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Musya8 [376]
2 years ago
13

Suppose a scheduled flight must average at least 60% occupancy to be profitable. A sample of 120 flights on a particular route w

as studied: the mean occupancy rate per flight was 58% with a sample standard deviation of 11%. We would like to find out if there is sucient evidence to support a claim that this flight route is unprofitable. a) In terms of this problem, what is the Type I error? b) Define and find ↵. c) In terms of this problem, what is the Type II error? d) Define and find when the true mean occupancy rate is 0.62. e) Find the p-value associated with this test. What would you conclude at the 0.10 level

Mathematics
1 answer:
garik1379 [7]2 years ago
8 0

Answer: see the graphic

Step-by-step explanation:

A. Type I error helps us to conclude that the flight is not profitable, when in fact it is profitable.

B. a = 0.05

C. Type II error does not show that the flight is profitable

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24/2=12

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Merrill Lynch Securities and Health Care Retirement Inc. are two large employers in downtown Toledo, Ohio. They are considering
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Answer:

Step-by-step explanation:

We would determine the mean an standard deviation first

Mean = (107 + 92 + 97 + 95 + 105 + 101 + 91 + 99 + 95 + 104)/10 = 98.6

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n = 10

Summation(x - mean) = (107 - 98.6)^2 + (92 - 98.6)^2 + (97 - 96.6)^2 + (95 - 98.6)^2 + (105 - 98.6)^2 + (101 - 98.6)^2 + (91 - 98.6)^2 + (99 - 98.6)^2 + (95 - 98.6)^2 + (104 - 98.6)^2 = 274

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The confidence interval is used to determine the range of values that could possibly contain a population parameter (population mean)

Confidence interval is written in the form,

(Point estimate - margin of error, Point estimate + margin of error)

The sample mean, x is the point estimate for the population mean.

We will use the t distribution because the sample size is small and the population standard deviation is not given.

Degree of freedom, df = 10 - 1 = 9

α = 1 - 0.99 = 0.01

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The lower boundary of the confidence interval is

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The upper boundary of the confidence interval is

98.6 + 5.38 = 103.98

We estimate with 99% confidence that the mean weekly child care cost of their employees is between $93.22 and $103.98

Also,

99% of the confidence intervals constructed in this way would contain the true value for the population mean of mean weekly child care cost of their employees.

6 0
2 years ago
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Answer:

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