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Alik [6]
2 years ago
11

Which of the following would increase the width of a confidence interval for a population​ mean? Choose the correct answer below

. A. Increase the level of confidence B. Decrease the sample standard deviation. C. Increase the sample size D. All of the above
Mathematics
1 answer:
Lelechka [254]2 years ago
7 0

Answer:

A. Increase the level of confidence

Step-by-step explanation:

The margin of error is given by:

The margin of error is:

M = \frac{Ts}{\sqrt{n}}

In which T is related to the level of confidence(the higher the level of confidence, the higher T is), s is the standard deviation of the sample and n is the size of the sample.

Increase the width:

That is, increasing the margin of error, as the width is twice the margin of error, the possible options are:

Increase T -> increase confidence level.

Increase s -> Increase the standard deviation of the sample.

Decrease n -> Decrease the sample size.

Thus, the correct answer is given by option A.

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The conversion rate for US Dollars to Costa Rican Colones is 1 USD = 518 Colones.
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The cost of 0.5 kg of bananas is 393.60 Colones as per the given conversion rates

Conversion rate of 1 USD to Costa Rican Colones = 518 Colones

The conversion rate of kg to pounds given in the question: 1 kg = 2.2025 lbs

Cost of one pound of bananas = $0.69

Bananas required to be purchased = 0.5kg

Converting 0.5kg bananas to pounds = 0.5*2.2025 = 1.10125 pounds

Cost of 1.10125 pound of bananas in dollars = 1.10125*0.69 = 0.7598

Cost of 1.1025 pounds of bananas in Colones = 0.7598*518 = 393.60 Colones

Hence, the cost is 393.60

Therefore, the cost of 0.5 kg bananas in Colones is 393.60 Colones

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brainly.com/question/2274822

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