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finlep [7]
3 years ago
15

Classify the scenarios based on whether they are observational studies, experimentsor surveys. Scenarlo A: Twenty employees of a

company are randomly selected and asked if they would like the company to revise the existing dress code Scenarlo B: Analysts at an energy research compare the frequency of power outages caused by snowfall in different states over the past decade. Scenarlo To determine and compare the accuracles of pedometer X and pedometer Y researchers randomly ask Consumers to use one of the two pedometers and then analyze the results. D: At a coffee shop, ten customers are randomly selected each day and asked how many times a week they visit the shop.
Mathematics
1 answer:
Alex777 [14]3 years ago
6 0

Answer:

Scenario A: Survey

Scenario B: Observational Studies

Scenario C: Experiment

Scenario D: Survey

If it not the answer you can correct me

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In constructing a 95 percent confidence interval, if you increase n to 4n, the width of your confidence interval will (assuming
Damm [24]

Answer:

about 50 percent of its former width.

Step-by-step explanation:

Let's assume that our parameter of interest is given by \theta and in order to construct a confidence interval we can use the following formula:

\hat \theta \pm ME(\hat \theta)

Where \hat \theta is an estimator for the parameter of interest and the margin of error is defined usually if the distribution for the parameter is normal as:

ME = z_{\alpha} SE

Where z_{\alpha/2} is a quantile from the normal standard distribution that accumulates \alpha/2 of the area on each tail of the distribution. And SE represent the standard error for the parameter.

If our parameter of interest is the population proportion the standard of error is given by:

SE= \frac{\hat p (1-\hat p)}{n}

And if our parameter of interest is the sample mean the standard error is given by:

SE = \frac{s}{\sqrt{n}}

As we can see the standard error for both cases assuming that the other things remain the same are function of n the sample size and we can write this as:

SE = f(n)

And since the margin of error is a multiple of the standard error we have that ME = f(n)

Now if we find the width for a confidence interval we got this:

Width = \hat \theta + ME(\hat \theta) -[\hat \theta -ME(\hat \theta)]

Width = 2 ME (\hat \theta)

And we can express this as:

Width =2 f(n)

And we can define the function f(n) = \frac{1}{\sqrt{n}} since as we can see the margin of error and the standard error are function of the inverse square root of n. So then we have this:

Width_i= 2 \frac{1}{\sqrt{n}}

The subscript i is in order to say that is with the sample size n

If we increase the sample size from n to 4n now our width is:

Width_f = 2 \frac{1}{\sqrt{4n}} =2 \frac{1}{\sqrt{4}\sqrt{n}} =\frac{2}{2} \frac{1}{\sqrt{n}} =\frac{1}{\sqrt{n}} =\frac{1}{2} Width_i

The subscript f is in order to say that is the width for the sample size 4n.

So then as we can see the width for the sample size of 4n is the half of the wisth for the width obtained with the sample size of n. So then the best option for this case is:

about 50 percent of its former width.

7 0
4 years ago
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