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Arlecino [84]
3 years ago
9

A high school student named David Merrell did a study of how music affects the ability of rats to run a maze. He trained 71 rats

to run a maze. Next he randomly divided the rats to three groups. Each group had 23 or 24 rats. For the next month, one group of rats listened all day to heavy metal music by the group Anthrax. One group listened all day to Mozart. One group never heard music. The rats ran the same maze three times a week for a month. The response variable is the time in seconds to complete the maze. By the end of the month he compared the average run time for each treatment group to the average run time for the control group. The Anthrax group was much slower at running the maze. The Mozart group was much faster. The differences are statistically significant at the 1% level. Run times are not heavily skewed in any group. What conclusion can we draw from these results?
Mathematics
1 answer:
ladessa [460]3 years ago
3 0
I don’t understand what you are saying...
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An electricity post is supported by a 12 m long strut(wire). If the strut makes an angle of 53° with the ground level, how far f
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Step-by-step explanation:

I do hope that you understand through the steps in the attachment, if not kindly reach out!

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9514 1404 393

Answer:

  f^-1(x) = 4+∛((x-6)/5)

Step-by-step explanation:

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  x = f(y)

then write the answer in functional form.

  \displaystyle x=f(y)\\\\x=5(y-4)^3+6\\\\x-6=5(y-4)^3\\\\\frac{x-6}{5}=(y-4)^3\\\\\sqrt[3]{\frac{x-6}{5}}=y-4\\\\y=4+\sqrt[3]{\frac{x-6}{5}}\\\\\boxed{f^{-1}(x)=4+\sqrt[3]{\frac{x-6}{5}}}

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3 0
3 years ago
A private and a public university are located in the same city. For the private university, 1038 alumni were surveyed and 647 sa
Snezhnost [94]

Answer:

The difference in the sample proportions is not statistically significant at 0.05 significance level.

Step-by-step explanation:

Significance level is missing, it is  α=0.05

Let p(public) be the proportion of alumni of the public university who attended at least one class reunion  

p(private) be the proportion of alumni of the private university who attended at least one class reunion  

Hypotheses are:

H_{0}: p(public) = p(private)

H_{a}: p(public) ≠ p(private)

The formula for the test statistic is given as:

z=\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}} where

  • p1 is the sample proportion of  public university students who attended at least one class reunion  (\frac{808}{1311}=0.616)
  • p2 is the sample proportion of private university students who attended at least one class reunion  (\frac{647}{1038}=0.623)
  • p is the pool proportion of p1 and p2 (\frac{808+647}{1311+1038}=0.619)
  • n1 is the sample size of the alumni from public university (1311)
  • n2 is the sample size of the students from private university (1038)

Then z=\frac{0.616-0.623}{\sqrt{{0.619*0.381*(\frac{1}{1311} +\frac{1}{1038}) }}} =-0.207

Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.  

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4 years ago
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