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marissa [1.9K]
3 years ago
10

A statistician selected a random sample of 125 observations from a population with a known standard deviation equal to 16 and co

mputed a sample mean equal to 77. (SHOW WORK for CREDIT; round to 3 decimal places) a) Estimate the population mean with 93% confidence. b) Repeat Part a) using a 89% level of confidence. c) Repeat Part a) using a population variance equal to 441. d) Repeat Part a) using a sample size equal to 625. Any changes to C.I.?
Mathematics
1 answer:
serious [3.7K]3 years ago
8 0

Answer:

Step-by-step explanation:

Given that a statistician selected a random sample of 125 observations from a population with a known standard deviation equal to 16 and computed a sample mean equal to 77

We can use Z critical values since population std deviation is known. Also sample size >30

We find std error of mean = \frac{16}{\sqrt{125} } \\=1.4311

Margin of error = Z critical value * 1.4311

Z critical values for 93% = 1.81

for 89% = 1.60

n Std error Z critical  Conf interval  

            89%    

     

125 1.4311     1.6  (74.71024 79.28976 )

     

     

          93%    

            1.81  (74.409709 79.590291 )

We find that when confidence level increases interval width increses.

c) When sigma changes to 441, std error changes to \frac{16}{\sqrt{441} } \\=0.7619

So we get

n Std error Z critical  Conf interval  

           89%    

     

441  0.7619     1.6  (75.781 78.219 )

     

     

         93%    

           1.81  (75.621 78.37904)

d) When n = 625, std error changes to 16/25 = 0.64

n Std error Z critical  Conf interval  

        89%    

     

441 0.64 1.6        (75.976 78.024 )

     

     

       93%    

        1.81      (75.842 78.1584)

When sample size increases, confidence interval width decreases.

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Answer:

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Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane n=(-2,-2,1), then r will have the next parametric equations:

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The distance from Po to the plane

d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u

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