Question

Find the value of E, the margin of error, for c = 0.90, n = 16 and s = 2.2.

Answer #1

b )Given that,

s =2.2

n = 16

Degrees of freedom = df = n - 1 =16 - 1 = 15

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,15 = ( using student t table)

Margin of error = E = t/2,df * (s /n)

= **1.753** * ( 2.2/
16) = 0.964

Find the value of E, the margin of error, for c equals 0.99,
n equals 10 and s equals 3.6.

Find the margin of error for the given values of c, σ, and
n.
c=0.90, σ=3.3, n=81.
Eequals= _ (Round to three decimal places as needed.)

Find the Margin of Error for each problem:
1) c = .95; n =
30; s = 3.5 Margin of Error=
2) c = .85; n =
18; s = 0.49 Margin of Error=
3) c = .95; n =
12; s = 12.6 Margin of Error=
4) c = .99; n =
26; s = 18.9 Margin of Error=
Round answers to three decimal places.

Find the margin of error for the given values of c, s, and
n.
c=0.80, s=2.3, n=10

Find the margin of error for the given values of c, s, and
n.
c=0.80,
s=3.4,
n=14

If n=16, x¯(x-bar)=37, and s=6, find the margin of error at a
80% confidence level

If n=27, ¯ x (x-bar)=46, and s=9, find the margin of error at a
95% confidence level (use at least two decimal places)
If n=19, ¯xx¯ (x-bar)=48, and s=6, find the margin of error at a
95% confidence level (use at least two decimal places)
If n=16, ¯xx¯ (x-bar)=48, and s=6, find the margin of error at a
99% confidence level (use at least three decimal places)

Find the margin of error for the given values of c, s, and
n.
cequals=0.8080,
sequals=66,
nequals=2626
LOADING...
Click the icon to view the t-distribution table.
The margin of error is?
(Round to one decimal place as needed.)

if n=27, x bar =39, s=3, find the margin of error at a 98%
confidence interval.

Find the margin of error for the given confidence level and
values of x and n. x = 80, n = 146, confidence level 99%

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