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Tatiana [17]
3 years ago
11

A company employs 400 salespeople. Of these, 83 received a bonus last year, 100 attended a special sales training program at the

beginning of last year, and 42 both attended the special sales training program and received a bonus. (Note: the bonus was based totally on sales performance.)
d) What proportion of the salespeople who attended the special sales training program received a bonus?
Mathematics
2 answers:
andrey2020 [161]3 years ago
8 0

Answer:

21/50

Step-by-step explanation:

Total Number of salespeople in company = 400

Of these salespeople, number of people recieved bonus last year = 83

Number of salesperson attended special sales training = 100

And, Number of salesperson attended both special sales training and received a bonus = 42

Then proportion of salesperson who attended special sales training program and received bonus = Number of salesperson attended both special sales program and received a bonus / total number of salesperson attended special sales training

                                   =  42/100

                                  = 21/50

Nady [450]3 years ago
3 0

Answer:

a) What proportion of the 400 salespeople received a bonus last year?

b) What proportion of the 400 salespeople attended the special sales training program at the beginning of last year?

c) What proportion of the 400 salespeople both attended the special sales training program and received a bonus?

d) What proportion of the salespeople who attended the special sales training program received a bonus?

e) Based on your answers to parts a and d, does the special sales training program seem to have been effective? Explain your answer.

Step-by-step explanation:

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A machine, when working properly, produces 5% or less defective items. Whenever the machine produces significantly greater than
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Answer:

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.

Step-by-step explanation:

A machine, when working properly, produces 5% or less defective items.

This means that the null hypothesis is:

H_0: p \leq 0.05

Test if the percentage of defective items produced by this machine is greater than 5%.

This means that the alternate hypothesis is:

H_a: p > 0.05

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.05 is tested at the null hypothesis:

This means that \mu = 0.05, \sigma = \sqrt{0.05*0.95}

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This means that n = 300, X = \frac{27}{300} = 0.09

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.09 - 0.05}{\frac{\sqrt{0.05*0.95}}{\sqrt{300}}}

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Looking at the Z-table, Z = 3.18 has a pvalue of 0.9993

1 - 0.9993 = 0.0007

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.

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