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natulia [17]
1 year ago
15

A recent gallup study reported that firms with top engagement scores had _____ higher profitability.

Mathematics
1 answer:
Neko [114]1 year ago
7 0

A recent gallup study reported that firms with top engagement scores had 18% higher profitability.

<h3>What is gallup study?</h3>

Gallup defines engaged employees as those who are involved in, enthusiastic about and committed to their work and workplace.

The Gallup survey or study aims to identify the 12 elements of good management.

Gallup can run it for you, or you can manage it yourself.

It gives you information about you where you need to focus your effort and it's invaluable for building high-performing teams and positive, productive working environments.

Thus, a recent gallup study reported that firms with top engagement scores had 18% higher profitability.

Learn more about gallup study here: brainly.com/question/17116621

#SPJ4

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7−5+3−1  

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7−(5+3)−1

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A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 420 gram setting. It is
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Answer:

t=\frac{424-420}{\frac{26}{\sqrt{61}}}=1.202    

p_v =2*P(t_{(60)}>1.202)=0.234  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from  420. So the specification is satisfied.

Step-by-step explanation:

Data given and notation  

\bar X=424 represent the sample mean

s=26 represent the sample standard deviation

n=61 sample size  

\mu_o =420 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 420 or not, the system of hypothesis would be:  

Null hypothesis:\mu = 420  

Alternative hypothesis:\mu \neq 420  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-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".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{424-420}{\frac{26}{\sqrt{61}}}=1.202    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=61-1=60  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(60)}>1.202)=0.234  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from  420. So the specification is satisfied.

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