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arlik [135]
3 years ago
8

A major airline is concerned that the waiting time for customers at its ticket counter may be exceeding its target average of 19

0 seconds. To test this, the company has selected a random sample of 100 customers and times them from when the customer first arrives at the checkout line until he or she is at the counter being served by the ticket agent. The mean time for this sample was 202 seconds with a standard deviation of 28 seconds. Given this information and the desire to conduct the test using an alpha level of 0.02, which of the following statements is true?
A) The chance of a Type II error is 1 - 0.02 = 0.98.
The test to be conducted will be structured as a two-tailed test.
The test statistic will be approximately t = 4.286, so the null hypothesis should be rejected.
The sample data indicate that the difference between the sample mean and the hypothesized population mean should be attributed only to sampling error.
Mathematics
1 answer:
sertanlavr [38]3 years ago
3 0

Answer:

a) Statement is true

b) Statement incorrect. Company goal is to attend customer at 190 second at the most. So test should be one tail test  (right)

c) We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error

Step-by-step explanation:

a)  as the significance level is = 0,02  that means  α = 0,02 (the chance of error type I )  and the β (chance of type error II   is

1  -  0.02  =  0,98

b) The company establishe in 190 seconds time for customer at its ticket counter (if this time is smaller is excellent ) company is concerned about bigger time because that could be an issue for customers. Therefore the test should be a one test-tail to the right

c) test statistic

Hypothesis test should be:

null hypothesis                H₀   =  190

alternative hypothesis    H₀   >  190

t(s)  =  ( μ  -  μ₀ ) / s/√n      ⇒   t(s)  = ( 202 - 190 )/(28/√100 )

t(s)  = 12*10/28

t(s)  = 4.286

That value is far away of any of the values found for 99 degree of fredom and between  α  ( 0,025 and 0,01 ). We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error

If we look table  t-student we will find that for 99 degree of freedom and α = 0.02.

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Answer:

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Step-by-step explanation:

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3 years ago
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marishachu [46]

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Step-by-step explanation:

4 0
3 years ago
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ikadub [295]
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6 0
4 years ago
Read 2 more answers
In 2010 Singapore welcome 11600000 overseas visitors.This value has been rounded off to the nearest 100000.What are the largest
spayn [35]

Answer:

The largest possible number of overseas visitors would be 11,640,000.

The smallest possible number of overseas visitors would be 11,550,000.

Step-by-step explanation:

The number 11,600,000 was rounded off to the nearest 100,000.

This means that either 0 or 1 was added to the figure with place value of 100,000.

It would be zero (0) in the case where the number after the 100,000th placed value number is lower than 5 and it would be one (1) in the case where the number after the 100,000th placed value is greater than or equal to 5.

Therefore, the largest possible number of overseas visitors would be 11,640,000.

The smallest possible number of overseas visitors would be 11,550,000.

7 0
4 years ago
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Lana71 [14]

Step-by-step explanation:

I think you mean you want to find out 40% of a number,, when 60% of it is 180, so I'll answer that:

60% = 180

÷6

10% = 30

x4

40% = 120

(if you want to find 100%, multiply 30 by 10)

Hopefully this helps :)

4 0
2 years ago
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