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kow [346]
2 years ago
15

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks

that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.
Mathematics
1 answer:
kiruha [24]2 years ago
3 0

The standard normal distribution is a special type of normal distribution where the mean is 0, and the standard deviation is 1. The hypothesis can not be concluded.

<h3>What is Standard normal distribution?</h3>

The standard normal distribution is a special type of normal distribution where the mean is 0, and the standard deviation is 1.

For the given problems, the Hypothesis mean is 4.5. The sample mean was 4.75 hours with a normal standard deviation of 2.0.

Now, calculating the z-score,

z = \dfrac{\bar x - \mu_o}{\dfrac{\sigma}{\sqrt n}}\\\\\\z = \dfrac{4.75-4.5}{\dfrac{2}{\sqrt{15}}}\\\\\\z = 0.48

Now, as per the standard normal table, the p-value is 0.6844. Since the p-value is greater than the significance level, the hypothesis can not be concluded.

Hence, The hypothesis can not be concluded.

Learn more about Standard normal distribution:

brainly.com/question/25447725

#SPJ1

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Each day, X arrives at point A between 8:00 and 9:00 a.m., his times of arrival being uniformly distributed. Y arrives independe
astraxan [27]

Answer:

Y will arrive earlier than X one fourth of times.

Step-by-step explanation:

To solve this, we might notice that given that both events are independent of each other, the joint probability density function is the product of X and Y's probability density functions. For an uniformly distributed density function, we have that:

f_X(x) = \frac{1}{L}

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Now, as  X is distributed over a 1 hour interval, and Y is distributed over a 0.5 hour interval, we have:

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Now, the probability of an event is equal to the integral of the density probability function:

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It's useful to draw a diagram here, I have attached one in which you can see the integration region.

You can see there a box, that represents all possible outcomes for Y and X. There's a diagonal coming from the box's upper right corner, that diagonal represents the cases in which both X and Y arrive at the same time, under that line we have that Y arrives before X, that is our integration region.

Let's set up the integration:

\iint_A f_{X,Y} (x,y) dx\, dy\\\\\iint_A f_{X} (x) \, f_{Y} (y) dx\, dy\\\\2 \iint_A  dx\, dy

We have used here both the independence of the events and the uniformity of distributions, we take the 2 out because it's just a constant and now we just need to integrate. But the function we are integrating is just a 1! So we can take the integral as just the area of the integration region. From the diagram we can see that the region is a triangle of height 0.5 and base 0.5. thus the integral becomes:

2 \iint_A  dx\, dy= 2 \times \frac{0.5 \times 0.5 }{2} \\\\2 \iint_A  dx\, dy= \frac{1}{4}

That means that one in four times Y will arrive earlier than X. This result can also be seen clearly on the diagram, where we can see that the triangle is a fourth of the rectangle.

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