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lilavasa [31]
1 year ago
7

A random sample of 87 eighth grade students’ scores on a national mathematics assessment

Mathematics
1 answer:
Finger [1]1 year ago
8 0

There is not enough evidence to support the administrator’s claim and the true mean is not significantly greater than 280.

<h3>What is a statistical hypothesis?</h3>

A hypothesis to test the given parameters requires that we determine if the mean score of the eighth graders is more than 283, thus:

The null hypothesis:

\mathbf{H_o \leq 283}

The alternative hypothesis:

\mathbf{H_i > 283}

From the population deviation, the Z test for the true mean can be computed as:

\mathbf{Z = \dfrac{\hat X - \mu _o}{\dfrac{\sigma}{\sqrt{n}}}}

\mathbf{Z = \dfrac{283 -280}{\dfrac{37}{\sqrt{87}}}}

Z = 0.756

Note that, since we are carrying out a right-tailed test, the p-value for the test statistics is expressed as follows:

P(z > 0.756)

P = 0.225

Since the P-value is greater than the significance level at α = 0.14, we can conclude that there is not enough evidence to support the administrator’s claim and the true mean is not significantly greater than 280.

Learn more about hypothesis testing here:
brainly.com/question/16251072

#SPJ1

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Suppose EG=3​, EB=8​, A(F)=7​, m∠EBG=23​, m∠EGF=32​, and m∠CAE=51. Find m∠CAF.
ASHA 777 [7]

Answer:

m∠CAF is 28°

Step-by-step explanation:

The given parameters are;

EG = 3, EB = 8, A_F = 7, m∠EBG = 23°, ∠EGF 32°, and m∠CAE = 51°

From the diagram, we have;

m∠EBG = m∠EAD ;                {}                                    Given

Therefore, m∠EAD = 23°  

m∠CAE = m∠CAF + m∠EAD {}                                     Angle addition postulate

Therefore, we have;

51° = m∠CAF + 23° {}                        

m∠CAF = 51° - 23°                            {}              

m∠CAF = 28°

8 0
3 years ago
A briefcase has a three- digit lock code that doesn't include zero as a digit. What is the probability that the lock code consis
storchak [24]

Answer:

1/21.

Step-by-step explanation:

There are 9 digits and the total number of permutations of 3 from 9 is 9P3

= 9!/6!  = 504.

There are 4 even digits so the number of permutations of 3 from these 4 is 4!  (4-3)!  = 4*3*2 = 24.

So the required probability =  24/504

= 1/21.

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When you add negative, it would be the same as subtracting
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