A random sample of n (shown on right) business students was taken from a normal

Question

A random sample of n (shown on right) business students was taken from a normal population with unknown standard deviation resulting in a sample mean, x (shown on right,) and sample standard deviation, s (shown on right). What would the Margin of Error be for a (1-)100% [ is shown on right] two tailed confidence interval be? ROUND THE t or Z STATISTIC OFF TO 2 DECIMAL PLACES [1.645 rounds off to 1.64]. n = 49 s = 4 = 0.1 x = 20 a. 2.286 b. 0.937 c. 0.960 d. 0.057 e. none A random sample of n (shown on right) business students was taken from a normal population with unknown standard deviation resulting in a sample mean, x (shown on right,) and sample standard deviation, s (shown on right). What would the Margin of Error be for a (1-)100% [ is shown on right] two tailed confidence interval be? ROUND THE t or Z STATISTIC OFF TO 2 DECIMAL PLACES [1.645 rounds off to 1.64]. n = 49 s = 4 = 0.1 x = 20 a. 2.286 b. 0.937 c. 0.960 d. 0.057 e. none

Explanation / Answer

given that,
sample mean, x =20
standard deviation, s =4
sample size, n =49
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4/ sqrt ( 49) )
= 0.571
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 48 d.f is 1.677
margin of error = 1.677 * 0.571
= 0.958 ~ 0.960

ANSWER: Option C

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