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GalinKa [24]
3 years ago
11

The director of research and development is testing a new medicine. She wants to know if there is evidence at the 0.05 level tha

t the medicine relieves pain in more than 325 seconds.
After performing a hypothesis test, she decides to reject the null hypothesis. What is the conclusion?

a. There is sufficient evidence at the 0.05 level of significance that the medicine relieves pain in more than 325 seconds.

b. There is not sufficient evidence at the 0.05 level of significance that the medicine relieves pain in more than 325 seconds.
Mathematics
1 answer:
Nataly [62]3 years ago
7 0

Answer:

a. There is sufficient evidence at the 0.05 level of significance that the medicine relieves pain in more than 325 seconds.

Step-by-step explanation:

With the info provided let the parameter of interest \mu who represent the mean that the medicine relieves pain, the system of hypothesis would be:

Null Hyp: \mu \leq 325

Alternative Hyp : \mu >325

If we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

If the deviation is not known we can use the t test with the following statistic:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}

On this case the decision was REJECT the null hypothesis. That means that we obtain a significant result with a p value lower than the signficance level

p_v

So then we can conclude this:

a. There is sufficient evidence at the 0.05 level of significance that the medicine relieves pain in more than 325 seconds.

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Answer:

3.3333 (repeating)

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He has 50 missing assignments and 15 days to do them.

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50 / 15 = 3.33 (repeating)

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3 years ago
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The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation
liraira [26]

Answer:

d. 76.98%

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 40000, \sigma = 5000

What percentage of MBA's will have starting salaries of $34,000 to $46,000?

This is the pvalue of Z when X = 46000 subtracted by the pvalue of Z when X = 34000. So

X = 46000

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

Z = \frac{46000 - 40000}{5000}

Z = 1.2

Z = 1.2 has a pvalue of 0.8849

X = 34000

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

Z = \frac{34000 - 40000}{5000}

Z = -1.2

Z = -1.2 has a pvalue of 0.1151

0.8849 - 0.1151 = 0.7698

So the correct answer is:

d. 76.98%

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Please help me on this question and explain your answer thanks
Arte-miy333 [17]
Each variable used is the first letter of each of their names.

t+b+c=87
c=2t
b=t+7

Substitute for b and c
t+ (t+7)+2t= 87
4t+7=87
4t=80
t=20

Plug in the known t value.
c=2(20)
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b=20+7
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Final answer: Tammy=$20, Boris= $27, Carlos=$40
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Answer:

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Step-by-step explanation:

Hope this helps

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