Question
In: Statistics and Probability
Independent random samples of times served for 10 released prisoners for fraud offense showed a mean...
Independent random samples of times served for 10 released prisoners for fraud offense showed a mean of 10.12 months and a standard deviation of 4.90 months whereas times served for 10 released prisoners for firearms offense showed a mean of 18.78 months and a standard deviation of 4.64 months. a) At = 0.05, do the data suggest that mean time served for fraud is less than that of firearm offenses? Assume that times are normally distributed in both groups and have same variances. (Use the critical value approach and the P-value approach) b) Find a 90% confidence interval for the difference between the mean times served by prisoners in the fraud and firearms offense categories. c) Use the Mann-Whitney test at = 0.05 using the data below: Fraud: 3.6 5.3 10.7 8.5 11.8 17.9 5.9 7.0 13.9 16.6 Firearms: 25.5 10.4 18.4 19.6 20.9 23.8 17.9 21.9 13.3 16.1
Solutions
orchestra answered 2 years agoRelated Solutions
Independent random samples of released prisoners in the fraud and firearms offense categories yielded the given...
Independent random samples of released prisoners in the fraud
and firearms offense categories yielded the given information on
time served, in months. At the 1% significance level, do the data
provide sufficient evidence to conclude that the mean time served
for fraud is less than that for firearms offenses? Assume that
populations standard deviations are same for both group. (i.e., we
assume σ1 = σ2 .) What is the parameter of interest? What is the
underlying distribution?
Fraud: 15.2, 11.2,...
URGENT!! Independent random samples of the selling prices (in 10 000 TL ) of houses in...
URGENT!!
Independent random samples of the selling prices (in 10 000 TL )
of houses in three popular districts of four big cities were taken.
The selling prices are as shown below. Assume that the populations
from which the sample drawn are normally distributed. (use
2 digits after decimal point)
District A
District B
District C
Ankara
45
52
54
İstanbul
46
51
58
İzmir
40
60
50
Bursa
37
49
46
1. Complete the following analysis of
variance table....
In a controlled laboratory environment, independent random samples of 10 adults and 10 children were tested...
In a controlled laboratory environment, independent random
samples of 10 adults and 10 children were tested by a psychologist
to determine the room temperature that each person finds most
comfortable. The study provided the following results:
Adults Children
Sample size 10 10
Sample mean (in degrees)
77.5
74.5
Sample variance 4.5 2.5
Which one of the following is the correct 99% confidence
interval for the true difference in population mean temperatures
that adults and children find most comfortable?
Given two independent random samples with the following results:
Given two independent random samples with the following results: n1=170x1=36 n2=123x2=65 Use this data to find the 90% confidence
interval for the true difference between the population
proportions.Step 1 of 4 : Find the values of the two sample proportions, pˆ1
and pˆ2. Round your answers to three decimal places.Step 2 of 4: Find the margin of error. Round your answer to six
decimal places.Step 3 of 4: Construct the 90% confidence interval. Round your
answers to three decimal places.
Table 2 shows two random independent samples of stock return over 10 years’ time. Assume the...
Table 2 shows two random independent samples of stock return
over 10 years’ time. Assume the rate of return ?? for
stock A and ?? for stock B are normally distributed with
population standard deviation ?? = 0.01 and ?? = 0.02 respectively.
To investigate whether the average rate of return for stock A
different from the average rate of return for stock B, please
answer the following questions.
State the null hypothesis and the alternative
hypothesis.
Select an appropriate...
Two random samples are selected from two independent populations. A summary of the samples sizes, sample...
Two random samples are selected from two independent
populations. A summary of the samples sizes, sample means, and
sample standard deviations is given below:
n1=37,n2=44,x¯1=58.9,x¯2=74.7,s1=5.5s2=10.1 n 1 =37, x ¯ 1 =58.9, s
1 =5.5 n 2 =44, x ¯ 2 =74.7, s 2 =10.1 Find a 95.5% confidence
interval for the difference μ1−μ2 μ 1 − μ 2 of the means, assuming
equal population variances. Confidence Interval
Two random samples are selected from two independent populations. A summary of the samples sizes, sample...
Two random samples are selected from two independent
populations. A summary of the samples sizes, sample means, and
sample standard deviations is given below:
n1=39,n2=48,x¯1=52.5,x¯2=77.5,s1=5s2=11 Find a 97.5% confidence
interval for the difference μ1−μ2 of the means, assuming equal
population variances. Confidence Interval =
A random sample of 20 purchases showed the amounts in the table (in $). The mean...
A random sample of 20 purchases showed the amounts in the table
(in $). The mean is $50.76 and the standard deviation is
$19.48
a) Construct a 80% confidence interval for the mean purchases of
all customers, assuming that the assumptions and conditions for
the confidence interval have been met.
b) How large is the margin of error?
c) How would the confidence interval change if you had assumed
that the standard deviation was known to be $20?
60.91
64.3...
A random sample of 20 purchases showed the amounts in the table (in $). The mean...
A random sample of 20 purchases showed the amounts in the table
(in $). The mean is $49.88 and the standard deviation is
$20.86.
B) What is the margin of error ?A) Construct a 80% confidence
interval for the mean purchases of all customers, assuming that
assumptions and conditions for confidence interval have been met.
(What is the confidence interval) ?
C) How would the confidence interval change if you had assumed
that the standard deviation was known to be...
A random sample of 2020 purchases showed the amounts in the table (in $). The mean...
A random sample of
2020
purchases showed the amounts in the table (in $). The mean
is
$49.8149.81
and the standard deviation is
$21.9521.95.
a) Construct a
8080%
confidence interval for the mean purchases of all customers,
assuming that the assumptions and conditions for the confidence
interval have been met.
b) How large is the margin of error?
c) How would the confidence interval change if you had assumed
that the standard deviation was known to be
$2222?
A) The...
ADVERTISEMENT
ADVERTISEMENT
Latest Questions
- In the article from Forbes notes that the goal of a leader is not to clone...
- Many people keep time using a 24 hour clock (11 is 11am and 23 is 11pm,...
- UNIX ONLY -- DIFFICULT LAB ASSIGNMENT This is an extremely difficult subject for me. Could you...
- Do you know why your credit card can be refused even though you are profitable?
- The Following Java expression: S/10 * 10 + 10 – S = C Show that this...
- Year Cash Flow 0 −$9,700 1 5,800 2 2,500 3 3,400 &nb
- In a two – story building, an exterior column is to be designed for a service...
ADVERTISEMENT