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Marina CMI [18]
4 years ago
12

How would changes in sample size affect the margin of error, assuming all esle remained constant?

Mathematics
1 answer:
klasskru [66]4 years ago
7 0

Answer:

A. A larger sample size would cause the interval to narrow.

Step-by-step explanation:

Sampling Error is error caused by observing sample instead of population. It is denoted by difference between population parameter (population mean) & sample statistic (sample mean).

Margin Error is a statistical measure of random sampling error in a survey results. High margin error implies less reliability of sample results to draw conclusion about population ; and less margin error implies more reliability of sample results to draw conclusion about population.

Margin error is inversely related to sample size : large sample size implies low margin error, and small sample size implies  high margin error. So, A larger sample size would cause the margin error interval to narrow.

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arsen [322]
<h2>⚘Your Answer:------</h2>

<h3><u>Given</u><u> </u><u>Information</u><u>:</u></h3>

  • <u>Diameter of can</u>:- 3.6 cm
  • <u>Height of</u><u> </u><u>can</u><u>:</u><u>-</u> 6.4 cm

<h3><u>To</u><u> </u><u>Find</u><u> </u><u>Out</u><u>:</u></h3>

  • <u>Volume</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>Can</u><u>.</u>

<h3><u>Solution</u><u>:</u></h3>

Radius = (Diameter/2)

ㅤㅤㅤㅤ3.6 cm/2

ㅤㅤㅤㅤ1.8 cm

<u>So</u><u>,</u><u> </u><u>Radius</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>soup</u><u> </u><u>can</u><u> </u><u>is</u> 1.8 cm.

Volume øf cylinder = Volume øf can

Volume of cylinder = π r²h

(π = 22/7)

(r = radius = 1.8 cm)

(h = height = 6.4 cm

ㅤㅤㅤㅤㅤㅤㅤㅤㅤ= 22/7×(1.8 cm)²× 6.4 cm

ㅤㅤㅤㅤㅤㅤㅤㅤㅤ= 22/7× 3.24 cm² × 6.4 cm

ㅤㅤㅤㅤㅤㅤㅤㅤㅤ= 22/7 × 20.736 cm³

ㅤㅤㅤㅤㅤㅤㅤㅤㅤ= (456.192/7) cm³

ㅤㅤㅤㅤㅤㅤㅤㅤㅤ= 65.17 cm³

So, the Volume of the soup can is 65.17 cm³

ㅤㅤㅤㅤㅤㅤㅤㅤ⚘Thank You

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Step-by-step explanation:

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