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Verizon [17]
2 years ago
14

Instructions: For each problem, identify the appropriate test statistic to be use (t test or z-test). Then compute z or t value.

Engineering
1 answer:
Natalka [10]2 years ago
6 0

The sample of 81 students was selected with a mean score of 90, this illustrates an example of a right tailed one sample z test.

<h3>How to illustrate the sample?</h3>

From the information given, the teacher claims that the mean score of students in his class is greater than 82 with a standard deviation of 20. In this case, the right tailed one sample z test is used.

The z value will be:

= (90 - 82)/20/✓81

= 3.6

Since 3.6 > 1.645, the null hypothesis will be rejected as there's enough evidence to support the teacher's claim.

When an online medicine shop claims that the mean delivery time for medicines is less than 120 minutes with a standard deviation of 30 minutes, the left tailed one sample test is used.

The z value will be:

= (100 - 120)/(30/✓49)

= -4.66

The null hypothesis is rejected as there is enough evidence to support the claim of the medicine shop.

Learn more about sampling on:

brainly.com/question/17831271

#SPJ1

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This test should be performed on all cord sets, receptacles that aren't part of a building or structure's permanent wiring, and
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A continuity test

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3 years ago
Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical
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Answer:

The theoretical density for Niobium is 1.87 g/cm^3.

Explanation:

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density  of the unit cell

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

We have :

Z = 2 (BCC)

M = 92.91 g/mol ( Niobium)

Atomic radius for niobium = r = 0.143 nm

Edge length of the unit cell = a

r = 0.866 a (BCC unit cell)

a=\frac{0.143 nm}{0.866}=0.165 nm=0.165 \times 10^{-7} cm

1 nm = 10^{-7} cm

On substituting all the given values , we will get the value of 'a'.

\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}

\rho =1.87 g/cm^3

The theoretical density for Niobium is 1.87 g/cm^3.

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3 years ago
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