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Alex787 [66]
3 years ago
12

Banks provide a number of services including all of the following except __________.

Mathematics
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:

A. Payday Lending

Explanation:

Payday Lending is common for <em>small lending companies</em> and <u>not banks</u>. They assist clients who have a <em>minimal amount of salary</em> to borrow <u>a small amount of money with a high interest.</u> The amount of money he can borrow will depend on the amount of salary he receives per month. <em>The client will not be required to give any form of collateral</em> for the borrowed money, thus, the client is said to be <em>high-risk</em>. However, he will have to return the money over a<em> short period of time</em> with an interest rate that is high. Though it is easy to get a loan with this kind of service, you have to make sure that you are employed.

<em>Banks prefer to offer loans to people who have a steady income and a certain amount of salary.  </em>

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Divide 40 by 4 to remove the 1/4

10

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

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= 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

b

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

c)

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 = \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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