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Deffense [45]
3 years ago
14

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between 2000

0 and 35000 . Assume that a confidence interval estimate of the population mean annual starting salary is desired.
What is the planning value for the population standard deviation?
Mathematics
1 answer:
ehidna [41]3 years ago
7 0

Answer:

The planning value for the population standard deviation is of 4330.

Step-by-step explanation:

Uniform probability distribution:

The uniform probability distribution has two bounds, a and b. The standard deviation is given by:

S = \sqrt{\frac{(b-a)^2}{12}}

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between 20000 and 35000.

Uniform in this interval, so a = 20000, b = 35000

What is the planning value for the population standard deviation?

S = \sqrt{\frac{(35000 - 20000)^2}{12}} = 4330

The planning value for the population standard deviation is of 4330.

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A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean
elena55 [62]

Answer:

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}). So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*0.058 = 0.171

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg

The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

3 0
3 years ago
PLEASE HELP I WILL GIVE BRAINLIEST MUST USE WORK :)
Nikolay [14]

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m =(-2-7)/(8-0)

m =(-9/8)

slope is -9/8

the y intercept is 7   (0,7)

y = mx +b

y = -9/8 x+7

Answer: y = -9/8x +7

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127.5 is what percent of 51?
poizon [28]
Multiply 1.275 with 51 to get the answer (65.025)
8 0
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What is 12/2? Answer for 10 points!!!
Mkey [24]

Answer:

6

Step-by-step explanation:

12/2=6/1=6

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Equilibrium Price 
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3 years ago
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