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5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g

anygoal [31]

Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{1117}{10} = 111.7$

Sum of squares of differences = 642.1

$S.D = \sqrt{\frac{642.1}{49}} = 8.44$

We are given the following in the question:

Population mean, μ = 106

Sample mean, $\bar{x}$ = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 106\\H_A: \mu > 106$

We use one-tailed(right) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135$

Now,

$t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833$

Since,

$t_{stat} > t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

4 0

3 years ago

Solve x Log 10×=5.9 ×=

Alja [10]

Answer:

x=0.59

Step-by-step explanation:

because if u divide 5.9 by 10 it will give u 0.59

and isolate x

8 0

3 years ago

Complex numbers are often used when dealing with alternating current (AC) circuits. In the equation V = IZ, V is voltage, I is c

gizmo_the_mogwai [7]

$V=IZ\\\\1+i=I\cdot(2-i)\quad|:(2-i)\\\\\\I=\dfrac{1+i}{2-i}=\dfrac{(1+i)(2+i)}{(2-i)(2+i)}=\dfrac{2+i+2i+i^2}{2^2-i^2}=\dfrac{2+3i-1}{4-(-1)}=\dfrac{1+3i}{5}\\\\\\\\\boxed{I=\frac{1+3i}{5}=\frac{1}{5}+\frac{3}{5}i}$

4 0

3 years ago

Pleaseeeeeeeee help me

wlad13 [49]

The answer is OH and IH

3 0

3 years ago

Read 2 more answers

field has a width of 70m and a length of 105m, each correct to the nearest five metres. What is the lower bound for the area of

Zina [86]

Answer: Find the best estimate of the perimeter of the field.

8 0

2 years ago