mathxl.com Find The Variance For The Given Data: 13. Homework Do Homework - Mird
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mathxl.com Find The Variance For The Given Data: 13. Homework Do Homework - Mirdiana Bahtiri STA2023-Fall 2017, Class Nbr. 6533rdiana Bahtiri 12/17/17 7:54 PM Homework: HW Ch 7 Score: 0 of 1 pt 7.4.9 Save 9 of 12 (3 complete) HW Score: 25%, 3 of 12 pts Question Help * Use the given confidence level and sample data to find a confidence interval for the population standard deviation o. Assume that a simple random sample has been selected from a population that has a normal distribution. Salaries of college graduates who took a geology course in college 80% confidence; n-41, x-$70,800, s-s16247 click the icon to view the table of Chi-Square critical values. Round to the nearest dollar as needed.)Explanation / Answer
7.4.9
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s = standard deviation
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since alpha =0.05
^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 40 df are 59.3417 , 24.433
s.d( s^2 )=16247
sample size(n)=41
confidence interval for ^2= [ 40 * 263965009/59.3417 < ^2 < 40 * 263965009/24.433 ]
= [ 10558600360/59.3417 < ^2 < 10558600360/24.433 ]
[ 177928848.6848 < ^2 < 432145064.462 ]
and confidence interval for = sqrt(lower) < < sqrt(upper)
= [ sqrt (177928848.6848) < < sqrt(432145064.462), ]
= [ 13338.9973 < < 20788.0991 )
= $13339< < 20788$
7.3.11
TRADITIONAL METHOD
given that,
sample mean, x =1.55
standard deviation, s =1.8993
sample size, n =6
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.8993/ sqrt ( 6) )
= 0.775
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 5 d.f is 2.571
margin of error = 2.571 * 0.775
= 1.994
III.
CI = x ± margin of error
confidence interval = [ 1.55 ± 1.994 ]
= [ -0.444 , 3.544 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =1.55
standard deviation, s =1.8993
sample size, n =6
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 5 d.f is 2.571
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 1.55 ± t a/2 ( 1.8993/ Sqrt ( 6) ]
= [ 1.55-(2.571 * 0.775) , 1.55+(2.571 * 0.775) ]
= [ -0.444 , 3.544 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ -0.444 , 3.544 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
7.3.38
TRADITIONAL METHOD
given that,
standard deviation, =19626
sample mean, x =62800
population size (n)=50
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 19626/ sqrt ( 50) )
= 2776
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 2776
= 4566
III.
CI = x ± margin of error
confidence interval = [ 62800 ± 4566 ]
= [ 58234,67366 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =19626
sample mean, x =62800
population size (n)=50
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 62800 ± Z a/2 ( 19626/ Sqrt ( 50) ) ]
= [ 62800 - 1.645 * (2776) , 62800 + 1.645 * (2776) ]
= [ 58234,67366 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [58234 , 67366 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 62800
standard error =2776
z table value = 1.645
margin of error = 4566
confidence interval = [ 58234 , 67366 ]
7.2.29
assume sample proportion =0.5
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.1 is = 1.645
Sample Proportion = 0.5
ME = 0.05
n = ( 1.645 / 0.05 )^2 * 0.5*0.5
= 270.6025 ~ 271
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