Question
In: Statistics and Probability
Calculate 80%, 90% and 99% confidence intervals for the player weight standard deviation. Sx:17.22429337 and n=40...
Calculate 80%, 90% and 99% confidence intervals for the player weight standard deviation. Sx:17.22429337 and n=40
A) (chi-square right tail)= (chi-square left tail)= 80% confidence interval and interpretation:
b) chi-square right tail)= (chi-square left tail)= 90% confidence interval and interpretation:
c) chi-square right tail)= (chi-square left tail)= 99% confidence interval and interpretation:
d) How do these confidence intervals relate to the population standard deviation?
Solutions
orchestra answered 2 years agoRelated Solutions
Calculate 80%, 90% and 99% confidence intervals for the player weight standard deviation. Include here the...
Calculate 80%, 90% and 99% confidence intervals for the player
weight standard deviation. Include here the sample standard
deviation and sample size -30pts-
S= 16.47 n= 50 〖
χ²〗_R=(chi-square right tail)=
〖χ²〗_L=(chi-square left tail)=
80% confidence interval and interpretation:
〖χ²〗_R=
〖χ²〗_L=
90% confidence interval and interpretation:
〖χ²〗_R=
〖χ²〗_L=
99% confidence interval and interpretation:
How do these confidence intervals relate to the population
standard deviation?
Calculate 80%, 90% and 99% confidence intervals for the standard deviation. Include here the sample standard...
Calculate 80%, 90% and 99% confidence intervals for the standard
deviation. Include here the sample standard deviation and sample
size -30pts-
S=
n=
〖χ²〗_R=(chi-square right tail)=
〖χ²〗_L=(chi-square left tail)=
80% confidence interval:
Interpretation:
〖χ²〗_R= 〖χ²〗_L= 90% confidence interval:
Interpretation:
〖χ²〗_R= 〖χ²〗_L= 99% confidence interval:
Interpretation:
How do these confidence intervals relate to the population
standard deviation you calculated in question #1, when the set of
data was treated as a population? -5 pts-
Use the sample you chose to perform...
Find the mean, standard deviation, 80%, 95%, and 99% confidence intervals, and margin of error for...
Find the mean, standard deviation,
80%, 95%, and 99% confidence intervals, and margin of error for
those intervals. Show all work.
$32,096
$29,716
$29,227
$27,209
$22,794
$19,306
$19,238
$19,164
18,994
$18,329
$17,978
$17,626
$16,823
$16,709
$15,967
$15,921
$15,908
$15,905
$15,638
$15,629
$14,414
$14,229
$14,137
$14,051
$13,907
$13,776
$13,281
$12,739
$12,300
$11,994
$11,262
$11,112
$10,984
$10,824
$10,823
$10,448
$10,081
$9,883
$9,668
$9,663
$9,337
$8,855
$8,741
$8,596
$8,385
$7,204
$7,072
$6,503
$5,653
$4,602
mean 94.6129 n= 31 standard deviation is 45.0612 find the 95% and 99% confidence intervals of...
mean 94.6129
n= 31
standard deviation is 45.0612
find the 95% and 99% confidence intervals of the standard
devaition
find the 95% and 99% confidence interval for population
mean
data set
10
19
35
39
41
56
67
67
69
73
75
78
81
82
83
83
84
91
92
94
123
125
137
138
140
142
150
152
161
163
183
uConstruct confidence intervals for the population mean of 80%, 90%, 95%, 99% using the following data...
uConstruct confidence intervals for the population mean of 80%,
90%, 95%, 99% using the following data and a population standard
deviation of 900:
un = 100
u?x ̅ = 425
Calculate and interpret 90%, 95%, and 99% confidence intervals for both groups of approaches in question...
Calculate and interpret 90%, 95%, and 99% confidence intervals
for both groups of approaches in question 5. Also provide 90%, 95%,
and 99% confidence intervals for both groups when the sample size
is increased from 10 to 150. Interpret all statistics. a. A random
sample of 15 families representing three social classes has been
observed for the frequency with which the parents administer
physical punishment to the children over a period of a week. Are
the differences significant? Use the...
Age of College Students Find the 80% confidence intervals for the variance and standard deviation of...
Age of College Students Find the 80% confidence
intervals for the variance and standard deviation of the ages of
seniors at Oak Park College if a random sample of 23 students has a
standard deviation of 2.9 years. Assume the variable is normally
distributed. Use the chi-square distribution table to find any
chi-square values to three decimal places. Round the final answers
to one decimal place.
Mean=33,820.21 Standard Deviation=22,948.45 n=52 Calculate a 99% confidence interval, assuming that sigma is unknown.
Mean=33,820.21
Standard Deviation=22,948.45
n=52
Calculate a 99% confidence interval, assuming that sigma is
unknown.
Determine the mean and standard deviation of your sample. Find the 80%, 95%, and 99% confidence...
Determine the mean and standard deviation of your sample. Find
the 80%, 95%, and 99% confidence intervals. Make sure to list the
margin of error for the 80%, 95%, and 99% confidence interval.
Create your own confidence interval (you cannot use 80%, 95%, and
99%) and make sure to show your work. Make sure to list the margin
of error. What trend do you see takes place to the confidence
interval as the confidence level rises? Explain mathematically why
that...
Determine the mean and standard deviation of your sample. Find the 80%, 95%, and 99% confidence...
Determine the mean and standard deviation of your sample.
Find the 80%, 95%, and 99% confidence intervals.
Make sure to list the margin of error for the 80%, 95%, and 99%
confidence interval.
Create your own confidence interval (you cannot use 80%, 95%,
and 99%) and make sure to show your work. Make sure to list the
margin of error.
Problem Analysis—Write a half-page reflection.
What trend do you see takes place to the confidence interval as
the confidence level...
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