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aalyn [17]
3 years ago
12

On their birthdays, employees at a large company are permitted to take a 60-minute lunch break instead of the usual 30 minutes.

Data were obtained from 10 randomly selected company employees on the amount of time that each actually took for lunch on his or her birthday. The company wishes to investigate whether these data provide convincing evidence that the mean time is greater than 60 minutes. Of the following, which information would NOT be expected to be a part of the process of correctly conducting a hypothesis test to investigate the question, at the 0.05 level of significance?
Being willing to assume that the distribution of actual birthday lunch times for all employees at the company is approximately normal

Knowing that there are no outliers in the data as indicated by the normal probability plot and boxplot

Using a t-statistic to carry out the test

Using 9 for the number of degrees of freedom

Given that the p-value is greater than 0.05, rejecting the null hypothesis and concluding that the mean time was not greater than 60 minutes.
Mathematics
1 answer:
yanalaym [24]3 years ago
6 0

Answer:

Given that the p-value is greater than 0.05, rejecting the null hypothesis and concluding that the mean time was not greater than 60 minutes

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Answer: 28.8 ounces of red paint

Step-by-step explanation:

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3 years ago
Question is in the picture
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The value, V (t), in dollars, of a stock t months after it is purchased is modeled by the following equation.
11Alexandr11 [23.1K]

Answer:

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b) Find derivative for V

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c) When V(t) = 75

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