10.00 points A simple random sample of 30 observations is derived from a nermall

Question

10.00 points A simple random sample of 30 observations is derived from a nermally distributed population with a known standard deviation of 2.6. Use Table 1 a. Is the condition that X is normally distributed satisfied? O Yes No b. Compute the margin of error with 99% confidence. (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and final answer to 2 decimal places.) Margin of error c. Compute the margin of error with 95% confidence. (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and final answer to 2 decimal places.) d. Which of the two margins of error will lead to a wider interva The margin of error with 95% confidence. o The margin of error with 99% confidence. Hints eBook & Resources

Explanation / Answer

a.
it is approximately since the size large enough
b.
given that,
standard deviation, =2.6
sample mean, x =58.3
population size (n)=30
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 2.6/ sqrt ( 30) )
= 0.4747
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.4747
= 1.2228
c.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.4747
= 0.9304
d.
margin of error with99% CI

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