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amm1812
2 years ago
15

QUESTION 17 - 1 POINT

SAT
1 answer:
Triss [41]2 years ago
6 0

Answer:

PREGUNTA 17 - 1 PUNTO

Se sabe que leer un nombre en voz alta puede mejorar la memoria. A un psicólogo le gustaría probar la afirmación de que los adultos pueden

recordar más del 55 por ciento de los nombres de una lista, en promedio, después de leer la lista en voz alta. Para probar esta afirmación en el

nivel de significación, el psicólogo recopila los siguientes datos en una muestra de 50 participantes y registra el porcentaje de

nombres recordados en la lista después de leerla en voz alta. Los siguientes son los datos de este estudio:

Tamaño de la muestra = 50 participantes

Media de la muestra = 60 por ciento

A partir de datos anteriores, se sabe que la desviación estándar de la población es del 15 por ciento.

Trabaje con el procedimiento de prueba e interprete los resultados de la prueba de hipótesis. ¿Cuál es tu conclusión?

No se puede rechazar la hipótesis nula. NO hay pruebas sólidas para concluir que los adultos pueden recordar más

del 55 % de los nombres de una lista, en promedio, después de leer la lista en voz alta.

No se puede rechazar la hipótesis nula. Existe una fuerte evidencia para concluir que los adultos pueden recordar más que

los nombres en una lista, en promedio, después de leer la lista en voz alta.

Rechazar la hipótesis nula. NO hay evidencia sólida para concluir que los adultos pueden recordar más de 5

nombres en una lista, en promedio, después de leer la lista en voz alta.

Rechazar la hipótesis nula. Existe una fuerte evidencia para concluir que los adultos pueden recordar más del 55%

de los nombres en una lista, en promedio, después de leer la lista en voz alta.

Explanation:

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Answer:

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2 years ago
Which character in hamilton are you?<br> I am Angelica
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3 years ago
The graph of y=f(x) is shown above. Which of the following could be the graph of y=f′′(x)?.
AfilCa [17]

The true statement about the graph of f(x) is f(x) = \frac 35x + 3

The graph of f(x) is a straight line that passes through the points (-5,0) and (0,3).

Start by calculating the slope (m)

m = \frac{y_2 - y_1}{x_2 -x_1}

So, we have:

m = \frac{3-0}{0--5}

Simplify

m = \frac{3}{5}

The equation of the line is then calculated as:

y = m(x -x_1) + y_1

This gives

y = \frac 35(x -0) + 3

Open bracket

y = \frac 35x + 3

Express y as a function of x

f(x) = \frac 35x + 3

Hence, the true statement about the graph of f(x) is f(x) = \frac 35x + 3

Read more about linear models at:

brainly.com/question/25603458

6 0
2 years ago
company manufactured six television sets on a given day, and these TV sets were inspected for being good or defective. The resul
lana [24]

Sampling distribution involves the proportions of a data element in a given sample.

  • <em>The proportion of Good TV set is 0.67</em>
  • <em>The number of ways of selecting 5 from 6 TV sets is 6</em>
  • <em>The number of ways of selecting 4 from 6 TV sets is 15</em>

<em />

Given

n = 6

Sample Space = Good, Good, Defective, Defective, Good, Good

<u>(a) Proportion that are good</u>

From the sample space, we have:

Good = 4

So, the proportion (p) that are good are:

p = \frac{Good}{n}

p = \frac{4}{6}

p = 0.67

<u>(b) Ways to select 5 samples (without replacement)</u>

This is calculated using:

^nC_r = \frac{n!}{(n - r)!r!}

Where

r = 5

So, we have:

^6C_5 = \frac{6!}{(6 - 5)!5!}

^6C_5 = \frac{6!}{1!5!}

^6C_5 = \frac{6 \times 5!}{1 \times 5!}

^6C_5 = \frac{6}{1}

^6C_5 = 6

Hence, there are 6 ways

<u>(c) All possible sample space of 4</u>

First, we calculate the number of ways to select 4.

This is calculated using:

^nC_r = \frac{n!}{(n - r)!r!}

Where

r = 4

So, we have:

^6C_4 = \frac{6!}{(6 - 4)!4!}

^6C_4 = \frac{6!}{2!4!}

^6C_4 = \frac{6 \times 5 \times 4}{2 \times 1 \times 4!}

^6C_4 = \frac{30}{2}

^6C_4 = 15

So, the table is as follows:

\left[\begin{array}{ccc}TV&Good&Proportion\\1,2,3,4&2&0.5&2,3,4,5&2&0.5&3,4,5,6&2&0.5\\4,5,6,1&3&0.75&5,6,1,2&4&1&6,1,2,3&3&0.75\\1,2,3,5&3&0.75&3,5,6,2&3&0.75&1,3,4,5&2&0.5\\1,3,4,6&2&0.5&1,4,5,2&3&0.75&2,4,6,1&3&0.75\\2,4,6,3&2&0.5&2,4,6,5&3&0.75&3,5,6,1&3&0.75\end{array}\right]

The proportion column is calculated by dividing the number of Good TVs by the total selected (4) i.e.

p = \frac{Good}{n}

<u>(d) The sampling distribution</u>

In (a), we have:

p = 0.67 --- proportion of Good TV

The sampling error is calculated as follows:

SE_n = |p - p_n|

So, we have:

\left[\begin{array}{ccc}TV&Good&SE\\1,2,3,4&2&0.17&2,3,4,5&2&0.17&3,4,5,6&2&0.17\\4,5,6,1&3&0.08&5,6,1,2&4&0.33&6,1,2,3&3&0.08\\1,2,3,5&3&0.08&3,5,6,2&3&0.08&1,3,4,5&2&0.17\\1,3,4,6&2&0.17&1,4,5,2&3&0.08&2,4,6,1&3&0.08\\2,4,6,3&2&0.17&2,4,6,5&3&0.08&3,5,6,1&3&0.08\end{array}\right]

Read more about sampling distributions at:

brainly.com/question/10554762

3 0
3 years ago
A 100-n weight is hung at the middle of a 4-m long horizontal clothesline. If the middle of the clothesline sags by 1 m, what is
Elena L [17]

The tension on each segment of the clothesline is :  110 N

<u>Given data : </u>

mass of object = 100 n  = 10 kg

Horizontal distance of clothesline = 4 m

middle of clothesline sag s by : 1m

<h3 /><h3>Determine the tension on each segment of clothesline</h3>

<u>First step</u> : calculate the horizontal angle made by the sagging

β = arctan ( 1 m / 2m )  ----- ( 1 )

  = arctan ( 0.5 )

  ≈ 26.57°

Note : Tension in th y axis ( Ty ) = Tsinβ

Therefore :

Tension on each segment can be calculated using the formula below

2Tsinβ - mg = 0

solve for T

T = mg / 2sinβ

  = ( 10 * 9.8 ) / 2 * sin 26.57°

  = 98 / 0.89

  = 110 N

Hence we can conclude that the tension on each segment of the clothesline is : 110 N

Learn more about Tension calculations : brainly.com/question/24994188

7 0
2 years ago
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