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dusya [7]
3 years ago
14

A large software company gives job applicants a test of programming ability and the mean for that test has been 160 in the past.

twenty-five job applicants are randomly selected from one large university and they produce a mean score and standard deviation of 183 and 12, respectively. use a 0.05 level of significance to test the claim that this sample comes from a population with a mean score greater than 160.

Mathematics
2 answers:
Nezavi [6.7K]3 years ago
6 0
See attachment for answer.

grandymaker [24]3 years ago
3 0

Step-by-step explanation:

Mean of the test of programming ability was 160 in the past.

Again now, twenty-five job applicants are randomly selected from one university and they produce a mean score and standard deviation of 183 and 12.

So in this case, the null hypothesis will be,

H_o: \mu = 160

and alternate hypothesis will be,

H_a: \mu > 160

Mean of the sample = \overline{x} = 183

Standard deviation of the sample = \sigma = 12

sample size = n = 25

Using t distribution,

t=\dfrac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}

=\dfrac{183-160}{\frac{12}{\sqrt{25}}}

=9.58

P-value =P(t > 9.58)\approx 0

As the obtained value is less than 0.05 or 5% level of significance, so we have to reject the null hypothesis.

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Annette [7]

Given:

$184500 are to be divided between 2 people.

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To find:

The amount each one receive in $.

Solution:

Let $x be the amount received by first person. Then,

Amount received by the second person = 80% of x

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Total amount = Amount received by first person + by second person

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Total amount is 184500. So,

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So, the amount received by the first person is $102,500.

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So, the amount received by the second person is $82,000.

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3 years ago
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Answer:

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Step-by-step explanation:

x + 24. 5 = 34.8

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Check your answer:

x + 24. 5 = 34.8

10.3 + 24.5 = 34.8

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Hope this helps!

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