A researcher studied the relationship between the salary of a working woman with school-aged children and the number of children she had. The results are shown in the following frequency table:
Number of Children
Salary |
2 or fewer children |
more than 2 children |
high salary |
13 |
2 |
medium salary |
20 |
10 |
low salary |
30 |
25 |
If a working woman has more than 2 children, what is the probability she has a low or medium salary?
A. 0.79 B. 0.45 C. 0.35 D. 0.95
In: Math
Q 1.An online retailer, Mr Collins Ndhlovu, has two adverts posted in different parts of a well-known social networking website, Advertisement A and Advertisement B. An average of 2 ‘clicks’ are generated by Advertisement A during the period Monday 10.00 to 10.05am. There are on average 5 ‘clicks’ generated by Advertisement B during the same period. Calculate the probability that on a particular Monday between 10.00 and 10.05 am: i)Advertisement A generates at most 3 clicks. ii)Advertisement A generates at least 4 clicks. ii)Advertisement B generates no more than 4 clicks. iv)Advertisement A generates exactly 2 clicks and Advertisement B exactly 2 clicks. v)At least 3 clicks are generated in total by the two advertisements. (5marks)
In: Math
Explain the difference between a confidence interval and a prediction interval?
In: Math
The National Association of Home Builders provided data on the
cost of the two most popular home remodeling projects. Sample data
on cost in thousands of dollars for two types of remodeling
projects are as follows.
Kitchen | Master Bedroom | Kitchen | Master Bedroom | |
27.0 | 18.0 | 23.0 | 17.8 | |
17.4 | 21.1 | 19.7 | 24.6 | |
22.8 | 26.4 | 16.9 | 22.0 | |
21.9 | 24.8 | 21.8 | ||
21.0 | 25.4 | 19.0 |
Using Kitchen as population 1 and Master Bedroom as population
2, develop a point estimate of the difference between the
population mean remodeling costs for the two types of projects (to
1 decimal).
$ thousand
Develop a 90% confidence interval for the difference between the
two population means (to 1 decimal). Use z-table.
( , )
In: Math
A recent survey reported that 39% of 18- to 29-year-olds in a certain country own tablets. Using the binomial distribution, complete parts (a) through (e) below.
a. What is the probability that in the next six 18- to 29-year-olds surveyed, four will own a tablet?
The probability is ? (Type an integer or a decimal. Round to four decimal places as needed.)
b. What is the probability that in the next six 18- to 29-year-olds surveyed, all six will own a tablet?
c. What is the probability that in the next six 18- to 29-year-olds surveyed, at least four will own a tablet?
d. What are the mean and standard deviation of the number of 18- to 29-year-olds who will own a tablet in a survey of six?
e. What assumption(s) do you need to make in (a) through (c)?
In: Math
The computer that controls a bank's automatic teller machine crashes a mean of 0.4 0.4 times per day. What is the probability that, in any seven-day week, the computer will crash less than 4 4 times? Round your answer to four decimal places.
In: Math
Describe the kind of data that are collected for an independent-measures t-test and the hypotheses that the test evaluates. The key to helping formulate your explanation would be to include the assumptions of this statistical model, the type of sample used in this model, and a statement about the null hypothesis.
In: Math
Calculate a geometric series
1.For v greater than 0 and less than 1, what is the Sum(v^i) for i = 1 to 100?
2.For v greater than 0 and less than 1, what is the Sum(v^i) for i = 1 to infinity?
3.Let v = 1/(1+r). State the answer to question 2 in terms of r.
In: Math
Question 1 contains the actual values for 12 periods (listed in order, 1-12). In Excel, create forecasts for periods 6-13 using each of the following methods: 5 period simple moving average; 4 period weighted moving average (0.63, 0.26, 0.08, 0.03); exponential smoothing (alpha = 0.23 and the forecast for period 5 = 53); linear regression with the equation based on all 12 periods; and quadratic regression with the equation based on all 12 periods. Round all numerical answers to two decimal places.
1. The actual values for 12 periods (shown in order) are:
(1)
45 (2)
52
(3)
48
(4)
59 (5)
55 (6)
54 (7)
64 (8)
59 (9)
72 (10)
66 (11)
67 (12)
78
Using a 5 period simple moving average, the forecast for period 13
will be:
2. Using
the 4 period weighted moving average, the forecast for period 13
will be:
3. With
exponential smoothing, the forecast for period 13 will be
4.
With linear regression, the forecast for period 13 will be:
5. With
quadratic regression, the forecast for period 13 will be:
6. Considering
only the forecasts for period 6-12, what is the lowest MAD value
for any of the methods?
In: Math
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