Express the confidence interval (0.432, 0.52 in the form of ± e.

1 answer:

4 0

For a known standard deviation, the confidence interval for sample size = n is
$(x-z \frac{ \sigma }{ \sqrt{n}},x+x \frac{ \sigma }{ \sqrt{n} } )$
where
x = average
n = sample size
$\sigma$ = stad. deviation
z = contant that reflects confidence interval

Let a = x
Let b = $z \frac{ \sigma }{ \sqrt{n} }$

From the given information,
a - b = 0.432 (1)
a + b = 0.52 (2)

Add (1) and (2): 2a = 0.952 => a = 0.476
Subtract (2) from (1): -2b = -0.088 => b = 0.044

Therefore, the confidence interval may be written as
(0.476 - 0.044, 0.476 + 0.044), or as
(0.476 $\pm$ 0.044)

You might be interested in

If the integral of f(x)dx = 2.3 from 1 to 3 and F'(x) = f(x), what is the value of F(3) - F(0). The graph will be inserted as a

lana66690 [7]

F(3)−F(0)=∫30f(x)dx
so it looks like 2.3+1=3.3

4 0

3 years ago

Read 2 more answers

The circumference of the equator of a sphere was measured to be 82 82 cm with a possible error of 0.5 0.5 cm. Use linear approxi

True [87]

Answer:

The maximum error in the calculated surface area is approximately 8.3083 square centimeters.

Step-by-step explanation:

The circumference ( $s$ ), in centimeters, and the surface area ( $A_{s}$ ), in square centimeters, of a sphere are represented by following formulas:

$A_{s} = 4\pi\cdot r^{2}$ (1)

$s = 2\pi\cdot r$ (2)

Where $r$ is the radius of the sphere, in centimeters.

By applying (2) in (1), we derive this expression:

$A_{s} = 4\pi\cdot \left(\frac{s}{2\pi} \right)^{2}$

$A_{s} = \frac{s^{2}}{\pi^{2}}$ (3)

By definition of Total Differential, which is equivalent to definition of Linear Approximation in this case, we determine an expression for the maximum error in the calculated surface area ( $\Delta A_{s}$ ), in square centimeters: