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Gnoma [55]
1 year ago
5

31. The director of a marketing department wants to estimate the proportion of people who purchase a certain product online. The

director originally planned to obtain a random sample of 2.500 people who purchased the product. However, because of budget concerns, the sample site will be reduced to 1.500 people. Which of the following describes the effect of reducing the number of people in the sample? (A) The variance of the sample will increase
(B) The variance of the population will decrease.
(C) The variance of the sampling distribution of the estimator will increase
(D) The variance of the sampling distribution of the estimator will decrease
(E) The variance of the sampling distribution of the estimator will remain the same
Mathematics
1 answer:
vivado [14]1 year ago
6 0

Answer:

C

Step-by-step explanation:

AP Classroom says so

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Helen [10]

Hey, Melany. To answer this problem, simply set up a proportion: \frac{18.75}{100} = \frac{x}{128}. Now let's solve for x.

- Cross multiply:\frac{18.75*128}{100*x} --> 2400 = 100x

- Isolate x:

\frac{2400}{100} = \frac{100x}{100}

This means that x = 24.

There are 24 9th graders in the band.


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What number is 10 times greater than 12,052
AlekseyPX

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120,520

Step-by-step explanation:

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Step-by-step explanation:

1/4

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1 year ago
One of the solutions to x2 − 2x − 15 = 0 is x = −3. What is the other solution?
Bond [772]

Answer:x=5

Step-by-step explanation: x^2−2x−15=0

(x−5)=0                                                                                                                        x−5=0                                                                                                                           x=5

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son4ous [18]
There is a relationship between confidence interval and standard deviation:
\theta=\overline{x} \pm \frac{z\sigma}{\sqrt{n}}
Where \overline{x} is the mean, \sigma is standard deviation, and n is number of data points.
Every confidence interval has associated z value. This can be found online.
We need to find the standard deviation first: 
\sigma=\sqrt{\frac{\sum(x-\overline{x})^2}{n}
When we do all the calculations we find that:
\overline{x}=123.8\\ \sigma=11.84
Now we can find confidence intervals:
($90\%,z=1.645): \theta=123.8 \pm \frac{1.645\cdot 11.84}{\sqrt{15}}=123.8 \pm5.0\\($95\%,z=1.960): \theta=123.8 \pm \frac{1.960\cdot 11.84}{\sqrt{15}}=123.8 \pm 5.99\\ ($99\%,z=2.576): \theta=123.8 \pm \frac{2.576\cdot 11.84}{\sqrt{15}}=123.8 \pm 7.87\\
We can see that as confidence interval increases so does the error margin. Z values accociated with each confidence intreval also get bigger as confidence interval increases.
Here is the link to the spreadsheet with standard deviation calculation:
https://docs.google.com/spreadsheets/d/1pnsJIrM_lmQKAGRJvduiHzjg9mYvLgpsCqCoGYvR5Us/edit?usp=sharing
6 0
2 years ago
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