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WITCHER [35]
3 years ago
9

How big is the home field advantage in the National Football League (NFL)? To investigate, we select a sample of 80 games from t

he 2011 regular season1 and find the home team scored an average of 25.16 points with a standard deviation 10.14 points. In a separate sample of 80 different games, the away team scored an average of 21.75 points with a standard deviation of 10.33 points. Use this summary information to find a 90%c onfidence interval for the mean home field advantage, µH- µA, in points scored.
The 90% confidence interval is______ to________ .
Mathematics
1 answer:
ElenaW [278]3 years ago
5 0

Answer:

0.74 to 6.06

Step-by-step explanation:

The groups are independnet,

SE(xh bar-xa bar)=sqrt [sh^2/nh+sa^2/na]=sqrt [10.1^2/80+10.3^2/80]=1.61

At df=157, the t critical is 1.65

90%c.i=(xh bar-xa bar)+-tcritical SE(xh bar-xa bar)

=(25.2-21.8)+-1.65*1.61

=0.74 to 6.06

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2. Suppose over several years of offering AP Statistics, a high school finds that final exam scores are normally distributed wit
nirvana33 [79]

Answer:

By the Central Limit Theorem, the mean is 78, the standard deviation is s = \frac{6}{\sqrt{n}} and the shape is approximately normal.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 78 and a standard deviation of 6

This means that \mu = 78, \sigma = 6

Samples of n:

This means that the standard deviation is:

s = \frac{\sigma}{\sqrt{n}} = \frac{6}{\sqrt{n}}

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6 0
3 years ago
Evaluate 8-2.<br>A. -16<br>B. -64<br>C. -1/64<br>D. 1/64<br> ​
makvit [3.9K]

I'm assuming you meant to write 8^(-2) or 8^{-2} where the -2 is the exponent over the 8.

If my assumption is correct, then we use the rule a^{-b} = \frac{1}{a^b}

So,

a^{-b} = \frac{1}{a^b}\\\\8^{-2} = \frac{1}{8^2}\\\\8^{-2} = \frac{1}{64}

<h3>Answer: Choice D.  1/64</h3>
8 0
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