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Tcecarenko [31]
3 years ago
7

A recent national survey found that high school students watched an average (mean) of 7.1 movies per month with a population sta

ndard deviation of 1.0. The distribution of number of movies watched per month follows the normal distribution. A random sample of 33 college students revealed that the mean number of movies watched last month was 6.2. At the 0.05 significance level, can we conclude that college students watch fewer movies a month than high school students? State the null hypothesis and the alternate hypothesis.
Mathematics
1 answer:
muminat3 years ago
3 0

Answer:

H0: μc ≤ μs    Ha :μc > μs    

Step-by-step explanation:

The null and alternate hypotheses can be stated as

H0: μc ≤ μs    Ha :μc > μs    one tailed test

Where

μc =  Mean of college students watching movies in a month

μs  =  Mean of school students watching movies in a month

For one tailed test of α =0.05   the value of Z= ± 1.645

The critical region will be Z >  ± 1.645

It is of importance to note that by rejecting the null hypothesis and accepting the alternate hypothesis we are automatically rejecting all values of mean that are greater than 7.1

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OverLord2011 [107]

Answer:

The volume of the ball with the drilled hole is:

\displaystyle\frac{8000\pi\sqrt{2}}{3}

Step-by-step explanation:

See attached a sketch of the region that is revolved about the y-axis to produce the upper half of the ball. Notice the function y is the equation of a circle centered at the origin with radius 15:

x^2+y^2=15^2\to y=\sqrt{225-x^2}

Then we set the integral for the volume by using shell method:

\displaystyle\int_5^{15}2\pi x\sqrt{225-x^2}dx

That can be solved by substitution:

u=225-x^2\to du=-2xdx

The limits of integration also change:

For x=5: u=225-5^2=200

For x=15: u=225-15^2=0

So the integral becomes:

\displaystyle -\int_{200}^{0}\pi \sqrt{u}du

If we flip the limits we also get rid of the minus in front, and writing the root as an exponent we get:

\displaystyle \int_{0}^{200}\pi u^{1/2}du

Then applying the basic rule we get:

\displaystyle\frac{2\pi}{3}u^{3/2}\Bigg|_0^{200}=\frac{2\pi(200\sqrt{200})}{3}=\frac{400\pi(10)\sqrt{2}}{3}=\frac{4000\pi\sqrt{2}}{3}

Since that is just half of the solid, we multiply by 2 to get the complete volume:

\displaystyle\frac{2\cdot4000\pi\sqrt{2}}{3}

=\displaystyle\frac{8000\pi\sqrt{2}}{3}

5 0
3 years ago
The average weight of the entire batch of the boxes of cereal filled today was 20.5 ounces. A random sample of four boxes was se
zhenek [66]

Answer:

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And replacing we got:

s= 0.286

And then the estimator for the standard error is given by:

SE= \frac{0.286}{\sqrt{4}}= 0.143

Step-by-step explanation:

For this case we have the following dataset given:

20.05, 20.56, 20.72, and 20.43

We can assume that the distribution for the sample mean is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for this case would be:

SE= \frac{\sigma}{\sqrt{n}}

And we can estimate the deviation with the sample deviation:

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And replacing we got:

s= 0.286

And then the estimator for the standard error is given by:

SE= \frac{0.286}{\sqrt{4}}= 0.143

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3 years ago
0.12 word form decimal
7nadin3 [17]

Answer:

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Step-by-step explanation:

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3 years ago
Estimate the area under the curve f(x) = x2 from x = 1 to x = 5 by using four inscribed (under the curve) rectangles. Answer to
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A jar contains 2 orange, 4 green, 2 white and 2 black balls. What is the probability of drawing an orange ball without putting i
KonstantinChe [14]

Answer:

2/45

Step-by-step explanation:

We are told that:

A jar contains 2 orange, 4 green, 2 white and 2 black balls.

The total number of balls in the jar is calculated as:

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The probability of drawing an orange ball = P(Orange) = 2/10

The probability of drawing a black ball = P(Black) = 2/10

Therefore, the probability of drawing an orange ball without putting it back, then drawing a black ball is calculated as:

2/10 × 2/9 = 4/90

= 2/45

7 0
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