Estimate 94'903+49'995

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

Convert 1/2 yard in inches

Tom [10]

1/2 yard =18 inches

(multiply value by 36)

8 0

3 years ago

Three components are randomly sampled, one at a time, from a large lot. As each component is selected, it is tested. If it passe

satela [25.4K]

Answer:

The probability of SFS and SSF are same, i.e. P (SFS) = P (SSF) = 0.1311.

Step-by-step explanation:

The probability of a component passing the test is, P (S) = 0.79.

The probability that a component fails the test is, P (F) = 1 - 0.79 = 0.21.

Three components are sampled.

Compute the probability of the test result as SFS as follows:

P (SFS) = P (S) × P (F) × P (S)

$=0.79\times0.21\times0.79\\=0.131061\\\approx0.1311$

Compute the probability of the test result as SSF as follows:

P (SSF) = P (S) × P (S) × P (F)

$=0.79\times0.79\times0.21\\=0.131061\\\approx0.1311$

Thus, the probability of SFS and SSF are same, i.e. P (SFS) = P (SSF) = 0.1311.

7 0

3 years ago

Plz help me :^(( tyy

Whitepunk [10]

Plug in 1 for x
f(1) = |1-3| + 7
|1-3| = |-2| = 2
f(1) = 2 + 7
Solution: f(1) = 9

5 0

3 years ago

A company that produces cranberries knows that about 7%7\%7%7, percent of its cranberries are bruised in the bagging process. Th

lozanna [386]

Answer:

B. Yes, because 0.004<0.010.004<0.010, point, 004, is less than, 0, point, 01.

Step-by-step explanation:

Khan Academy

3 0

3 years ago