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BabaBlast [244]
3 years ago
15

Ten years ago, an organization reported that teenagers spent an average of 4.5 hours per week on the phone. The organization thi

nks that the current mean is higher. To test the claim, 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. Suppose that an appropriate hypothesis test is conducted to test the claim that the mean hours is more than 4.5 hours per week. What is the Type I error
Mathematics
1 answer:
lora16 [44]3 years ago
5 0

Answer: Type  I error would be a conclusion that average time spent on phone by teenager is higher than 4.5 hours per week which is actually the same.

Step-by-step explanation:

Type 1 error occurs when a null hypothesis is true but researcher rejects it.

Let \mu be the average time spent on the phone by teenagers .

Here, Null hypothesis: H_0:\mu=4.5

Alternative hypothesis : H_a: \mu>4.5

So, Type  I error would be a conclusion that average time spent on phone by teenager is higher than 4.5 hours per week which is actually the same.

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Answer:

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Step-by-step explanation:

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Which of the following describes the end behavior of y= -x^2 + bc + c as x approaches either positive or negative infinity?
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3 years ago
A coffee machine dispenses normally distributed amounts of coffee with a mean of 12 ounces and a standard deviation of 0.2 ounce
NemiM [27]

Answer:

0.9332 = 93.32% probability that the mean of the sample will be less than 12.1 ounces.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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 12 ounces and a standard deviation of 0.2 ounces.

This means that \mu = 12, \sigma = 0.2

Sample of 9:

This means that n = 9, s = \frac{0.2}{\sqrt{9}}

Find the probability that the mean of the sample will be less than 12.1 ounces.

This is the pvalue of Z when X = 12.1. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{12.1 - 12}{\frac{0.2}{\sqrt{9}}}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

0.9332 = 93.32% probability that the mean of the sample will be less than 12.1 ounces.

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3 years ago
ROUND 753 TO THE NEAREST HUNDRED.​
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it would be 800 because you moving it up 1 so yea

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