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Nikitich [7]
3 years ago
10

Suppose the results indicate that the null hypothesis should be rejected; thus, it is possible that a type I error has been comm

itted. Given the type of error made in this situation, what could researchers do to reduce the risk of this error? Increase the sample size. Choose a 0.01 significance level instead of a 0.05 significance level.
Mathematics
1 answer:
marissa [1.9K]3 years ago
8 0

Answer:

You need to Choose a 0.01 significance level , instead of 0.05 significance level.

Step-by-step explanation:

Consider the provided information.

It is given that type I error has been committed.

Type I error occur, if we rejected the null hypothesis even null hypothesis is true.

We reject the null hypothesis if p-value < α

Accept the null hypothesis if p-value ≥ α

According to given condition the value of α is 0.05  that means 95% chance that your assumption is correct and 5% chance that assumption is wrong.

In order to reduce the risk of this error you need to increase the significance level so that the value of α becomes lower than the p value.

lower α means you will be less likely to detect a true difference if one really exists.

Hence, you need to Choose a 0.01 significance level , instead of 0.05 significance level.

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(X^4-81)/x^3+3x^2+9x+27
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Answer:

x-3

Step-by-step explanation:

(x^2-9)(x^2-9)/x^2*(x+3)+9(x+3)

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Which figure takes a rotation of 360° to carry it onto itself? (consider each figure a regular polygon)
Mademuasel [1]

Answer:

B) trapezoid is the right answer

Step-by-step explanation:

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5 0
3 years ago
An electrician disconnected 5 wires of different colors from their respective connections and forgot the order in which they wer
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Answer:

120 tries

Step-by-step explanation:

You can try 5 different wires in the first connection.

Then you have 4 wires left, so you try 4 wires in the second connection. Remember, you are now trying 4 wires for each of the first 5, so up to here you already made 4 * 5 tries = 20 tries.

Next you try 3, then 2, then you have 1 left.

Number of tries: 5 * 4 * 3 * 2 * 1 = 120

Answer: 120 tries

8 0
3 years ago
A popular soft drink is sold in 2-liter (2000 milliliter) bottles. Because of variation in the filling process, bottles have a m
andrezito [222]

Answer:

a) 0.27% probability that the mean is less than 1995 milliliters.

b) 2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

Step-by-step explanation:

To solve this problem, it is important to understand the normal probability distribution and the central limit theorem

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2000, \sigma = 18

A) If the manufacturer samples 100 bottles, what is the probability that the mean is less than 1995 milliliters?

So n = 100, s = \frac{18}{\sqrt{100}} = 1.8

This probability is the pvalue of Z when X = 1995. So

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

By the Central Limit Theorem

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

Z = \frac{1995 - 2000}{1.8}

Z = -2.78

Z = -2.78 has a pvalue of 0.0027

So 0.27% probability that the mean is less than 1995 milliliters.

B) What mean overfill or more will occur only 10% of the time for the sample of 100 bottles?

This is the value of X when Z has a pvalue of 1-0.1 = 0.9.

So it is X when Z = 1.28

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

1.28 = \frac{X - 2000}{1.8}

X - 2000 = 1.8*1.28

X = 2002.3

2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

3 0
3 years ago
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