 |
60 |
|
80 |
47 |
|
77 |
44 |
|
41 |
33 |
1. What is the population for this survey?
2. Find the mode and the range for each data set:
Female and Male
3. Find the 5 number summary for each data set.
For male and female:
Minimum, 1st Quartile, Median, 3rd Quartile, Maximum
4. Find the mean for each data set:
Female:_________ Male:_________
5. Find the standard deviation for each data set.
Female:_________Male:______ 6. Assuming the population standard deviations are σ = 15 mg/dL for females and σ = 12 mg/dL for males, construct the 95% confidence intervals for HDL cholesterol for each group using the data. Write a sentence that explains the correct interpretation of each confidence interval.
7. Use your confidence intervals to decide if it is possible that the population mean HDL
cholesterol is the same for females and males. Briefly explain the logic behind your decision.
In: Math
For patients who have been given a diabetes test, blood-glucose readings are approximately normally distributed with mean 128 mg/dl and a standard deviation 10 mg/dl. Suppose that a sample of 4 patients will be selected and the sample mean blood-glucose level will be computed.
Enter answers rounded to three decimal places. According to the empirical rule, in 95 percent of samples the SAMPLE MEAN blood-glucose level will be between the lower-bound of _____ and the upper-bound of _____
In: Math
Fat contents (in percentage) for 10 randomly selected hot dogs
were given in the article "Sensory and Mechanical Assessment of the
Quality of Frankfurters". Use the following data to construct a 90%
confidence interval for the true mean fat percentage of hot dogs:
(Give the answers to two decimal places.)
( , )
|
26.0 |
22.1 |
23.6 |
17.0 |
30.6 |
21.8 |
26.3 |
16.0 |
20.9 |
19.5 |
In: Math
create a histogram of with the data. One relatively easy way to do this is to divide the counts into 10 groups, say, each of length: (max length - min length)/10. Then compute the frequency of the data in each bin, and plot.
data: 143.344, 178.223, 165.373, 154.768, 155.56, 163.88, 178.99, 145.764, 174.974, 136.88, 173.84, 174.88, 197.091, 183.222, 138.233
please show work
In: Math