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melomori [17]
3 years ago
6

A large software development firm recently relocated its facilities. Top management is interested in fostering good relations wi

th their new local community and has encouraged their professional employees to engage in local service activities. They believe that the​ firm's professionals volunteer an average of more than 21 hours per month. If this is not the​ case, they will institute an incentive program to increase community involvement. A random sample of 24 professionals yields a mean of 22 hours and a standard deviation of 2.22 hours.
12 13 14 14 15 15 15 16 16 16 16 16
17 17 17 18 18 18 18 19 19 19 20 21

The sample has a mean of 16.6 hours and a standard deviation of 2.22 credit hours.

At α = 0.05,

I. we reject the null hypothesis.
II. we fail to reject the null hypothesis.
III. the firm shouldn't need to institute an incentive program because the evidence indicates that professional employees volunteer an average of more than 15 hours per month in their local community.

a. II and III
b. I and III
c. II only
d. I only
e. III only
Mathematics
1 answer:
Pachacha [2.7K]3 years ago
5 0

Answer:

B. i and iii

Step-by-step explanation:

There are mistakes in your question. But I solve this problem using the average mean of 15 because that is what is contained in the solution

Hypothesis

H0: u <= 15

H1: u > 15

Mean of x = 16.6

S = 2.22

Test statistic used is t test

16.6 - 15 / 2.22/√24

= 3.531

We find the critical value

Degree of freedom = 24-1 = 23

This is a one tailed test

Critical value = 1.714

We find p value using Ms excel T distribution function

= 0.000894

To make the decision

3.531 > 1.714 so we conclude that

I.) Reject null hypothesis

Iii) conclude an average if more than 15 hours is volunteered per month. So no need to institute the program.

The answer option Is B

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Answer:

a) a=225 +0.674*16.5=236.121

So the value of height that separates the bottom 75% of data from the top 25% is 236.121.  

b) P(X \geq 3) = 1-P(X

c) P(\bar X \geq 225)=1- P(\bar X

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Part a

Let X the random variable that represent the cuts of a population, and for this case we know the distribution for X is given by:

X \sim N(225,16.5)  

Where \mu=225 and \sigma=16.5

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.25   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.674=\frac{a-225}{16.5}

And if we solve for a we got

a=225 +0.674*16.5=236.121

So the value of height that separates the bottom 75% of data from the top 25% is 236.121.  

Part b

For this case we know that the individual probability of select one wheel with a cutting rate higher than the calculated value in part a is 0.25, and we select n =10 so then we can use the binomial distribution for this case:

X\sim Bin(n=10, p=0.25)

And we want this probability:

P(X \geq 3) = 1-P(X

We can find the individual probabilities like this:

P(X=0)=(10C0)(0.25)^0 (1-0.25)^{10-0}=0.0563

P(X=1)=(10C1)(0.25)^1 (1-0.25)^{10-1}=0.1877

P(X=2)=(10C2)(0.25)^2 (1-0.25)^{10-2}=0.2816

P(X \geq 3) = 1-P(X

Part c

For this case we know that the distribution for the sample mean is given by:

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

And we want this probability:

P(\bar X \geq 225)

And for this case we can use the complement rule and the z score given by:

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

And if we replace we got:

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