Assume that a sample is used to estimate a population proportion p. Calculate th
ID: 3319201 • Letter: A
Question
Assume that a sample is used to estimate a population proportion p. Calculate the margin of error E that corresponds to the statistics given and the level of confidence. Round the margin of error to four decimals. 95% confidence, the sample size is 10,000, of which 40% are success
0.0072
0.0110
0.0126
0.0096
Use the confidence level given and the sample data for the construction of a confidence interval for the mu population mean. Assume that the population has a normal distribution. Thirty students selected at random took the final calculation. If the mean of the sample was 83 and the standard deviation was 13.5, build a confidence interval of 99% for the average score of all students.
76.21 <mu <89.79
76.23 <mu <89.77
76.93 <mu <89.07
78.81 <mu <87.19
In a normal distribution:
The median and the fashion are different
The average, the median and the fashion are the same
The mean and the standard deviation are the same
The mean is greater than the deviation
standard
Resolve the problem. The following confidence interval is obtained for a proportion of the population, p: (0.707, 0.745). Use these confidence interval limits to find the margin of error, E.
0.038
0.017
0.019
0.020
Explanation / Answer
QUESTION 2.
TRADITIONAL METHOD
given that,
sample mean, x =83
standard deviation, s =13.5
sample size, n =30
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 13.5/ sqrt ( 30) )
= 2.465
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 29 d.f is 2.756
margin of error = 2.756 * 2.465
= 6.793
III.
CI = x ± margin of error
confidence interval = [ 83 ± 6.793 ]
= [ 76.21 , 89.793 ]
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DIRECT METHOD
given that,
sample mean, x =83
standard deviation, s =13.5
sample size, n =30
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 29 d.f is 2.756
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 = [ 83 ± t a/2 ( 13.5/ Sqrt ( 30) ]
= [ 83-(2.756 * 2.465) , 83+(2.756 * 2.465) ]
= [ 76.21 , 89.793 ]
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interpretations:
1) we are 99% sure that the interval [ 76.207 , 89.793 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
ANSWER: OPTION A(76.21 <mu <89.79)
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