Question

Find the margin of error for the given confidence level and values of x and n. x=79, n=187, 98% confidence level

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 79 / 187 = 0.422

1 - = 1 - 0.422 = 0.578

Z/2 = 2.326

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.326 * (((0.422 * 0.578) / 187)

Margin of error = E = **0.084**

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

For the given confidence level and values of x and n, find the
following.
x=47,n=98 , confidence level 99.9%
Part: 0 / 30 of 3 Parts Complete
Part 1 of 3
(a) Find the point estimate. Round the answers to at least four
decimal places, if necessary.
The point estimate for the given data is _?
.

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)

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

If n=18, ¯x(x-bar)=47, and s=4, find the margin of error at a
80% confidence level
Give your answer to two decimal places.

If n=29, ¯x(x-bar)=39, and s=5, find the margin of error at a
80% confidence level
Give your answer to two decimal places.

If n=18, ¯ x x ¯ (x-bar)=30, and s=2, find the margin of error
at a 98% confidence level.
Assume the t distribution must be used. Give your answer to two
decimal places.

1. If n=23, x¯ (x-bar)=32, and s=5, find the
margin of error at a 95% confidence level (use at least two decimal
places)
2. If n=24, ¯x¯ (x-bar)=47, and s=4, find the
margin of error at a 99% confidence level (use at least three
decimal places)

If n=26, ¯xx¯ (x-bar)=50, and s=6, find the margin of error at a
99% confidence level (use at least three decimal places)

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