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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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taurus [48]

Answer:

82

Step-by-step explanation:

49 - 3 + 72/2

= 46 + 36

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7 0
3 years ago
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About % of the area under the curve of the standard normal distribution is between z = − 0.9 z = - 0.9 and z = 0.9 z = 0.9 (or w
prisoha [69]

Using the normal distribution, it is found that 63.18% of the area under the curve of the standard normal distribution is between z = − 0.9 z = - 0.9.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The area within 0.9 standard deviations of the mean is the <u>p-value of Z = 0.9(0.8159) subtracted by the p-value of Z = -0.9(0.1841)</u>, hence:

0.8159 - 0.1841 = 0.6318 = 63.18%.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

3 0
1 year ago
Solve the equation-4(x+10)-6=-3(x-2)<br>​
marta [7]

Answer:

x = -52

Step-by-step explanation:

-4x-46+46=-3x+6+46

-4x=-3x+52

-4x+3x=-3x+52+3x

8 0
2 years ago
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Joelle walked 2/5 of the way from her house to school. Franco also walked 2/5 of the way from his house to school. Joelle and Fr
gulaghasi [49]

Answer:

I think they both have the same amount walked, therefore they are both considered as they have walked the same distance

Step-by-step explanation:

3 0
2 years ago
To the Pythagorean Theorem
Aloiza [94]

Answer:

x = 29

Step-by-step explanation:

c^{2} = a^{2} + b^{2}

c^{2} = 20^{2} + 21^{2}

c^{2} = 400 + 441

c^{2} = 841

\sqrt{c^2} = \sqrt{841}

c = 29

4 0
2 years ago
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