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Nezavi [6.7K]
3 years ago
12

A manager wants to gauge employee satisfaction at a company. She hands out a survey questionnaire to everyone in the human resou

rces
department who were hired in the past two years. The employees must respond to the questionnaire within five days.
What type of bias are the survey results at risk for?
OA. The survey results are not at risk for any bias.
OB.
The survey results are most at risk for undercoverage.
Ос. The survey results are at risk for nonresponse bias and undercoverage.
OD
The survey results are most at risk for nonresponse bias.
Mathematics
1 answer:
denis-greek [22]3 years ago
7 0
B. Undercoverage? Because the survey was only given to a certain group of people
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Step-by-step explanation:

A reduced interval width means that the data is more accurate. This can only be achieved if the sample size is increased because a larger sample size is able to capture more of the characteristics of the variables being tested.

A smaller confidence interval will also lead to a reduced interval width because it means that the chances of the prediction being correct have increased.

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Step-by-step explanation:

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The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people.
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Answer:

The z-distribution should be used for this problem.

Step-by-step explanation:

The population distribution is assumed to be normal. Which distribution to use?

If we have the standard deviation for the sample, we use the t-distribution.

If we use the standard deviation for the population, we use the z-distribution.

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3 years ago
At a local university, a sample of 49 evening students was selected in order to determine whether the average age of the evening
pav-90 [236]

Answer:

z=\frac{23-21}{\frac{3.5}{\sqrt{49}}}=4    

p_v =2*P(z>4)=0.0000633  

When we compare the significance level \alpha=0.1 we see that p_v so we can reject the null hypothesis at 10% of significance. So the  the true mean is difference from 21 at this significance level.

Step-by-step explanation:

Data given and notation  

\bar X=23 represent the sample mean

\sigma=3.5 represent the population standard deviation

n=49 sample size  

\mu_o =21 represent the value that we want to test

\alpha=0.1 represent the significance level for the hypothesis test.  

z 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 average age of the evening students is significantly different from 21, the system of hypothesis would be:  

Null hypothesis:\mu = 21  

Alternative hypothesis:\mu \neq 21  

The statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Calculate the statistic

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

z=\frac{23-21}{\frac{3.5}{\sqrt{49}}}=4    

P-value

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

p_v =2*P(z>4)=0.0000633  

Conclusion  

When we compare the significance level \alpha=0.1 we see that p_v so we can reject the null hypothesis at 10% of significance. So the  the true mean is difference from 21 at this significance level.

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