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3 years ago

Guys please help ;w; and plz explain what in the world is this

Mathematics

1 answer:

Artyom0805 [142]3 years ago

4 0

Answer:

Question 5C

Question 6 D

Question 7 D

Question 8 25.13

Step-by-step explanation:

to find the circumfrance of a circle using pi witch would be 3.14 you need to use the formula 2 x pi (3.14) x r (radius)

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Consider the following hypothesis test. H0: μ ≥ 55 Ha: μ < 55 A sample of 36 is used. Identify the p-value and state your con

Ket [755]

Answer:

Step-by-step explanation:

Given that:

$H_o: \mu \ge 55 \ \\ \\ H_1 : \mu < 55$

(a) For x = 54 and s = 5.3

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{54-55}{\dfrac{5.3}{\sqrt{36}} }$

$Z = \dfrac{-1}{\dfrac{5.3}{6} }$

Z = -1.132

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-1.132,35,1) = 0.1326

Decision: p-value is greater than significance level; do not reject $\mathbf{H_o}$

$Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \$ $\mu < 55$

For x = 53 and s = 4.6

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{53-55}{\dfrac{4.6}{\sqrt{36}} }$

$Z = \dfrac{-2}{\dfrac{4.6}{6} }$

Z = -2.6087

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-2.6087,35,1) =0.0066

Decision: p-value is < significance level; we reject the null hypothesis.

$Conclusion: \ There \ is \ sufficient \ evidence \ to \ conclude \ that$ $\mu < 55$

For x = 56 and s = 5.0

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }$

Z = 1.2

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(1.2,35,1) = 0.88009

Decision: p-value is greater than significance level; do not reject $\mathbf{H_o}$

$Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \$ $\mu < 55$

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Answer:

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